C++14 中的统一初始化双分配给浮点变量不会产生缩小错误

Question

根据我对统一初始化的理解，以下代码应该会产生缩小错误，但它不会：

#include<iostream>
int main() {
    double d;
    float f = {d}; // should produce a narrowing error
return 0;

}

我认为这是一个错误，因为编译器必须考虑 d 的所有可能值，换句话说，d 可能包含的值的范围大于 float 可以包含的范围。

gcc version 4.9.3

Answer 1

这确实会使用 g++ -std=c++11 发出警告，但是如果您省略 c++11 编译器标志（使用 g++ 4.9.2）则不会。

使用 clang++ -std=c++14 -stdlib=libc++ -Wall -pedantic -Wdeprecated -Wextra 这不会编译，但会给出缩小错误而不是仅警告（使用 clang++ 3.6.0）。

Answer 2

需要收缩转换才能发出诊断。警告或错误满足。

运行你的代码 here（gcc 4.9.2 是我可以 link 使用的最接近的版本）发出警告

warning: narrowing conversion of 'd' from 'double' to 'float' inside { } [-Wnarrowing]

这满足了标准的要求。

如果你真的想要一个错误，你总是可以使用 -Werror 或将所有警告视为错误或 -pedantic-errors 将缩小转换视为错误。

Answer 3

标准没有区分"errors"和"warnings"。您的代码是 ill-formed，但标准是这样说的：

N4140 [intro.compliance]/2.2: If a program contains a violation of any diagnosable rule or an occurrence of a construct described in this Standard as “conditionally-supported”when the implementation does not support that construct, a conforming implementation shall issue at least one diagnostic message.

[intro.compliance]/8: A conforming implementation may have extensions (including additional library functions), provided they do not alter the behavior of any well-formed program. Implementations are required to diagnose programs that use such extensions that are ill-formed according to this International Standard. Having done so, however, they can compile and execute such programs.

GCC 为 ill-formed 代码发出诊断，因此它在这方面是兼容的。请注意，如果您传递 -pedantic-errors.

，它将发出错误

Answer 4

来自此处的错误：https://gcc.gnu.org/bugzilla/show_bug.cgi?id=55783

The standard only requires that "a conforming implementation shall issue at least one diagnostic message" so compiling the program with a warning is allowed. As Andrew said, -Werror=narrowing allows you to make it an error if you want.

C++14 中的统一初始化双分配给浮点变量不会产生缩小错误

Uniform initialization in C++14 double assigned to float variable does not produce narrowing error

c++

initialization

c++14