Why integer zero does not equal long zero?

Question

A strange piece of code I\'ve just discovered in C# (should also be true for other CLI languages using .NET\'s structs).

Answer 1

There is also the issue of narrowing or widening conversion. A long zero is always equal to an int zero, but not the other way around.

When a long is compared to an int, only the least significant 32-bits are compared and the rest are ignored, thus the int.Equals(long) operation cannot guarantee equality even if the lower bits match.

int a = 0;
long b = 0;

Trace.Assert(a.Equals((int)b));     // True   32bits compared to 32bits
Trace.Assert(a.Equals((long)b));    // False  32bits compared to 64bits (widening)
Trace.Assert(b.Equals((long)a));    // True   64bits compared to 64bits
Trace.Assert(b.Equals((int)a));     // True   64bits compared to 32bits (narrowing)

Also consider the case where the lower 32-bits are equal, but the upper ones are not.

uint a = 0;
ulong b = 0xFFFFFF000000;
Trace.Assert((uint)a == (uint)b);  // true because of a narrowing conversion
Trace.Assert((ulong)a == (ulong)b);  // false because of a widening conversion