Behaviour of Sizeof in C

I have learnt that when we pass the array name to sizeof, the name of the array does not decay to the pointer to base address. The code below verifies this fact by giving answer 10.

#include <stdio.h> 

int main(){  
    int arr[10];  
    printf("Size of array is %d" , sizeof(arr)/sizeof(int));  
    return 0;  
}

However when I run the code below, the answer comes 1. Irrespective of whether a dimension is written in prototype or not , the answer is 1. Why is it so ?

#include <stdio.h>

void dimension(int arr[]){  
    p开发者_如何学Gorintf("Sizof array is %d" , sizeof(arr)/sizeof(int));  
}


int main(){  
    int arr[10];  
    dimension(arr);  
    return 0;  
}  

This signature

void dimension(int arr[])

is absolutely equivalent to

void dimension(int *arr)

See also Question 6.4

Because you pass an array of unknown size which is equivalent to a pointer in this context. sizeof is calculated at compile time, not runtime.

When array is passed to a function, it is passed as a pointer, not an array, so the sizeof(arr) will return sizeof(int *)

In

void dimension(int arr[]){  
    printf("Sizof array is %d" , sizeof(arr)/sizeof(int));  
}

arr[] decays to a pointer, therefore you have the equivalent of

printf("Sizof array is %d" , sizeof(int*)/sizeof(int));

and because on your platform, sizeof(int*) == sizeof(int), you receive 1 as the result.

Note however, that for variable length arrays, sizeof becomes a runtime operation:

int main () {
    int i = ...;
    int x[i];
    printf("number of elements: %d", sizeof (x) / size(*x));
}

Arrays as function arguments do decay to pointer, though. Since this happens before sizeof() is called, you can't prevent it.

Just think about it: how can sizeof() know the size of an array if any size array can be passed and no extra info is available? You get sizeof(pointer), and that seems to be the same size as an int, in your setup.

because arr is a pointer, which is an integer .

Because you're passing an unknown size array.