How to create an n sized array in c

Question

I am very new to C but I am having trouble on what seems a very trivial problem. All I am trying to do is create a n sized array, such that at the time of running I don't know its size and can't specify it. The following code works perfectly well.

int main()
{
        
    int LuhnsArray[15] = {0};

    for(int i = 0; i < 15; i++)
    {
        printf("%i ", LuhnsArray[i]);
        printf("\n");
    }
}

However when I try and so something like this:

int main()
{
    
    int length = 15;
    
    int LuhnsArray[length] = {0};

    for(int i = 0; i < length; i++)
    {
        printf("%i ", LuhnsArray[i]);
        printf("\n");
    }
}

Something which I would deem logical (something which I think would work in python for example) it comes up with an error attached. Could someone please point me in the right direction. Any chance of help from a newbie?

hyde · Accepted Answer · 2022-07-21 20:06:38Z

5

The error is quite self-explanatory. Remove the initialization in definition:

int LuhnsArray[length]; /* = {0}; nope */

Instead, use memset if you want it zeroed out:

memset(LuhnsArray, 0, sizeof(LuhnsArray));

Note how memset takes byte size, which we can get conveniently for an array (but not for a pointer!) with sizeof operator.

Or just a for-loop:

for(int i = 0; i < length; i++)
{
    LuhnsArray[i] = 0;        
}

answered Jul 21, 2022 at 20:06

hyde

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Comments

Vlad from Moscow · Accepted Answer · 2022-07-21 21:37:14Z

As the error message says you may not initialize a variable length array in its declaration

int length = 15;

int LuhnsArray[length] = {0};

Note: A variable length array is an array the size of which is not specified with integer constant expression.

From the C Standard (6.7.9 Initialization)

3 The type of the entity to be initialized shall be an array of unknown size or a complete object type that is not a variable length array type

Instead you could write

#include <string.h>

//...

int length = 15;

int LuhnsArray[length];
memset( LuhnsArray, 0, sizeof( LuhnsArray ) );

Or instead of calling the function memset you could initialize the array in a loop as for example

for ( size_t i = 0; i < length; i++ )
{
    LuhnsArray[i] = 0;
}

Also you could allocate the array dynamically as for example

#include <stdlib.h>

//...

int length = 15;
int *LuhnsArray = malloc( length * sizeof( *LuhnsArray ) );
//..

or instead of malloc you could use calloc to zero initialize the array

#include <stdlib.h>

//...

int length = 15;
int *LuhnsArray = calloc( length, sizeof( *LuhnsArray ) );
//..

Pay also attention to that you may not declare a variable length array in a file scope. Variable length arrays must have automatic storage duration.

John Bode · Accepted Answer · 2022-07-22 02:06:23Z

1

The second snippet defines LuhnsArray as a variable-length array, meaning its size isn't determined until runtime.

Because of this, you may not use an initializer when you declare a VLA, because the size of the thing being initialized has to be known at compile time.

Similarly, a VLA may not be declared static or at file scope, since storage for such objects is allocated when the program is first loaded, even before main is executed.

answered Jul 22, 2022 at 2:06

John Bode

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