隐式转换没有警告 [英] No warning on implicit conversion

问题描述

  // g ++ sizeofint.cpp --std = c ++ 11 -Wconversion -Wall -Wextra -Werror -pedantic-errors 
 #include< iostream> 
 #include< utility> 
 
 int main（int argc，char ** argv）{
（void）argc; 
（void）argv; 
 int a = 0x12345678; 
 std :: cout<< sizeof（int）< ...< sizeof（uint16_t）<< std :: endl; 
 std :: pair< uint16_t，uint16_t> p {a，a}; // !!!!没有警告或转换错误！ 
 std :: cout<< p.first<< ：<< p.second<< std :: endl; 
 uint16_t b = a; // !!!!正确的行为：-Wconversion触发警告，其中-Werror转为错误
 std :: cout< b<< std :: endl; 
 return 0; 
} 
  

使用上面的代码，您可以清楚地看到<$ c当构建 p 时，$ c> int 到 uint16_t 。但是，从版本4.9.1开始的g ++不会在使用注释中提供的参数时抱怨任何转换。

稍后，g ++会抱怨隐式在构建 b 时转换为uint16_t。

我试图确保 p 的建设将导致至少一个警告（但最好是一个错误）。

任何想法？是否有一个标志我不知道要触发正确的行为？

解决方案

如果你的代码使用了 constexpr pair（const uint16_t& x，const uint16_t& y）; 构造函数 std :: pair< uint16_t，uint16_t> 得到警告和/或错误。您甚至不需要 -Wconversion - 在括号中缩小转换会导致程序格式错误。

但是，重载分辨率选择 std :: pair ' template 构造函数，这是更好的匹配。结果，转换发生，不是在你写的代码内，而是在构造函数内部。由于该构造函数是在系统标头中定义的（请参阅 GCC的文档）尽管您可以使用 -Wsystem-headers 启用警告，但是GCC会停止警告。

<系统标题的警告，该选项将产生许多不相关的警告，因此与非常严重的互动 -Werror 。

// g++ sizeofint.cpp --std=c++11 -Wconversion -Wall -Wextra -Werror -pedantic-errors
#include <iostream>
#include <utility>

int main(int argc, char **argv) {
    (void)argc;
    (void)argv;
    int a = 0x12345678;
    std::cout << sizeof(int) << "..." << sizeof(uint16_t) << std::endl;
    std::pair<uint16_t, uint16_t> p{a,a}; // !!!! no warning or error on conversion !!!!
    std::cout << p.first << ":" << p.second << std::endl;
    uint16_t b = a; // !!!! correct behavior: -Wconversion triggers warning, which -Werror turns to an error
    std::cout << b << std::endl;
    return 0;
}

With the above code, you can see clearly a implicit conversion from int to uint16_t when constructing p. However, g++ as of version 4.9.1 does not complain about any conversions when using the parameters provided in the comment at the beginning.

Later, g++ does complain about the implicit conversion to uint16_t when constructing b.

I'm trying to make sure that p's construction will result in at least a warning (but preferably an error).

Any thoughts? Is there a flag I don't know about to trigger the correct behavior?

解决方案

If your code had used the constexpr pair(const uint16_t& x, const uint16_t& y); constructor of std::pair<uint16_t, uint16_t>, you'd get a warning and/or error. You wouldn't even need -Wconversion for this - narrowing conversions inside braces render a program ill-formed.

But instead, overload resolution selected std::pair's template<class U, class V> constexpr pair(U&& x, V&& y); constructor, which is a better match. As a result, the conversion happens, not inside the code you wrote, but inside that constructor. Since that constructor is defined in a system header (see GCC's documentation, hat tip @quantdev), the warning is suppressed by GCC.

While you could use -Wsystem-headers to enable warnings from system headers, that option will produce lots of unrelated warnings and so interacts very badly with -Werror.

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