没有警告将布尔值隐式转换为浮动类型? [英] No warning for implicit cast of bool to floating type?

问题描述

看起来像这样的代码片段即使没有-Weverything也可以在不发出警告的情况下用clang编译:

Looks like this snippet compiles in clang without warning, even with -Weverything:

double x;
...
if (fabs(x > 1.0)) {
   ...
}

我错过了什么吗?还是编译器和C ++标准认为将 bool 转换为 double 是有意义的事情?

Am I missing something? Or do the compiler and C++ standard think that casting bool to double is something that makes sense?

推荐答案

这是使 bool 成为整数类型的结果.根据C ++标准，第3.9.1.6节

This is a consequence of making bool an integral type. According to C++ standard, section 3.9.1.6

布尔类型的值是 true 或 false (注意:没有 signed ， unsigned ， short 或 long bool 类型或值.—尾注) bool类型的值参与积分促销.(重点已添加)

Values of type bool are either true or false (Note: There are no signed, unsigned, short, or long bool types or values. — end note) Values of type bool participate in integral promotions. (emphasis is added)

这使得 bool 表达式的值被提升为 float ，其方式与提升 int 的方式相同，而不会发出警告，例如在4.5.6节中进行了介绍:

This makes values of bool expressions to be promoted to float in the same way the ints are promoted, without a warning, as described in section 4.5.6:

可以将类型为 bool 的prvalue转换为类型为 int 的prvalue，其中 false 变为零，而 true 成为一个.

A prvalue of type bool can be converted to a prvalue of type int, with false becoming zero and true becoming one.

从C ++ 11开始， fabs 为整数类型提供了额外的重载，因此升级直接从 bool 扩展到 int ，并在那里停止，因为它可以使用 fabs 的重载.

EDIT : Starting with C++11 fabs offers additional overloads for integral types, so the promotion goes directly from bool to int, and stops there, because an overload of fabs is available for it.

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