难道一元+运算符做的类型转换？ [英] Does Unary + operator do type conversions?

问题描述

到现在为止，我认为是没有用的一元 + 运营商。

Till now I was believing that there is no use of unary + operator.

但后来我用下面的例子中遇到：

But then I came across with following example:

char ch;
short sh;
int i;

printf("%d %d %d",sizeof(ch),sizeof(sh),sizeof(i)); // output: 1 2 4

printf("%d %d %d",sizeof(+ch),sizeof(+sh),sizeof(i)); // output: 4 4 4

这是否意味着 + 正在这里做什么类型的转换？

Does it mean + is doing type conversion here?

由于它表现下同

printf("%d %d %d",sizeof((int)ch),sizeof((int)sh),sizeof(i)); // output: 4 4 4

这迫使我觉得 + 正在做类型转换。

This forces me to think + is doing type conversion.

但后来我尝试在双击

double f;
printf("%d %d",sizeof(+f),sizeof((int)f),sizeof(f));  // output: 8 4 8

这迫使我重新思考一元 + 运营商。

This forces me to rethink about unary + operator.

所以我的第二个问题是：一元 + 操作员的sizeof 运营商特殊效果？

So my second question is: does unary + operator has special effect in sizeof operator?

推荐答案

的单目+ 的其操作数执行整型的提升，我们可以通过转到C99标准草案看到 6.5.3.3 的单目算术运算符的它说（的重点煤矿前进的）：

Unary + performs integer promotions on its operand, we can see this by going to the draft C99 standard section 6.5.3.3 Unary arithmetic operators which says (emphasis mine going forward):

一元+运算符的结果是价值的（推动）
  操作数。 整数促销活动操作数执行，并且
  结果具有提升的类型。

The result of the unary + operator is the value of its (promoted) operand. The integer promotions are performed on the operand, and the result has the promoted type.

和部分 6.3.1 的算术运算的说道：

如果int可以重新present原始类型的所有值，该值
  转换为int类型;否则，将其转换为一个unsigned int。
  这些被称为整数促销活动 ^48） 所有其它类型
  由整数提升不会改变。

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.⁴⁸⁾ All other types are unchanged by the integer promotions.

请注意，所有其他的类型是整数促销不变，thereofore的双的客人提供的双的。这也将持有的浮动的为好，这将不会被提升到的双的

Note that all other types are unchanged by the integer promotions and thereofore double stays a double. This would also hold for float as well, which would not be promoted to double.

另外请注意，使用％d个为中的sizeof 的结果是不确定的行为，因为结果的为size_t 的。正确的格式说明会％祖。

Also note that using %d for the result of sizeof is undefined behavior since the result is size_t. the proper format specifier would be %zu.

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