Juan Sanchez wrote:
Ok,
But cpp doesn't discriminate between ${BAR} and BAR.
#include <iostream>
using namespace std;
#define foo(x) x = 3; cout << x << "\n";
int main()
{
int y = 1;
foo(y);
cout << y << endl;
}
I said like cpp, not exactly cpp. The difference is you have to use ${} to
expand and argument not just its name. Anyway, I don't think
it will get changed because it would break backward compatibility
way too much.
-Bill
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