Why is every value not initialized to 0?

Question

#include <stdio.h>

int main(void) {
    int memo[1000];
    for (int i = 0; i < 1000; i++) {
        printf("%d\t", memo[i]);
    }
    return 0;
}

I thought everything should be initialized to 0 but it is not. Is there any reason for this? Thank you so much for your help.

Answer 1

Objects defined locally with automatic storage in a function body are uninitialized in C. The values can be anything, including trap values on some architectures that would cause undefined behavior just by reading them.

You can initialize this array as int memo[1000] = { 0 };. The first element is explicitly initialized to 0, and all the remaining ones will also initialized to 0 because any element for which the initializer is missing will be set to 0.

For completeness, int memo[1000] = { 42 }; will have its first element set to 42 and all remaining ones to 0. Similarly, the C99 initializer int memo[1000] = { [42] = 1 }; will have its 43rd element set to 1 and all others set to 0.

Answer 2

In C, when you do

int memo[1000];

you are allocating 1000 little spots in memory. Those spots may or may not currently hold some garbage data already. So it is often considered good practice to initialize all your variables. It will make debugging easier.

You can replace the above line with,

int memo[1000] = {0};

to initialize every element of the array to be 0;

Edit: There are definitely deeper reasons for the behaviour you're experiencing, as talked about by the other answers here. But, if you are a beginner to C or programming in general. This is sufficient for you to start using arrays with initialization without being overwhelmed by more complicated details. If you are more advanced then please defer to the other answers.

Answer 3

From the C Standard (6.7.9 Initialization)

10 If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:

— if it has pointer type, it is initialized to a null pointer;

— if it has arithmetic type, it is initialized to (positive or unsigned) zero;

— if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

— if it is a union, the first named member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

and

19 ... all subobjects that are not initialized explicitly shall be initialized implicitly the same as objects that have static storage duration.

Thus in this program

#include <stdio.h>

int main(void) {
    int memo[1000];
    for (int i = 0; i < 1000; i++) {
        printf("%d\t", memo[i]);
    }
    return 0;
}

the array memo has the automatic storage duration and according to the provided quote from the C Standard its elements have indeterminate values.

You could declare the array like

    int memo[1000] = { 0 };

In this case the first element of the array is explicitly initialized by 0 and all other elements are implicitly initialized also by 0.

You could select any element of the array to be explicitly initialized like for example

    int memo[1000] = { [999] = 0 };

If you would write

#include <stdio.h>

int memo[1000];

int main(void) {
    for (int i = 0; i < 1000; i++) {
        printf("%d\t", memo[i]);
    }
    return 0;
}

or

#include <stdio.h>

int main(void) {
    static int memo[1000];
    for (int i = 0; i < 1000; i++) {
        printf("%d\t", memo[i]);
    }
    return 0;
}

then elements of the array will be zero-initialized because a variable declared in a file scope or with the storage specifier static has the static storage duration.