C compile error: "Variable-sized object may not be initialized"

I am assuming that you are using a C99 compiler (with support for dynamically sized arrays). The problem in your code is that at the time when the compilers sees your variable declaration it cannot know how many elements there are in the array (I am also assuming here, from the compiler error that length is not a compile time constant).

You must manually initialize that array:

int boardAux[length][length];
memset( boardAux, 0, length*length*sizeof(int) );

You receive this error because in C language you are not allowed to use initializers with variable length arrays. The error message you are getting basically says it all.

6.7.8 Initialization

...

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

This gives error:

int len;
scanf("%d",&len);
char str[len]="";

This also gives error:

int len=5;
char str[len]="";

But this works fine:

int len=5;
char str[len]; //so the problem lies with assignment not declaration

You need to put value in the following way:

str[0]='a';
str[1]='b'; //like that; and not like str="ab";

After declaring the array

int boardAux[length][length];

the simplest way to assign the initial values as zero is using for loop, even if it may be a bit lengthy

int i, j;
for (i = 0; i<length; i++)
{
    for (j = 0; j<length; j++)
        boardAux[i][j] = 0;
}

The question is already answered but I wanted to point out another solution which is fast and works if length is not meant to be changed at run-time. Use macro #define before main() to define length and in main() your initialization will work:

#define length 10

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

Macros are run before the actual compilation and length will be a compile-time constant (as referred by David Rodríguez in his answer). It will actually substitute length with 10 before compilation.