explain the output please.

Question

#include <stdio.h>
#include <string.h>

void fun(char *arr)
{
int i;
unsigned int n = sizeof(arr);
printf("n = %d\n", n);
for (i=0; i<n; i++)
 printf("%c ", arr[i]);
}

// Driver program
int main()
{
char arr[] = {'k', 'h', 'u', 's', 'h', 'a', 'l'};
fun(arr);
return 0;
}

Answer 1

When passing an array as an argument to a function, a fixed array decays into a pointer, and the pointer is passed to the function.

#include <stdio.h>
#include <string.h>

void fun(char *arr)
{
int i;
unsigned int n = sizeof(arr); //This will give size of pointer. On different machines it can be a different size(on mine Its 4)
printf("n = %d\n", n);  // n=4
for (i=0; i<n; i++)
 printf("%c ", arr[i]);  
}

int main()
{
char arr[] = {'k', 'h', 'u', 's', 'h', 'a', 'l'};
fun(arr);
return 0;
}

 

output will be

4

k h u s

Comments

But if we write sizeof(arr) in main why it returns 7?

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I have edited!!

sizeof(arr) is the size of the array in memory, in this case 7 chars each being 1 byte equals 7

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Almost machines give 4 as answer if we take char array or integer

but still didn't satisfy with your answer

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My machine gives n=8. Maybe because in my machine char * takes 8 bytes.

Answer 2

Here in this snippet the main driver function is passing the pointer to the integers and thus its is returning the size of the pointer only not the array: the underlying concept is: " If we pass an array to a function its the address of first element plus the data size that is passed to function ".
But when you check the size of an array in main program itself then its the total size that is evaluated and thus it will return the total size.