在C ++中是文字0xffffffff int还是unsigned

Question

According to this , integer literals without type suffix are always int s. However, both gcc and clang interpret 0xffffffff (or any literal which explicitly sets the sign bit other than using the - ) as unsigned. Which is correct? (according to this the compilers are)

Answer 1

Per Paragraph 2.14.2/2 of the C++11 Standard,

The type of an integer literal is the first of the corresponding list in Table 6 in which its value can be represented.

Table 6 reports that for hexadecimal constants, the type should be:

• int ; or (if it doesn't fit)
• unsigned int ; or (if it doesn't fit)
• long int ; or (if it doesn't fit)
• unsigned long int ; or (if it doesn't fit)
• long long int ; or
• unsigned long long int .

Assuming your implementation has 32-bit int , since 0xffffffff does not fit in an int , its type should be unsigned int . For an implementation with a 64-bit int , the type would be int .

Notice, that if you had written the same literal as a decimal constant instead, the type could have only been:

• int ; or (if it doesn't fit)
• long int ; or (if it doesn't fit)
• long long int .