c++ - 隐式数值类型转换规则

Question

我有几个关于带符号和无符号变量的类型转换的问题。在这种情况下将选择哪种类型有符号整数 + 无符号整数 = ???
那么当我有常量而不是变量的情况下呢

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x1(unsigned int) = x1(unsigned int) - 0xFFFFFFFA

我读过类似常量将是 unsigned ints 的内容，除非它明确地写成 0xFFFFFFFA(UL)。然后

1) unsigned int - unsigned int = unsigned int
2) unsigned int = unsigned int

如果数字是 float 会怎样

 x1(unsigned int) = x2(signed short int) + x3(unsigned int) + x4(unsigned short int) 
 * 0.1(float);  
 1)float * unsigned short int = float
 2)float + unsigned int = float 
 3)float + signed short int = float
 4)unsigned int = (unsigned int)float

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这里我猜 'a' 将是 char

signed int = 'a' + signed short int - signed int 
1) 'a' + signed short int = int ???
2) int - signed int = int ???
3) signed int = (signed int) int ???

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还有一个

long double = signed int + wchar_t - unsigned int * 10 
1)unsigned int * 10(int) = int
2)wchar_t - int = int
3)signed int + int = int 
4)long double = (long double) int

Answer 1

你总是可以使用我评论的技巧:

struct {} _ = some_expression;

编译器错误会告诉您表达式的确切类型。

有关背景资料，请参阅 http://cppreference.com , 具体来说 Implicit Conversion

• Numeric Promotions

• Numeric Conversions

您还应该阅读 conv.prom和 conv.double在标准中。