Is unsigned char('0') legal C++

Question

The following compiles in Visual Studio but fails to compile under g++.

int main()
{
    int a = unsigned char('0');
    return 0;
}

Is unsigned char() a valid way to cast in C++?

Answer 1

No, it's not legal.

A function-style explicit type conversion requires a simple-type-specifier , followed by a parenthesized expression-list . (§5.2.3) unsigned char is not a simple-type-specifier ; this is related to a question brought up by James .

Obviously, if unsigned char was a simple-type-specifier , it would be legal. A work-around is to use std::identity :

template <typename T>
struct identity
{
    typedef T type;
};

And then:

int a = std::identity<unsigned char>::type('0');

std::identity<unsigned char>::type is a simple-type-specifier , and its type is simply the type of the template parameter.

Of course, you get a two-for-one with static_cast . This is the preferred casting method anyway.

Answer 2

在C ++中首选的方法是使用static_cast如下所示：

int a = static_cast<unsigned char>( '0' );

Answer 3

Try to add brackets int a = (unsigned char)('0');

or

typedef unsigned char uchar;

//inside main
int a = uchar('0');

Answer 4

No, it isn't - a function-style cast cannot have a space in its name.

A case for a C-style cast perhaps:

int main() {
    unsigned char c = (unsigned char) '0' ;
}

Answer 5

我很确定这是微软的扩展。

Answer 6

No, it isn't. But why the cast in the first place? This is perfectly valid,

int a = '0';

Answer 7

Why are you even trying to cast from char to unsigned char and assigning that to an int? You're putting an unsigned value into a signed int (which is legal, but uncool).

Writing '0' gives you a char with value 48. You can try

int i = (int) '0';

That way, you take the char, cast it to an int, and use it. You could even say

int i = '0';

And that would do the same thing. What exactly are you trying to do?

Answer 8

Are you trying to get the integer 0 or the character '0' into it? The character '0' on most implementations namely is just the integer 48 but put into 8 bits.

The only difference between a char and an int is that char must be smaller or equal to short int. and int must be larger or equal than short int accordingly, this usually makes char 8 bits, short in 16, and in 32 nowadays.

Stuf like 'a'+2 to get 'c' works namely. If you have an array that is long enough, you can also index it like array[' '] to get index 32.

If you're trying to cast it to the integer value 0, that would require an actual function that determines that.