what does "double a[size][size]" mean as a function argument?-CodePudding

I was coding a problem to print the elements on the diagonal and I do not understand why the first function is working and the second one is not.

Snippet 1:

void printDiag1(int size, double a[][101], int row, int col)
{
    if(size<=0) return;
    printDiag1(size-1,a,row,col);
    printf("%lf ",a[size-1][size-1]);
}

Snippet 2:

void printDiag2(int size, double a[size][size], int row, int col)
{
    if(size<=0) return;
    printDiag2(size-1,a,row,col);
    printf("%lf ",a[size-1][size-1]);
}

Can somebody explain me why and tell what

double a[size][size]

means in the arguments of the second function?

CodePudding user response：

In either function a is a 2D array (array of arrays). As per the rules of function parameters, arrays passed as parameters "decay" into a pointer to their first element. That's for example the very reason you are allowed to type double a[][101] with the left dimension empty.

So these two examples "decay" into pointers to arrays, double (*a)[101] and double (*)[size] respectively. The only difference is that double (*a)[101] can only point to an array of size 101 and double (*)[size] can only point to an array of size size. The latter is a pointer to a variable-length array (VLA) with flexible size.

VLA can be used for increased type safety which is what's happening here. printDiag2(size-1,a,row,col); doesn't work because a still expects to point at an array of type double [size][size], so you can't pass that to a function suddenly expecting double [size-1][size-1].

I would strongly recommend to stay away from recursion in general. It's a pox and some 99% of all use-cases where we spot it on SO are invalid. In this case a plain for loop would have been much faster, much more readable, less memory consuming and likely without bugs such as the one you got here.