C++ Syntax Puzzle

This is an interesting puzzle and possibly an interiew question:

Without attempting to compile it, is the following code compilable? Is t in scope in the if body or the else?

#include <iostream>

using namespace std;

int returnanint() {
 return 1;
}

using namespace std;
int main(int argc, char * argv[])
{
  if (int t=returnanint()) {
      cout << "t > 0 = " << t << endl;
    }
    else {
      cout << "t==0 " << endl;
      t++;
    }
  return 0;
}

Answer below
.
.
.
.
.
.
.
.
.
.
.
.
.
The answer is, yes it compiles. It's an interesting and certainly not intuitive gray area of C++. The int is declared outside of the braces following the if so is in scope for both those braces and those for the else branch.