Mutatók a C magyarázatában - nem olyan nehézek, mint gondolnád

A mutató vitathatatlanul a legnehezebben megérthető C jellemzője. De ezek egyike azoknak a tulajdonságoknak, amelyek a C-t kiváló nyelvvé teszik.

Ebben a cikkben a mutatók alapjaitól a tömbök, függvények és felépítés használatáig fogunk eljutni.

Tehát pihenjen, vegyen egy kávét, és készüljön fel arra, hogy mindent megismerjen a mutatókról.

Témák

A. Alapismeretek

Mi is pontosan a mutató?
Definíció és jelölés
Néhány speciális mutató
Mutató számtani

B. Tömbök és húrok

Miért pontok és tömbök?
1-D tömbök
2-D tömbök
Húrok
Mutatók tömbje
Mutató az Array-hez

C. Funkciók

Hívás érték v / s Hívás referencia alapján
Mutatók mint funkció-érvek
Mutatók mint függvény visszatér
Mutató a funkcióhoz
Funkciókra mutató mutatók tömbje
A funkcionális mutató mint érv

D. Felépítés

Mutató a szerkezethez
Struktúra tömb
A struktúra mutatója mint érv

E. Mutató a mutatóhoz

F. Következtetés

A. Meghatározás, jelölés, típusok és számtan

1. Pontosan mik a mutatók?

Mielőtt eljutnánk a mutatók definíciójához, értsük meg, mi történik, amikor a következő kódot írjuk:

int digit = 42;

A fordító fenntart egy memória blokkot egy intérték tárolására . Ennek a blokknak a neve és a blokkban digittárolt értéke az 42.

Most, hogy emlékezzünk a blokkra, egy címet vagy egy helyszámot (például 24650) rendelünk hozzá .

A helyszám értéke nem fontos számunkra, mivel ez egy véletlenszerű érték. De ehhez a címhez az operátor &(ampersand) vagy címével férhetünk hozzá .

printf("The address of digit = %d.",&digit); /* prints "The address of digit = 24650. */

A változó értékét digita címéből megkaphatjuk egy másik operátor *(csillag) segítségével, az úgynevezett indirection vagy dereferencing vagy value at address operátorral.

printf("The value of digit = %d.", *(&digit); /* prints "The value of digit = 42. */

2. Definíció és jelölés

Egy változó címe tárolható egy másik változóban, amelyet mutatóváltozóként ismerünk. A változó címének mutatóban történő tárolásának szintaxisa:

dataType *pointerVariableName = &variableName;

Mert a mi digitváltozó, ez lehet írni, mint ez:

int *addressOfDigit = &digit;

vagy így:

int *addressOfDigit; addressOfDigit= &digit;

Ez így olvasható: - Egy int(egész) mutató mutató addressOfDigittárolja a address of(&)digitváltozót.

Néhány megértendő pont:

dataType- El kell mondanunk a számítógépnek, hogy mi a változó adattípusa, amelynek címét tárolni fogjuk. Itt intvolt a digit.

Ez nem azt jelenti, hogy addressOfDigitegy típusú értéket tárol int. Egy egész mutató (például addressOfDigit) csak az egész típusú változók címét tárolhatja.

int variable1; int variable2; char variable3; int *addressOfVariables;

*- A mutató változó egy speciális változó abban az értelemben, hogy egy másik változó címének tárolására szolgál. Hogy megkülönböztessük azokat a változóktól, amelyek nem tárolnak címet, *szimbólumként használjuk a deklarációban.

Itt hozzárendelhetjük az egész mutató címét variable1és címét, de nem, mivel nem . Szükségünk lesz egy karaktermutató változóra a cím tárolásához.variable2addressOfVariablesvariable3char

A addressOfDigitmutatóváltozónk segítségével az alábbiak szerint nyomtathatjuk ki a címet és az értéket digit:

printf("The address of digit = %d.", addressOfDigit); /* prints "The address of digit = 24650." */ printf("The value of digit = %d.", *addressOfDigit); /*prints "The value of digit = 42. */

Itt *addressOfDigitolvasható értékként a tárolt címen addressOfDigit.

Figyeljük meg szoktuk %da formátum azonosító számára addressOfDigit. Nos, ez nem teljesen helyes. A helyes azonosító lenne %p.

Használatával %pa cím hexadecimális értékként jelenik meg. De a memória címe egészben, valamint oktális értékben is megjeleníthető. Ennek ellenére, mivel ez nem teljesen helyes , figyelmeztetés jelenik meg.

int num = 5; int *p = # printf("Address using %%p = %p",p); printf("Address using %%d = %d",p); printf("Address using %%o = %o",p);

Az általam használt fordító kimenete a következő:

Address using %p = 000000000061FE00 Address using %d = 6422016 Address using %o = 30377000

Ez a figyelmeztetés akkor jelenik meg, amikor a% d - "figyelmeztetést használja: a '% d' formátum az 'int' típusú argumentumot várja, de a 2. argumentumban az 'int *' típus van.

3. Néhány speciális mutató

A Vad Pointer

char *alphabetAddress; /* uninitialised or wild pointer */ char alphabet = "a"; alphabetAddress = &alphabet; /* now, not a wild pointer */

When we defined our character pointer alphabetAddress, we did not initialize it.

Such pointers are known as wild pointers. They store a garbage value (that is, memory address) of a byte that we don't know is reserved or not (remember int digit = 42;, we reserved a memory address when we declared it).

Suppose we dereference a wild pointer and assign a value to the memory address it is pointing at. This will lead to unexpected behaviour since we will write data at a memory block that may be free or reserved.

Null Pointer

To make sure that we do not have a wild pointer, we can initialize a pointer with a NULL value, making it a null pointer.

char *alphabetAddress = NULL /* Null pointer */

A null pointer points at nothing, or at a memory address that users can not access.

Void Pointer

A void pointer can be used to point at a variable of any data type. It can be reused to point at any data type we want to. It is declared like this:

void *pointerVariableName = NULL;

Since they are very general in nature, they are also known as generic pointers.

With their flexibility, void pointers also bring some constraints. Void pointers cannot be dereferenced as any other pointer. Appropriate typecasting is necessary.

void *pointer = NULL; int number = 54; char alphabet = "z"; pointer = &number; printf("The value of number = ", *pointer); /* Compilation Error */ /* Correct Method */ printf("The value of number = ", *(int *)pointer); /* prints "The value at number = 54" */ pointer = &alphabet; printf("The value of alphabet = ", *pointer); /* Compilation Error */ printf("The value of alphabet = ", *(char *)pointer); /* prints "The value at alphabet = z */

Similarly, void pointers need to be typecasted for performing arithmetic operations.

Void pointers are of great use in C. Library functions malloc() and calloc() which dynamically allocate memory return void pointers. qsort(), an inbuilt sorting function in C, has a function as its argument which itself takes void pointers as its argument.

Dangling Pointer

A dangling pointer points to a memory address which used to hold a variable. Since the address it points at is no longer reserved, using it will lead to unexpected results.

main(){ int *ptr; ptr = (int *)malloc(sizeof(int)); *ptr = 1; printf("%d",*ptr); /* prints 1 */ free(ptr); /* deallocation */ *ptr = 5; printf("%d",*ptr); /* may or may not print 5 */ }

Though the memory has been deallocated by free(ptr), the pointer to integer ptr still points to that unreserved memory address.

4. Pointer Arithmetic

We know by now that pointers are not like any other variable. They do not store any value but the address of memory blocks.

So it should be quite clear that not all arithmetic operations would be valid with them. Would multiplying or dividing two pointers (having addresses) make sense?

Pointers have few but immensely useful valid operations:

You can assign the value of one pointer to another only if they are of the same type (unless they're typecasted or one of them is void *).

int ManU = 1; int *addressOfManU = &ManU; int *anotherAddressOfManU = NULL; anotherAddressOfManU = addressOfManU; /* Valid */ double *wrongAddressOfManU = addressOfManU; /* Invalid */

2. You can only add or subtract integers to pointers.

int myArray = {3,6,9,12,15}; int *pointerToMyArray = &myArray[0]; pointerToMyArray += 3; /* Valid */ pointerToMyArray *= 3; /* Invalid */

When you add (or subtract) an integer (say n) to a pointer, you are not actually adding (or subtracting) n bytes to the pointer value. You are actually adding (or subtracting) n-times the size of the data type of the variable being pointed bytes.

int number = 5; /* Suppose the address of number is 100 */ int *ptr = &number; int newAddress = ptr + 3; /* Same as ptr + 3 * sizeof(int) */

The value stored in newAddress will not be 103, rather 112.

3. Subtraction and comparison of pointers is valid only if both are members of the same array. The subtraction of pointers gives the number of elements separating them.

int myArray = {3,6,9,12,15}; int sixthMultiple = 18; int *pointer1 = &myArray[0]; int *pointer2 = &myArray[1]; int *pointer6 = &sixthMuliple; /* Valid Expressions */ if(pointer1 == pointer2) pointer2 - pointer1; /* Invalid Expressions if(pointer1 == pointer6) pointer2 - pointer6

4. You can assign or compare a pointer with NULL.

The only exception to the above rules is that the address of the first memory block after the last element of an array follows pointer arithmetic.

Pointer and arrays exist together. These valid manipulations of pointers are immensely useful with arrays, which will be discussed in the next section.

B. Arrays and Strings

1. Why pointers and arrays?

In C, pointers and arrays have quite a strong relationship.

The reason they should be discussed together is because what you can achieve with array notation (arrayName[index]) can also be achieved with pointers, but generally faster.

2. 1-D Arrays

Let us look at what happens when we write int myArray[5];.

Five consecutive blocks of memory starting from myArray[0] to myArray[4] are created with garbage values in them. Each of the blocks is of size 4 bytes.

Thus, if the address of myArray[0] is 100 (say), the address of the rest of the blocks would be 104, 108, 112, and 116.

Have a look at the following code:

int prime[5] = {2,3,5,7,11}; printf("Result using &prime = %d\n",&prime); printf("Result using prime = %d\n",prime); printf("Result using &prime[0] = %d\n",&prime[0]); /* Output */ Result using &prime = 6422016 Result using prime = 6422016 Result using &prime[0] = 6422016

So, &prime, prime, and &prime[0] all give the same address, right? Well, wait and read because you are in for a surprise (and maybe some confusion).

Let's try to increment each of &prime, prime, and &prime[0] by 1.

printf("Result using &prime = %d\n",&prime + 1); printf("Result using prime = %d\n",prime + 1); printf("Result using &prime[0] = %d\n",&prime[0] + 1); /* Output */ Result using &prime = 6422036 Result using prime = 6422020 Result using &prime[0] = 6422020

Wait! How come &prime + 1 results in something different than the other two? And why are prime + 1 and &prime[0] + 1 still equal? Let's answer these questions.

prime and &prime[0] both point to the 0th element of the array prime. Thus, the name of an array is itself a pointer to the 0th element of the array.

Here, both point to the first element of size 4 bytes. When you add 1 to them, they now point to the 1st element in the array. Therefore this results in an increase in the address by 4.

&prime, on the other hand, is a pointer to an int array of size 5. It stores the base address of the array prime[5], which is equal to the address of the first element. However, an increase by 1 to it results in an address with an increase of 5 x 4 = 20 bytes.

In short, arrayName and &arrayName[0] point to the 0th element whereas &arrayName points to the whole array.

We can access the array elements using subscripted variables like this:

int prime[5] = {2,3,5,7,11}; for( int i = 0; i < 5; i++) { printf("index = %d, address = %d, value = %d\n", i, &prime[i], prime[i]); }

We can do the same using pointers which are always faster than using subscripts.

int prime[5] = {2,3,5,7,11}; for( int i = 0; i < 5; i++) { printf("index = %d, address = %d, value = %d\n", i, prime + i, *(prime + i)); }

Both methods give the output:

index = 0, address = 6422016, value = 2 index = 1, address = 6422020, value = 3 index = 2, address = 6422024, value = 5 index = 3, address = 6422028, value = 7 index = 4, address = 6422032, value = 11

Thus, &arrayName[i] and arrayName[i] are the same as arrayName + i and *(arrayName + i), respectively.

3. 2-D Arrays

Two-dimensional arrays are an array of arrays.

int marks[5][3] = { { 98, 76, 89}, { 81, 96, 79}, { 88, 86, 89}, { 97, 94, 99}, { 92, 81, 59} };

Here, marks can be thought of as an array of 5 elements, each of which is a one-dimensional array containing 3 integers. Let us work through a series of programs to understand different subscripted expressions.

printf("Address of whole 2-D array = %d\n", &marks); printf("Addition of 1 results in %d\n", &marks +1); /* Output */ Address of whole 2-D array = 6421984 Addition of 1 results in 6422044

Like 1-D arrays, &marks points to the whole 2-D array, marks[5][3]. Thus, incrementing to it by 1 ( = 5 arrays X 3 integers each X 4 bytes = 60) results in an increment by 60 bytes.

printf("Address of 0th array = %d\n", marks); printf("Addition of 1 results in %d\n", marks +1); printf("Address of 0th array =%d\n", &marks[0]); printf("Addition of 1 results in %d\n", &marks[0] + 1); /* Output */ Address of 0th array = 6421984 Addition of 1 results in 6421996 Address of 0th array = 6421984 Addition of 1 results in 6421996

If marks was a 1-D array, marks and &marks[0] would have pointed to the 0th element. For a 2-D array, elements are now 1-D arrays. Hence, marks and &marks[0] point to the 0th array (element), and the addition of 1 point to the 1st array.

printf("Address of 0th element of 0th array = %d\n", marks[0]); printf("Addition of 1 results in %d\n", marks[0] + 1); printf("Address of 0th element of 1st array = %d\n", marks[1]); printf("Addition of 1 results in %d\n", marks[1] + 1); /* Output */ Address of 0th element of 0th array = 6421984 Addition of 1 results in 6421988 Address of 0th element of 1st array = 6421996 Addition of 1 results in 6422000

And now comes the difference. For a 1-D array, marks[0] would give the value of the 0th element. An increment by 1 would increase the value by 1.

But, in a 2-D array, marks[0] points to the 0th element of the 0th array. Similarly, marks[1] points to the 0th element of the 1st array. An increment by 1 would point to the 1st element in the 1st array.

printf("Value of 0th element of 0th array = %d\n", marks[0][0]); printf("Addition of 1 results in %d", marks[0][0] + 1); /* Output */ Value of 0th element of 0th array = 98 Addition of 1 results in 99

This is the new part. marks[i][j] gives the value of the jth element of the ith array. An increment to it changes the value stored at marks[i][j]. Now, let us try to write marks[i][j] in terms of pointers.

We know marks[i] + j would point to the ith element of the jth array from our previous discussion. Dereferencing it would mean the value at that address. Thus, marks[i][j] is the same as *(marks[i] + j).

From our discussion on 1-D arrays, marks[i] is the same as *(marks + i). Thus, marks[i][j] can be written as *(*(marks + i) + j) in terms of pointers.

Here is a summary of notations comparing 1-D and 2-D arrays.

Expression	1-D Array	2-D Array
&arrayName	points to the address of whole array adding 1 increases the address by 1 x sizeof(arrayName)	points to the address of whole array adding 1 increases the address by 1 x sizeof(arrayName)
arrayName	points to the 0th element adding 1 increases the address to 1st element	points to the 0th element (array) adding 1 increases the address to 1st element (array)
&arrayName[i]	points to the the ith element adding 1 increases the address to (i+1)th element	points to the ith element (array) adding 1 increases the address to the (i+1)th element (array)
arrayName[i]	gives the value of the ith element adding 1 increases the value of the ith element	points to the 0th element of the ith array adding 1 increases the address to 1st element of the ith array
arrayName[i][j]	Nothing	gives the value of the jth element of the ith array adding 1 increases the value of the jth element of the ith array
Pointer Expression To Access The Elements	*( arrayName + i)	( ( arrayName + i) + j)

4. Strings

A string is a one-dimensional array of characters terminated by a null(\0). When we write char name[] = "Srijan";, each character occupies one byte of memory with the last one always being \0.

Similar to the arrays we have seen, name and &name[0] points to the 0th character in the string, while &name points to the whole string. Also, name[i] can be written as *(name + i).

/* String */ char champions[] = "Liverpool"; printf("Pointer to whole string = %d\n", &champions); printf("Addition of 1 results in %d\n", &champions + 1); /* Output */ Address of whole string = 6421974 Addition of 1 results in 6421984 printf("Pointer to 0th character = %d\n", &champions[0]); printf("Addition of 1 results in %d\n", &champions[0] + 1); /* Output */ Address of 0th character = 6421974 Addition of 1 results in a pointer to 1st character 6421975 printf("Pointer to 0th character = %d\n", champions); printf("Addition of 1 results in a pointer to 1st character %d\n", champions + 1); /* Output */ Address of 0th character = 6421974 Addition of 1 results in 6421975 printf("Value of 4th character = %c\n", champions[4]); printf("Value of 4th character using pointers = %c\n", *(champions + 4)); /* Output */ Value of 4th character = r Value of 4th character using pointers = r

A two-dimensional array of characters or an array of strings can also be accessed and manipulated as discussed before.

/* Array of Strings */ char top[6][15] = { "Liverpool", "Man City", "Man United", "Chelsea", "Leicester", "Tottenham" }; printf("Pointer to 2-D array = %d\n", &top); printf("Addition of 1 results in %d\n", &top + 1); /* Output */ Pointer to 2-D array = 6421952 Addition of 1 results in 6422042 printf("Pointer to 0th string = %d\n", &top[0]); printf("Addition of 1 results in %d\n", &top[0] + 1); /* Output */ Pointer to 0th string = 6421952 Addition of 1 results in 6421967 printf("Pointer to 0th string = %d\n", top); printf("Addition of 1 results in %d\n", top + 1); /* Output */ Pointer to 0th string = 6421952 Addition of 1 results in 6421967 printf("Pointer to 0th element of 4th string = %d\n", top[4]); printf("Pointer to 1st element of 4th string = %c\n", top[4] + 1); /* Output */ Pointer to 0th element of 4th string = 6422012 Pointer to 1st element of 4th string = 6422013 printf("Value of 1st character in 3rd string = %c\n", top[3][1]); printf("Same using pointers = %c\n", *(*(top + 3) + 1)); /* Output */ Value of 1st character in 3rd string = h Same using pointers = h

5. Array of Pointers

Like an array of ints and an array of chars, there is an array of pointers as well. Such an array would simply be a collection of addresses. Those addresses could point to individual variables or another array as well.

The syntax for declaring a pointer array is the following:

dataType *variableName[size]; /* Examples */ int *example1[5]; char *example2[8];

Following the operators precedence, the first example can be read as - example1 is an array([]) of 5 pointers to int. Similarly, example2 is an array of 8 pointers to char.

We can store the two-dimensional array to string top using a pointer array and save memory as well.

char *top[] = { "Liverpool", "Man City", "Man United", "Chelsea", "Leicester", "Tottenham" };

top will contain the base addresses of all the respective names. The base address of "Liverpool" will be stored in top[0], "Man City" in top[1], and so on.

In the earlier declaration, we required 90 bytes to store the names. Here, we only require ( 58 (sum of bytes of names) + 12 ( bytes required to store the address in the array) ) 70 bytes.

The manipulation of strings or integers becomes a lot easier when using an array of pointers.

If we try to put "Leicester" ahead of "Chelsea", we just need to switch the values of top[3] and top[4] like below:

char *temporary; temporary = top[3]; top[3] = top[4]; top[4] = temporary;

Without pointers, we would have to exchange every character of the strings, which would have taken more time. That's why strings are generally declared using pointers.

6. Pointer to Array

Like "pointer to int" or "pointer to char", we have pointer to array as well. This pointer points to whole array rather than its elements.

Remember we discussed how &arrayName points to the whole array? Well, it is a pointer to array.

A pointer to array can be declared like this:

dataType (*variableName)[size]; /* Examples */ int (*ptr1)[5]; char (*ptr2)[15];

Notice the parentheses. Without them, these would be an array of pointers. The first example can be read as - ptr1 is a pointer to an array of 5 int(integers).

int goals[] = { 85,102,66,69,67}; int (*pointerToGoals)[5] = &goals; printf("Address stored in pointerToGoals %d\n", pointerToGoals); printf("Dereferncing it, we get %d\n",*pointerToGoals); /* Output */ Address stored in pointerToGoals 6422016 Dereferencing it, we get 6422016

When we dereference a pointer, it gives the value at that address. Similarly, by dereferencing a pointer to array, we get the array and the name of the array points to the base address. We can confirm that *pointerToGoals gives the array goals if we find its size.

printf("Size of goals[5] = %d, *pointerToGoals); /* Output */ Size of goals[5] = 20

If we dereference it again, we will get the value stored in that address. We can print all the elements using pointerToGoals.

for(int i = 0; i < 5; i++) printf("%d ", *(*pointerToGoals + i)); /* Output */ 85 102 66 69 67

Pointers and pointer to arrays are quite useful when paired up with functions. Coming up in the next section!

C. Functions

1. Call by Value vs Call by Reference

Have a look at the program below:

#include  int multiply(int x, int y){ int z; z = x * y; return z; } main(){ int x = 3, y = 5; int product = multiply(x,y); printf("Product = %d\n", product); /* prints "Product = 15" */ }

The function multiply() takes two int arguments and returns their product as int.

In the function call multiply(x,y), we passed the value of x and y ( of main()), which are actual arguments, to multiply().

The values of the actual arguments are passed or copied to the formal argumentsx and y ( of multiply()). The x and y of multiply() are different from those of main(). This can be verified by printing their addresses.

#include  int multiply(int x, int y){ printf("Address of x in multiply() = %d\n", &x); printf("Address of y in multiply() = %d\n", &y); int z; z = x * y; return z; } main(){ int x = 3, y = 5; printf("Address of x in main() = %d\n", &x); printf("Address of y in main() = %d\n", &y); int product = multiply(x,y); printf("Product = %d\n", product); } /* Output */ Address of x in main() = 6422040 Address of y in main() = 6422036 Address of x in multiply() = 6422000 Address of y in multiply() = 6422008 Product = 15

Since we created stored values in a new location, it costs us memory. Wouldn't it be better if we could perform the same task without wasting space?

Call by reference helps us achieve this. We pass the address or reference of the variables to the function which does not create a copy. Using the dereferencing operator *, we can access the value stored at those addresses.

We can rewrite the above program using call by reference as well.

#include  int multiply(int *x, int *y){ int z; z = (*x) * (*y); return z; } main(){ int x = 3, y = 5; int product = multiply(&x,&y); printf("Product = %d\n", product); /* prints "Product = 15" */ }

2. Pointers as Function Arguments

In this section, we will look at various programs where we give int, char, arrays and strings as arguments using pointers.

#include  void add(float *a, float *b){ float c = *a + *b; printf("Addition gives %.2f\n",c); } void subtract(float *a, float *b){ float c = *a - *b; printf("Subtraction gives %.2f\n",c); } void multiply(float *a, float *b){ float c = *a * *b; printf("Multiplication gives %.2f\n",c); } void divide(float *a, float *b){ float c = *a / *b; printf("Division gives %.2f\n",c); } main(){ printf("Enter two numbers :\n"); float a,b; scanf("%f %f",&a,&b); printf("What do you want to do with the numbers?\nAdd : a\nSubtract : s\nMultiply : m\nDivide : d\n"); char operation = '0'; scanf(" %c",&operation); printf("\nOperating...\n\n"); switch (operation) { case 'a': add(&a,&b); break; case 's': subtract(&a,&b); break; case 'm': multiply(&a,&b); break; case 'd': divide(&a,&b); break; default: printf("Invalid input!!!\n"); } }

We created four functions, add(), subtract(), multiply() and divide() to perform arithmetic operations on the two numbers a and b.

The address of a and b was passed to the functions. Inside the function using * we accessed the values and printed the result.

Similarly, we can give arrays as arguments using a pointer to its first element.

#include  void greatestOfAll( int *p){ int max = *p; for(int i=0; i  max) max = *(p+i); } printf("The largest element is %d\n",max); } main(){ int myNumbers[5] = { 34, 65, -456, 0, 3455}; greatestOfAll(myNumbers); /* Prints :The largest element is 3455" */ }

Since the name of an array itself is a pointer to the first element, we send that as an argument to the function greatestOfAll(). In the function, we traverse through the array using loop and pointer.

#include  #include  void wish(char *p){ printf("Have a nice day, %s",p); } main(){ printf("Enter your name : \n"); char name[20]; gets(name); wish(name); }

Here, we pass in the string name to wish() using a pointer and print the message.

3. Pointers as Function Return

#include  int* multiply(int *a, int *b){ int c = *a * *b; return &c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); }

The function multiply() takes two pointers to int. It returns a pointer to int as well which stores the address where the product is stored.

It is very easy to think that the output would be 15. But it is not!

When multiply() is called, the execution of main() pauses and memory is now allocated for the execution of multiply(). After its execution is completed, the memory allocated to multiply() is deallocated.

Therefore, though c ( local to main()) stores the address of the product, the data there is not guaranteed since that memory has been deallocated.

So does that mean pointers cannot be returned by a function? No!

We can do two things. Either store the address in the heap or global section or declare the variable to be static so that their values persist.

Static variables can simply be created by using the keywordstatic before data type while declaring the variable.

To store addresses in heap, we can use library functions malloc() and calloc() which allocate memory dynamically.

The following programs will explain both the methods. Both methods return the output as 15.

#include  #include  /* Using malloc() */ int* multiply(int *a, int *b){ int *c = malloc(sizeof(int)); *c = *a * *b; return c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); } /* Using static keyword */ #include  int* multiply(int *a, int *b){ static int c; c = *a * *b; return &c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); }

4. Pointer to Function

Like pointer to different data types, we also have a pointer to function as well.

A pointer to function or function pointer stores the address of the function. Though it doesn't point to any data. It points to the first instruction in the function.

The syntax for declaring a pointer to function is:

 /* Declaring a function */ returnType functionName(parameterType1, pparameterType2, ...); /* Declaring a pointer to function */ returnType (*pointerName)(parameterType1, parameterType2, ...); pointerName = &functionName; /* or pointerName = functionName; */

The below example will make it clearer.

int* multiply(int *a, int *b) { int *c = malloc(sizeof(int)); *c = *a * *b; return c; } main() { int a=3,b=5; int* (*p)(int*, int*) = &multiply; /* or int* (*p)(int*, int*) = multiply; */ int *c = (*p)(&a,&b); /* or int *c = p(&a,&b); */ printf("Product = %d",*c); }

The declaration for the pointer p to function multiply() can be read as ( following operator precedence) - p is a pointer to function with two integer pointers ( or two pointers to int) as parameters and returning a pointer to int.

Since the name of the function is also a pointer to the function, the use of & is not necessary. Also removing * from the function call doesn't affect the program.

5. Array of Pointers to Functions

We have already seen how to create an array of pointers to int, char, and so on. Similarly, we can create an array of pointers to function.

In this array, every element will store an address of a function, where all the functions are of the same type. That is, they have the same type and number of parameters and return types.

We will modify a program discussed earlier in this section. We will store the addresses of add(), subtract(), multiply() and divide() in an array make a function call through subscript.

#include  void add(float *a, float *b){ float c = *a + *b; printf("Addition gives %.2f\n",c); } void subtract(float *a, float *b){ float c = *a - *b; printf("Subtraction gives %.2f\n",c); } void multiply(float *a, float *b){ float c = *a * *b; printf("Multiplication gives %.2f\n",c); } void divide(float *a, float *b){ float c = *a / *b; printf("Division gives %.2f\n",c); } main(){ printf("Enter two numbers :\n"); float a,b; scanf("%f %f",&a,&b); printf("What do you want to do with the numbers?\nAdd : a\nSubtract : s\nMultiply : m\nDivide : d\n"); char operation = '0'; scanf(" %c",&operation); void (*p[])(float* , float*) = {add,subtract,multiply,divide}; printf("\nOperating...\n\n"); switch (operation) { case 'a': p[0](&a,&b); break; case 's': p[1](&a,&b); break; case 'm': p[2](&a,&b); break; case 'd': p[3](&a,&b); break; default: printf("Invalid input!!!\n"); } }

The declaration here can be read as - p is an array of pointer to functions with two float pointers as parameters and returning void.

6. Pointer to Function as an Argument

Like any other pointer, function pointers can also be passed to another function, therefore known as a callback function or called function. The function to which it is passed is known as a calling function.

A better way to understand would be to look at qsort(), which is an inbuilt function in C. It is used to sort an array of integers, strings, structures, and so on. The declaration for qsort() is:

void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void *));

qsort() takes four arguments:

a void pointer to the start of an array
number of elements
size of each element
a function pointer that takes in two void pointers as arguments and returns an int

The function pointer points to a comparison function that returns an integer that is greater than, equal to, or less than zero if the first argument is respectively greater than, equal to, or less than the second argument.

The following program showcases its usage:

#include  #include  int compareIntegers(const void *a, const void *b) { const int *x = a; const int *y = b; return *x - *y; } main(){ int myArray[] = {97,59,2,83,19,97}; int numberOfElements = sizeof(myArray) / sizeof(int); printf("Before sorting - \n"); for(int i = 0; i < numberOfElements; i++) printf("%d ", *(myArray + i)); qsort(myArray, numberOfElements, sizeof(int), compareIntegers); printf("\n\nAfter sorting - \n"); for(int i = 0; i < numberOfElements; i++) printf("%d ", *(myArray + i)); } /* Output */ Before sorting - 97 59 2 83 19 97 After sorting - 2 19 59 83 97 97

Since a function name is itself a pointer, we can write compareIntegers as the fourth argument.

D. Structure

1. Pointer to Structure

Like integer pointers, array pointers, and function pointers, we have pointer to structures or structure pointers as well.

struct records { char name[20]; int roll; int marks[5]; char gender; }; struct records student = {"Alex", 43, {76, 98, 68, 87, 93}, 'M'}; struct records *ptrStudent = &student;

Here, we have declared a pointer ptrStudent of type struct records. We have assigned the address of student to ptrStudent.

ptrStudent stores the base address of student, which is the base address of the first member of the structure. Incrementing by 1 would increase the address by sizeof(student) bytes.

printf("Address of structure = %d\n", ptrStudent); printf("Adress of member `name` = %d\n", &student.name); printf("Increment by 1 results in %d\n", ptrStudent + 1); /* Output */ Address of structure = 6421984 Adress of member `name` = 6421984 Increment by 1 results in 6422032

We can access the members of student using ptrStudent in two ways. Using our old friend * or using -> ( infix or arrow operator).

With *, we will continue to use the .( dot operator) whereas with -> we won't need the dot operator.

printf("Name w.o using ptrStudent : %s\n", student.name); printf("Name using ptrStudent and * : %s\n", ( *ptrStudent).name); printf("Name using ptrStudent and -> : %s\n", ptrStudent->name); /* Output */ Name without using ptrStudent: Alex Name using ptrStudent and *: Alex Name using ptrStudent and ->: Alex

Similarly, we can access and modify other members as well. Note that the brackets are necessary while using * since the dot operator(.) has higher precedence over *.

2. Array Of Structure

We can create an array of type struct records and use a pointer to access the elements and their members.

struct records students[10]; /* Pointer to the first element ( structure) of the array */ struct records *ptrStudents1 = &students; /* Pointer to an array of 10 struct records */ struct records (*ptrStudents2)[10] = &students;

Note that ptrStudent1 is a pointer to student[0] whereas ptrStudent2 is a pointer to the whole array of 10 struct records. Adding 1 to ptrStudent1 would point to student[1].

We can use ptrStudent1 with a loop to traverse through the elements and their members.

 for( int i = 0; i name, ( ptrStudents1 + i)->roll);

3. Pointer to Structure as an Argument

We can also pass the address of a structure variable to a function.

#include  struct records { char name[20]; int roll; int marks[5]; char gender; }; main(){ struct records students = {"Alex", 43, {76, 98, 68, 87, 93}, 'M'}; printRecords(&students); } void printRecords( struct records *ptr){ printf("Name: %s\n", ptr->name); printf("Roll: %d\n", ptr->roll); printf("Gender: %c\n", ptr->gender); for( int i = 0; i marks[i]); } /* Output */ Name: Alex Roll: 43 Gender: M Marks in 0th subject: 76 Marks in 1th subject: 98 Marks in 2th subject: 68 Marks in 3th subject: 87 Marks in 4th subject: 93

Note that the structure struct records is declared outside main(). This is to ensure that it is available globally and printRecords() can use it.

If the structure is defined inside main(), its scope will be limited to main(). Also structure must be declared before the function declaration as well.

A struktúrákhoz hasonlóan mutathatunk hivatkozásokat a szakszervezetekre, és a nyíl operátor ( ->) segítségével hozzáférhetünk a tagokhoz .

E. Mutató a mutatóhoz

Eddig különféle primitív adattípusokra, tömbökre, karakterláncokra, függvényekre, struktúrákra és szakszervezetekre mutattunk mutatót.

Az automatikus kérdés, ami eszembe jut: mi a helyzet a mutató mutatóval?

Nos, jó hír neked! Ők is léteznek.

int var = 6; int *ptr_var = &var; printf("Address of var = %d\n", ptr_var); printf("Address of ptr_var = %d\n", &ptr_var); /* Output */ Address of var = 6422036 Address of ptr_var = 6422024

A intváltozó címének tárolásához varmegvan a mutató intptr_var. Szükségünk lenne egy másik mutatóra a cím címének tárolásához ptr_var.

Mivel ptr_vara típusa int *, a címének tárolásához létre kell hoznunk egy mutatót int *. Az alábbi kód megmutatja, hogyan lehet ezt megtenni.

int * *ptr_ptrvar = &ptr_var; /* or int* *ppvar or int **ppvar */

Használhatjuk ptr_ptrvara cím eléréséhez, ptr_varés kettős kivezetéssel használhatjuk a var elérését.

printf("Address of ptr_var = %d\n", ptr_ptrvar); printf("Address of var = %d\n", *ptr_ptrvar); printf("Value at var = %d\n", *(*ptr_ptrvar)); /* Output */ Address of ptr_var = 6422024 Address of var = 6422036 Value at var = 6

It is not required to use brackets when dereferencing ptr_ptrvar. But it is a good practice to use them. We can create another pointer ptr_ptrptrvar, which will store the address of ptr_ptrvar.

Since ptr_ptrvar is of type int**, the declaration for ptr_ptrptrvar will be

int** *ptr_ptrptrvar = &ptr_ptrvar;

We can again access ptr_ptrvar, ptr_var and var using ptr_ptrptrvar.

printf("Address of ptr_ptrvar = %d\n", ptr_ptrptrvar); printf("Value at ptr_ptrvar = %d\n",*ptr_ptrptrvar); printf("Address of ptr_var = %d\n", *ptr_ptrptrvar); printf("Value at ptr_var = %d\n", *(*ptr_ptrptrvar)); printf("Address of var = %d\n", *(*ptr_ptrptrvar)); printf("Value at var = %d\n", *(*(*ptr_ptrptrvar))); /* Output */ Address of ptr_ptrvar = 6422016 Value at ptr_ptrvar = 6422024 Address of ptr_var = 6422024 Value at ptr_var = 6422036 Address of var = 6422036 Value at var = 6

If we change the value at any of the pointer(s) using ptr_ptrptrvar or ptr_ptrvar, the pointer(s) will stop pointing to the variable.

Conclusion

Phew! Yeah, we're finished. We started from pointers and ended with pointers (in a way). Don't they say that the curve of learning is a circle!

Try to recap all the sub-topics that you read. If you can recollect them, well done! Read the ones you can't remember again.

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Maradj otthon, maradj biztonságban.