使用额外括号时“错误：预期表达式”

Question

I just stumbled across a compiler error that I was not expecting:

std::cout << sizeof(int)   << std::endl;  // ---> this is valid (obviously)
std::cout << sizeof((int)) << std::endl;  // ---> this leads to
                                          // "error: expected expression"

Similarly, we have:

template <typename T>
struct Foo
{
    T value;
};
Foo<int>   f1;  // ---> this is valid (obviously)
Foo<(int)> f2;  // ---> this leads to "error: expected expression"

Apparently, the compiler interprets (T) as an explicit conversion and stops there. I understand the error, I understand what the compiler is expecting, what I do not get is why (T) cannot be treated as T if it is not in the context of (T)(exp) . I thought that the compiler would be able to see through that (and probably return a warning), so I suppose this means that there are ambiguous cases where allowing those extra parentheses would lead to errors.

My questions are: when would that be dangerous? Is that in the C++ standard?

This was tested with clang 3.3. and g++ 4.7.2.

Answer 1

sizeof expects either a value (an expression) or a type. (int) is neither one of them - you can't treat types as values and expect them to act correctly in cases where you would expect values to do so, because they are not (first-class) values. Therefore, you can't just parenthesize a type and have the same type come out.