Question

float b = 1.0f;
int i = (int)b;
int& j = (int&)b;

cout << i << endl;
cout << j << end;

Then the output of i was 1, and the output of j was 1065353216! It is a big surprise to me! So what is the true meaning of (int&) conversion?

Answer 1

This is the problem with a C-style cast. You have to look closely to see what you’re getting. In your case “(int)” was a normal static cast. The value is converted to an int via truncation. In your case “(int&)” was a reinterpret cast. The result is an lvalue that refers to the memory location of b but is treated as an int. It’s actually a violation of the strict aliasing rules. So, don’t be surprized if your code won’t work anymore after turning on all optimizations.

Equivalent code with C++ style casts:

float b = 1.0f;
int i = static_cast<int>(b);
int& j = reinterpret_cast<int&>(b);
cout<<i<<endl;
cout<<j<<end;

Check your favorite C++ book on these kinds of casts.