C initializes and re-initializes arrays to zero at runtime when asked

Question

#include <stdio.h>

int main(void) {
    for (int i = 0; i < 5; i++) {
        int arr[5] = {0};
        // arr gets zeroed at runtime, and on every loop iteration!
        printf("%d %d %d %d %d\n", arr[0], arr[1], arr[2], arr[3], arr[4]);
        // overwrite arr with non-zero crap!
        arr[0] = 3;
        arr[1] = 5;
        arr[2] = 2;
        arr[3] = 4;
        arr[4] = 1;
    }
    return 0;
}

Apparently this works:

> gcc -Wall -Wextra -pedantic -std=c99 -o test test.c;./test
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

But:

• What's going on under the hood?
• Is this guaranteed to work for arrays of any size?
• Have I found the most elegant way of zeroing an array on every iteration of a loop?

Answer 1

You are creating a new array at each iteration, so there is no re-initialization.

Answer 2

See according to me whenever you are iterating over a loop, then whatever variables you are declaring inside the loop are local to that loop. So at the end of the loop they are getting destroyed, and when the loop is again starting then they are being created again. So, 1. answered above; 2. yes this should work for array of any size; 3. no...

Answer 3

The variable arr is created and initialized on every iteration of the loop because that's what you wrote. In your example, the code does something analogous to memset() to set the array to zero, but possibly more efficiently and inline. You could write a non-zero initializer too.

Yes, it works for 'any' size of array. If the arrays are bigger, the loop will be slower.

It is a very elegant way to achieve the effect. The compiler will do it as fast as possible. If you can write a function call that's quicker, the compiler writer's have goofed.

Answer 4

This has been answered before:

• how does array[100] = {0} set the entire array to 0?

What's going on under the hood?

The compiler will generate code equivalent to memset (see Strange assembly from array 0-initialization for full details).

Is this guaranteed to work for arrays of any size?

Up to the limit of your available stack size, yes.

Have I found the most elegant way of zeroing an array on every iteration of a loop?

I suppose so. :)

Answer 5

Yes, the C Standard guarantees that your array elememts are zeroed on each loop iteration. If you only want them zerored on the first iteration, make the array static, i.e. with

static int arr[5] = {0}; 

You may leave out the = { 0 } in this case, as static data is required to be initialized as if all elements were assigned 0 in the absence of an initializer (or NULL for pointers, or 0.0 for floating point values, recursively for structs).

Note also, that if you use less initializers than the array dimension specifies, the remaining elements are assigned zeros. So

int arr[5] = {42, 0xdead};

is equivalent to

int arr[5] = {42, 0xdead, 0, 0, 0};

and using a single 0 is just a special case of this rule.

Answer 6

Not ignoring the logical error of the code (re initialization at each iteration) I will recommend using memcpy or memset whenever you want to assign all the elements of an array.. This is the tried and tested way, which fortunately works as a charm in almost all the compiler implementations.

Memset Example

MemCpy Example

So, Using Memset your code will become;;

int a[5]; 
memset(a,0,<number of bytes to be set to> sizeof(int)*5); 

Please read the documentation to find out why it is fast, you will enjoy...

EDIT Sorry and Thanks for all the down voters / commentors... I corrected the answer...