Question

I’m fairly new to C++ and I recently came across this problem.

This code will obviously work:

void setvalues(int *c, int *d)
{
    (*c) = 1;
    (*d) = 2;
}
int main()
{
    int a, b;
    setvalues(&a, &b);
    std::cout << a << b;
}

So why does this return an error? Visual C++ 2010 error:

C2664: 'setvalues' : cannot convert parameter 1 from 'int (*)[2]' to 'int *[]'

void setvalues(int *c[2], int *d[2])
{
   (*c[1]) = 1;
   (*d[1]) = 2;
}
int main()
{
    int a[2], b[2];
    setvalues(&a, &b);
    std::cout << a[1] << b[1];
}

What’s different about pointers to arrays? I searched around but no luck.

Answer 1

The type int *a[2] means array of 2 pointers to int, but the expression &a with the definition int a[2] means pointer to an array of 2 int. Both are different types and there is no conversion among them. As Vlad already mentioned, to provide the proper type you need to add parenthesis:

void setvalues( int (*c)[2] )

Or you could use actual references in C++:

void setvalues( int (&c)[2] )

In the later case you don’t need to use the address-of operator or dereference it inside the setvalue function:

int a[2];
setvalues(a); // this is a reference to the array

A simpler way to write the code is to use typedef:

typedef int twoints[2];
void setvalue( toints& c );
int main() {
   twoints a; // this is int a[2];
   setvalue(a);
}