comparison operator values C++

Question

We all know C++ (while not a superset) is pretty much derived from C.

In C++, the operators <, <=, >, >=, ==, and != all have boolean return values. However, in C, the same operators returned 1 or 0, since there was no 'bool' type in C.

Since all integer values except 0 are treated as "true", and 0 is "false", I want to know:

Does C++ still restrict the return values of the operators to be 1 vs 0, or does a 'true', from one of these operators, return any 1-byte value, so long as it isn't 0?

I want to know since using these return values as explicit 1 or 0 would be useful in bitwise operations without branching.

As a terrible example, take the following:

bool timesTwo;
int value;

//...

if(timesTwo)
    value << 1;

//vs

value << (int) timesTwo;

Answer 1

Does C++ still restrict the return values of the operators to be 1 vs 0, or does a 'true', from one of these operators, return any 1-byte value, so long as it isn't 0?

The comparison operators, assuming that they have not been overloaded, only ever return true and false.

int(true) is always 1.

int(false) is always 0.

So,

int one(1), two(2);
assert( (one<two) == 1 );
assert( (two<one) == 0 );

Answer 2

In fact, bool is automatically converted into integer in C++ (1 if true, 0 if false) when used in expressions and there's no need to cast.

value << (int) timesTwo;

The case is not necessary: if `timesTwo`` is true then

you can directly do without a casr:

value<<timesTwo;

which is equivalent to:

value<<1;