为什么不缩小转换使用与大括号分隔的初始化程序导致错误？ [英] Why doesn't narrowing conversion used with curly-brace-delimited initializer cause an error?

问题描述

我学习了C ++编程语言中的大括号分隔初始化器，第4版。 >

I learnt about curly-brace-delimited initializer in The C++ Programming Language, 4th ed. > Chapter 2: A Tour of C++: The Basics.

我在下面的书中引用。

The形式是传统的，可追溯到C，但如果有疑问，请使用general {} -list form（§6.3.5.2）。
如果没有别的，它可以避免失去信息的转化（缩小转化;§10.5）：

The = form is traditional and dates back to C, but if in doubt, use the general {} -list form (§6.3.5.2). If nothing else, it saves you from conversions that lose information (narrowing conversions; §10.5):

int i1 = 7.2;    // i1 becomes 7
int i2 {7.2};    // error : floating-point to integer conversion
int i3 = {7.2};  // error : floating-point to integer conversion (the = is redundant)

但是，我无法重现这些结果。

However, I am unable to reproduce these results.

我有以下代码。

#include <iostream>

int main()
{
    int i1 = 7.2;
    int i2 {7.2};
    int i3 = {7.2};

    std::cout << i1 << "\n";
    std::cout << i2 << "\n";
    std::cout << i3 << "\n";
}

当我编译并运行它时，我没有得到任何错误。我收到关于 std = c ++ 11 的警告，但没有错误。

When I compile and run it, I don't get any error. I get a warning about std=c++11 but no error.

$ g++ init.cpp 
init.cpp: In function ‘int main()’:
init.cpp:6:12: warning: extended initializer lists only available with -std=c++11 or -std=gnu++11
     int i2 {7.2};
            ^
$ ./a.out 
7
7
7

此外，警告只适用于第二次分配，但第三次分配没有警告。这似乎表明 = 不是真的多余的，如书中所提到的。如果 = 是冗余的，则第二个和第三个赋值都会产生警告，或者两者都不会产生警告。
然后我用 -std = c ++ 11 标志编译它们。

Further, the warning is only for the second assignment but there is no warning for the third assignment. This seems to indicate that the = is not really redundant as mentioned in the book. If = were redundant, either both the second and third assignments would have produced warnings or both would not have produced warnings. Then I compile them with the -std=c++11 flag.

$ g++ -std=c++11 init.cpp 
init.cpp: In function ‘int main()’:
init.cpp:6:16: warning: narrowing conversion of ‘7.2000000000000002e+0’ from ‘double’ to ‘int’ inside { } [-Wnarrowing]
     int i2 {7.2};
                ^
init.cpp:7:18: warning: narrowing conversion of ‘7.2000000000000002e+0’ from ‘double’ to ‘int’ inside { } [-Wnarrowing]
     int i3 = {7.2};
                  ^
$ ./a.out 
7
7
7

仍然没有错误。只有警告。尽管在这种情况下，第二个和第三个分配在生成警告时的行为相同。

Still no error. Only warnings. Although in this case the second and third assignments behave identically with respect to generating warnings.

所以我的问题是：虽然这本书提到第二和第三个赋值是错误，为什么这个代码不能编译？

So my question is: Although the book mentions that the second and third assignments are errors, why doesn't this code fail to compile?

推荐答案

这是错误的，应该有诊断，作为警告（您收到的 ）或错误。 gcc因为从C ++ 03移植问题而导致几个版本的警告：

This is ill-formed and there should be diagnostic, however it can either be a warning(which you received) or an error. gcc made this a warning for several versions due to porting issue from C++03:

标准只要求一致的实施应至少发出一个诊断消息警告。如安德鲁所说，-Werror = narrowing允许你使它成为一个错误，如果你想要的。

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.

G ++ 4.6给出一个错误，但它被改为一个警告故意为4.7，许多人（包括我自己）发现，缩小转换，其中最常遇到的问题之一，当试图编译大型C ++ 03代码库作为C ++ 11。以前格式良好的代码，如char c [] = {i，0}; （我只会在char的范围内）引起的错误，必须更改为char c [] = {（char）i，0}

G++ 4.6 gave an error but it was changed to a warning intentionally for 4.7 because many people (myself included) found that narrowing conversions where one of the most commonly encountered problems when trying to compile large C++03 codebases as C++11. Previously well-formed code such as char c[] = { i, 0 }; (where i will only ever be within the range of char) caused errors and had to be changed to char c[] = { (char)i, 0 }

，但现在最近的gcc和clang版本使此为错误，查看它的gcc 。

but now recent versions of gcc and clang make this an error, see it live for gcc.

参考 draft C ++ 11标准部分 8.5.4 [dcl.init.list] / em>说：

For reference the draft C++11 standard section 8.5.4 [dcl.init.list] says:

否则，如果初始化器列表有一个元素，对象或
引用从元件;如果需要缩小转换
（见下文）将元素转换为T，程序是
格式不正确。 [示例：

Otherwise, if the initializer list has a single element, the object or reference is initialized from that element; if a narrowing conversion (see below) is required to convert the element to T, the program is ill-formed. [ Example:

int x1 {2}; // OK
int x2 {2.0}; // error: narrowing

-end示例]

和：

缩小转换是隐式转换

A narrowing conversion is an implicit conversion

• 从浮点类型转换为整数类型，或

...]

[注意：如上所述，列表初始化中不允许在顶层进行此类转换。-end
note] ：

[ Note: As indicated above, such conversions are not allowed at the top level in list-initializations.—end note ] [ Example:

[...]

int ii = {2.0}; // error: narrows

[...]

因此，浮点数到整数转换是一个缩小的转换，并且是错误的。

So a floating point to integer conversion is a narrowing conversion and is ill-formed.

和 1.4 实施合规性[intro.compliance]说：

and section 1.4 Implementation compliance [intro.compliance] says:

虽然本国际标准仅声明对C ++实现，那些需求
通常更容易理解，如果他们被称为程序，部分程序或
执行程序的要求。此类要求具有以下含义：

Although this International Standard states only requirements on C++ implementations, those requirements are often easier to understand if they are phrased as requirements on programs, parts of programs, or execution of programs. Such requirements have the following meaning:

[...]

• 程序包含违反任何可诊断规则或本标准中的
  描述为有条件支持的结构，当实现不支持该结构时，
  a符合实现应发出至少一个诊断

[...]

告诉我们只需要诊断。

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