浮点升级：stroustrup与编译器-谁是对的？ [英] floating-point promotion : stroustrup vs compiler - who is right?

问题描述

在Stroustrup的新书 C ++编程语言-第四版的第10.5.1节中，他说，在执行算术运算之前，使用整数提升从较短的整数类型中创建整数，并且类似地，点奖励用于创建浮点数的双打。

In section 10.5.1 of Stroustrup's new book "The C++ Programming Language - Fourth Edition" he says, that before an arithmetic operation is performed, integral promotion is used to create ints out of shorter integer types, and similarly, floating-point promotion is used to create doubles out of floats.

我用以下代码确认了第一个要求：

I confirmed the first claim with the following code:

#include <iostream>
#include <typeinfo>

int main()
{
    short a;
    short b;
    std::cout << typeid(a + b).name() << std::endl;
}

此命令会在vc ++中输出 int，在gcc中输出 i。

This outputs "int" with vc++ and "i" with gcc.

但是用浮子而不是短裤进行测试，输出仍然是 float或 f：

But testing it with floats instead of shorts, the output is still "float" or "f":

#include <iostream>
#include <typeinfo>

int main()
{
    float a;
    float b;
    std::cout << typeid(a + b).name() << std::endl;
}

根据Stroustrup，浮点促销规则没有例外，所以我期望输出为 double或 d。

According to Stroustrup there are no exceptions to the floating-point promotion-rule, so I expected "double" or "d" as output.

上述提到的有关促销的部分是错误的还是不清楚？ C ++ 98和C ++ 11在类型提升方面有什么区别吗？

Is the mentioned section about promotions wrong or somehow unclear? And is there any difference in C++98 and C++11 regarding type promotions?

推荐答案

我不知道什么正是Stroustrup的书说的那样，但是根据标准，在这种情况下，浮动 s不会转换为 double s。在应用大多数算术二进制运算符之前，将应用5p9中描述的常规算术转换：

I don't know what exactly Stroustrup's book says, but according to the standard, floats will not be converted to doubles in this case. Before applying most arithmetic binary operators, the usual arithmetic conversions described in 5p9 are applied:

• 如果任何一个操作数都属于范围枚举类型（7.2），则不执行任何转换；如果另一个操作数的类型不同，则表达式格式错误。


• 如果其中一个操作数为long double类型，则另一个应转换为long double。


• 否则，如果其中一个操作数为double，则另一个


• 否则，如果其中一个操作数为float，则另一个操作数应转换为float。


• 否则，积分提升（ 4.5）应在两个操作数上执行。 [...]

• If either operand is of scoped enumeration type (7.2), no conversions are performed; if the other operand does not have the same type, the expression is ill-formed.
• If either operand is of type long double, the other shall be converted to long double.
• Otherwise, if either operand is double, the other shall be converted to double.
• Otherwise, if either operand is float, the other shall be converted to float.
• Otherwise, the integral promotions (4.5) shall be performed on both operands. [...]

积分促销是造成两个短缺的原因转换为 int s。但是，根据这些规则，两个 float 不会转换为 double s。如果将 float 添加到 double ，则 float 将会被转换为 double 。

The integral promotions are what causes two shorts to be converted to ints. But two floats will not be converted to doubles according to these rules. If you add a float to a double, the float will be converted to a double.

上面的内容来自C ++ 11。 C ++ 03包含相同的规则，只是其中一个引用了范围枚举。

The above is from C++11. C++03 contains the same rules, except for the one referring to scoped enumerations.

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