VC ++中的文字类型(Literals types in VC++)

 我知道标准如下： 
 
 
 
 以0开头的积分被解释为八进制。 
 
 
 以0x或0X开头的积分被解释为十六进制。 
 
 
 整数文字的类型取决于其值和符号： 
 
 
 
 默认情况下，小数是有符号的，并且具有值适合的最小类型int，long，long long。 
 
 
 十六进制和八进制可以是有符号或无符号的，并且具有最小类型的int，unsigned int，long，unsigned long，long long，unsigned long long，其中文字值适合。 
 
 
 没有类型为short的文字但这可以通过后缀覆盖。 
 
 
 但是VC ++怎么样？！ 它似乎将十进制，八进制和十六进制视为相同，并且小数也允许使用无符号类型。 
 
 类似下面的代码： 
 
cout << typeid(4294967295).name() << endl;
cout << typeid(4294967296).name() << endl;

cout << typeid(0xffffffff).name() << endl;
cout << typeid(0x100000000).name() << endl;
 
 得到： 
 
unsigned long
__int64
unsigned int
__int64
 
 这是预期的，为什么它与标准不同？ 

I know the standard is as follows:
 
 
 
Integrals starting with 0 are interpreted as octal.
 
 
Integrals starting with 0x or 0X are interpreted as hexadecimal.
 
 
The type of an integer literal depend on its value and notation:
 
 
 
Decimals are by default signed and has the smallest type of int, long, long long in which the value fits.
 
 
Hexadecimal and octal can be signed or unsigned and have the smallest type of int, unsigned int, long, unsigned long, long long, unsigned long long in which the literal value fits.
 
 
No literals of type short but this can be override by a suffix.
 
 
But what about VC++?! It seems to be treating decimal, octal and hexadecimal the same and unsigned types are also allowed for decimals.
 
something like the following code:
 
cout << typeid(4294967295).name() << endl;
cout << typeid(4294967296).name() << endl;

cout << typeid(0xffffffff).name() << endl;
cout << typeid(0x100000000).name() << endl;
 
gives:
 
unsigned long
__int64
unsigned int
__int64
 
Is this expected and why it is different from the standard?

原文：https://stackoverflow.com/questions/19167208