C programming pointers and array

Question

#include<stdio.h>

int main()
{
  //int *a={5641,5,98};
  char *s="this is a character array";
  printf("%s",s);
}

This method of INITIALIZING array works for character array but fails for interger array please tell me why..?

Read [this](https://www.geeksforgeeks.org/whats-difference-between-char-s-and-char-s-in-c/) — Ashish Kumar, Aug 19 '19 at 10:35

score 5 · Answer 1 · answered Aug 19 '19 at 10:39

5

For starters in your program s is not an array. It is a pointer.

char *s="this is a character array";

So the pointer s is initialized by the address of the first character of the string literal "this is a character array" that is indeed has type of a character array.

In C you can use a compound literal to initialize a pointer that will point to first element of an array.

For example

#include <stdio.h>

int main(void) 
{
    enum { N = 5 };
    int *p = ( int[N] ) { 1, 2, 3, 4, 5 };

    for ( size_t i = 0; i < N; i++ ) printf( "%d ", p[i] );
    putchar( '\n' );

    return 0;
}

answered Aug 19 '19 at 10:39

Vlad from Moscow

301,070
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Don't you need to call `free()` before going out of scope? – Robur_131 Aug 19 '19 at 10:49
@Robur_131 No, you need not because the array was not allocated dynamically. – Vlad from Moscow Aug 19 '19 at 10:50
Or is it because there is no associated ``malloc()``? – Robur_131 Aug 19 '19 at 10:50
So, `*p` is essentially a local variable for `main()` , will it be destroyed once `main()` goes out of scope? – Robur_131 Aug 19 '19 at 10:51
1

@Robur_131 The compound literal has the automatic storage duration because it is defined in the body of a function. – Vlad from Moscow Aug 19 '19 at 10:53
@SuvenPandey N is needed to keep the size of the array. Otherwise you will not ne able to calculate its size. In C const int N = 5; is evaluated at run-time. – Vlad from Moscow Aug 19 '19 at 10:57
How would you figure out the length of `int *a = (int []) { 5641, 5, 98};`? – Robur_131 Aug 19 '19 at 11:00
1

@Robur_131 As I already wrote you are unable to do this. Either define a variable that will keep the size of the array or use a sentinel value within the array. – Vlad from Moscow Aug 19 '19 at 11:01
@SuvenPandey Yes, the code will be compiled but how will you know the size of the array? – Vlad from Moscow Aug 19 '19 at 11:18
Well, the compiler can figure out the size for `int a[] = {1,2,3}` and `char s[] = "Hello"` so I see no reason why size would be an issue as long as size is constant. – Nevus Aug 19 '19 at 11:20
@SuvenPandey A compound literal creates an unnamed object. So you can not calculate its size. – Vlad from Moscow Aug 19 '19 at 11:25
Yes it is A pointer. But why whole string gets printed when we do so. It's a pointer pointing to address that can store value of a single character. It's size is 4 bytes still it prints such a large string. – Kshyamano Aug 19 '19 at 18:12
@Kshyamano Strings are sequences of characters terminated by a zero character. So output functions know how many characters they need to output. – Vlad from Moscow Aug 20 '19 at 10:12

score 0 · Answer 2 · answered Aug 19 '19 at 10:37

0

    #include<stdio.h> 

int main() 
{ 
  int arr[5] = { 1, 2, 3, 4, 5 }; 
  int *ptr = arr; 

  printf("%p\n", ptr); 
  return 0; 
}

answered Aug 19 '19 at 10:37

Avinash Sharma

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5
15

2

Need some explanation – nicomp Aug 19 '19 at 10:41

Robur_131 · Answer 3 · 2019-08-21T06:25:17.853

If you want to statically initialize array, the use the form int a[] = { ... } as mentioned in the comment. Following is the way to do that:

#include<stdio.h>

int main()
{

    int a[] = { 5641, 5, 98};

    char const *s = "this is a character array";
    printf("%s\n",s);

    int arrayLength = sizeof(a) / sizeof(a[0]);
    int i = 0;
    printf("Array length is %d\n", arrayLength);
    for(; i < arrayLength ; ++i){
        printf("%d ", a[i]);
    }
    printf("\n");
}

If you want to dynamically allocate array, which is reside in the heap memory, then you could do the following:

#include <stdio.h>
#include <stdlib.h>

int main()
{

    //int a[] = { 5641, 5, 98};
    int arrayLength = 3;
    int *a = (int*) malloc( (arrayLength + 1) * sizeof(a));
    a[0] = 5641;
    a[1] = 5;
    a[2] = 98;

    char const *s = "this is a character array";
    printf("%s\n",s);

    int i = 0;
    printf("Array length is %d\n", arrayLength);
    for(; i < arrayLength ; ++i){
        printf("%d ", a[i]);
    }
    printf("\n");
    free(a);
}

You have to call malloc() to allocate memory in the heap. Call free() when you're done with array, otherwise it will lead to memory leak.

C programming pointers and array

3 Answers3