c++ - C++ 运算符中的隐式类型转换规则

Question

我想更好地知道何时应该施放。 C++中加法、乘法等时的隐式类型转换规则是什么？例如

int + float = ?
int * float = ?
float * int = ?
int / float = ?
float / int = ?
int / int = ?
int ^ float = ?

诸如此类...

表达式是否总是被计算为更精确的类型？ Java 的规则是否有所不同？如果我对这个问题的措辞不准确，请纠正我。

Answer 1

在 C++ 中，运算符(对于 POD 类型)始终作用于相同类型的对象。
因此，如果它们不相同，则会将其提升为与另一个匹配。
运算结果的类型与操作数相同(转换后)。

if:
either is      long double       other is promoted >      long double
either is           double       other is promoted >           double
either is           float        other is promoted >           float
either is long long unsigned int other is promoted > long long unsigned int
either is long long          int other is promoted > long long          int
either is long      unsigned int other is promoted > long      unsigned int
either is long               int other is promoted > long               int
either is           unsigned int other is promoted >           unsigned int
either is                    int other is promoted >                    int

Otherwise:
both operands are promoted to int

注意。操作的最小大小为int。因此，在操作完成之前，short/char 会被提升为 int。

在所有表达式中，int 在执行操作之前都会提升为 float。运算结果是一个float。

int + float =>  float + float = float
int * float =>  float * float = float
float * int =>  float * float = float
int / float =>  float / float = float
float / int =>  float / float = float
int / int                     = int
int ^ float =>  <compiler error>