C ++运算符中的隐式类型转换规则
我想要更好地知道我什么时候应该投降。例如，
I want to be better about knowing when I should cast. What are the implicit type conversion rules in C++ when adding, multiplying, etc. For example,
int + float = ?
int * float = ?
float * int = ?
int / float = ?
float / int = ?
int / int = ?
int ^ float = ?
et cetera ...
et cetera...
表达式总是被评估为更精确的类型？ Java的规则有所不同吗？
Will the expression always be evaluated as the more precise type? Do the rules differ for Java? Please correct me if I have worded this question inaccurately.
在C ++操作符对同一类型的对象。
因此，如果它们不相同，那么将被提升为匹配另一个。
操作结果的类型与操作数相同（转换后）。
In C++ operators (for POD types) always act on objects of the same type.
Thus if they are not the same one will be promoted to match the other.
The type of the result of the operation is the same as operands (after conversion).
If either is long double the other is promoted to long double
If either is double the other is promoted to double
If either is float the other is promoted to float
If either is long long unsigned int the other is promoted to long long unsigned int
If either is long long int the other is promoted to long long int
If either is long unsigned int the other is promoted to long unsigned int
If either is long int the other is promoted to long int
If either is unsigned int the other is promoted to unsigned int
If either is int the other is promoted to int
Both operands are promoted to int
注意。操作的最小大小为 int 。因此，在操作之前， short / char 被提升为 int
Note. The minimum size of operations is int. So short/char are promoted to int before the operation is done.
在所有的表达式中， int 被提升为 code>，然后再执行操作。该操作的结果是 float 。
In all your expressions the int is promoted to a float before the operation is performed. The result of the operation is a 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>