How to initialize an array[int][int] with zeros in C?

Question

Picture this scenario. I need to store the quantity of products of a delivery. Each delivery and product has a number to identify it. I am using an array somewhat like this: int array[500][500];

The main issue is that I do not have idea how to initialize the whole array with 0s. Any left-out junk in the 500 slots can backfire eventually in the many other algorithms I still have to code. I used to do int array[500] = {0}; but in this case, there are two ID's related to each index.

Answer 1

I used to do int array[500] = {0}; but in this case there is two ID's related to each index.

Doesn't matter; that's still valid and appropriate in C11.

Try it and see!

Live demo

^{(Don't mind the ".cpp"; that's fixed for us by Coliru, but -xc tells GCC to compile the code as C.)}

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You do not need extra brackets. The standard spells this out explicitly:

[C11: 6.7.9/20]: If the aggregate or union contains elements or members that are aggregates or unions, these rules apply recursively to the subaggregates or contained unions. If the initializer of a subaggregate or contained union begins with a left brace, the initializers enclosed by that brace and its matching right brace initialize the elements or members of the subaggregate or the contained union. Otherwise, only enough initializers from the list are taken to account for the elements or members of the subaggregate or the first member of the contained union; any remaining initializers are left to initialize the next element or member of the aggregate of which the current subaggregate or contained union is a part.

And later provides an example:

[C11: 6.7.9/26]: EXAMPLE 3 The declaration

int y[4][3] = { { 1, 3, 5 }, { 2, 4, 6 }, { 3, 5, 7 }, };

is a definition with a fully bracketed initialization: 1, 3, and 5 initialize the first row of y (the array object y[0] ), namely y[0][0] , y[0][1] , and y[0][2] . Likewise the next two lines initialize y[1] and y[2] . The initializer ends early, so y[3] is initialized with zeros. Precisely the same effect could have been achieved by

int y[4][3] = { 1, 3, 5, 2, 4, 6, 3, 5, 7 };

The initializer for y[0] does not begin with a left brace, so three items from the list are used. Likewise the next three are taken successively for y[1] and y[2] .

Recall, by the way, that arrays are aggregates in C:

[C11: 6.2.5/21]: Arithmetic types and pointer types are collectively called scalar types. Array and structure types are collectively called aggregate types.

Now, for the case of providing only one value (the 0), the normal rule you're used to (that all remaining elements take on zero anyway) still applies:

[C11: 6.7.9/21]: If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.

You may still receive style warnings from your compiler about this (I don't), but you can ignore them.

If you're using an antique version of C, though, you may need to double-up on your braces; I'm not too familiar with the history of this feature.

Answer 2

If the sizes of the array are compile-time constants like the value 500 then it is enough to write

int array[500][500] = { 0 };

Or you can even use a designated initializer for example like

int a[500][500] = { [0][0] = 0 };

or like

int a[500][500] = { [0] = { 0 } };

or something similar.

In this case all elements of the array will be zero-initialized.

If it is a variable length array then you may not initialize it at declaration. In this case you can use the standard C function memset. For example

#include <string.h>

//...

int main( void )
{
    size_t m = 500;
    size_t n = 500;

    int array[m][n];

    memset( array, 0, m * n * sizeof( int ) );
    //...

Answer 3

Use memset to initialize the entire memory block to 0:

NAME         
   memset - fill memory with a constant byte

SYNOPSIS         
   #include <string.h>

   void *memset(void *s, int c, size_t n);

DESCRIPTION         
   The memset() function fills the first n bytes of the memory area
   pointed to by s with the constant byte c.

RETURN VALUE         
   The memset() function returns a pointer to the memory area s.

So for your example:

int array[500][500];
memset(&array, 0, sizeof(int) * 500 * 500);

Answer 4

If your array is defined as a global variable or a static variable local to a function, you don't actually need to initialize it, all elements will have a zero value at program start.

If the array is defined inside a function body with automatic storage, int array[500][500]; , you can initialize it at the point of definition with

int array[500][500] = { 0 };

If you mean to reset all values to zero at any point during program execution, you can use memset from <string.h> :

memset(array, 0, sizeof array);