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?
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
-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?
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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