> The result of the expression is the condition. Thus, in your example, the bool check would apply to "s", after the expression is evaluated. This is a contrad

> The result of the expression is the condition. Thus, in your example, the bool check would apply to "s", after the expression is evaluated. This is a contradiction. There is no expression in my code that evaluates to s. foo() is an expression, and then std::string s = ... is assignment initialisation, which is not an expression. Edit: I suppose that if I used another form of initialisation, the answer becomes a bit more obvious: if (std::string s('x', 3)) (Not that this makes sense but just the point is to use a constructor with more than one argument.) In this case it's clear the test has to be the just-initialised variable. In fact there could be no arguments at all!