Can I assume (bool)true == (int)1 for any C++ compiler?

Yes. The casts are redundant. In your expression:

true == 1

Integral promotion applies and the bool value will be promoted to an int and this promotion must yield 1.

Reference: 4.7 [conv.integral] / 4: If the source type is bool... true is converted to one.

Charles Bailey's answer is correct. The exact wording from the C++ standard is (§4.7/4): "If the source type is bool, the value false is converted to zero and the value true is converted to one."

Edit: I see he's added the reference as well -- I'll delete this shortly, if I don't get distracted and forget...

Edit2: Then again, it is probably worth noting that while the Boolean values themselves always convert to zero or one, a number of functions (especially from the C standard library) return values that are "basically Boolean", but represented as ints that are normally only required to be zero to indicate false or non-zero to indicate true. For example, the is* functions in <ctype.h> only require zero or non-zero, not necessarily zero or one.

If you cast that to bool, zero will convert to false, and non-zero to true (as you'd expect).

According to the standard, you should be safe with that assumption. The C++ bool type has two values - true and false with corresponding values 1 and 0.

The thing to watch about for is mixing bool expressions and variables with BOOL expression and variables. The latter is defined as FALSE = 0 and TRUE != FALSE, which quite often in practice means that any value different from 0 is considered TRUE.

A lot of modern compilers will actually issue a warning for any code that implicitly tries to cast from BOOL to bool if the BOOL value is different than 0 or 1.