从const字符串隐式强制转换为bool [英] Implicit cast from const string to bool
问题描述
我有以下代码:
#include< iostream>
#include< string>
void foo(bool a)
{
std :: cout<< bool<<的std :: ENDL;
}
void foo(long long int a)
{
std :: cout<< long long int<<的std :: ENDL;
}
void foo(const std :: string& a)
{
std :: cout<< string<<的std :: ENDL;
int main(int argc,char * args [])
{
foo(1);
返回0;
}
执行时,我得到这个输出:
bool
输出:
字符串
为什么g ++ 4.9将这个字符串隐式转换为bool?
您的编译器正在正确地解释标准。是的,这是一个棘手的角落案例,许多采访者要求他们看起来比他们真正聪明。$ b 路线
const char [2]
(文字1
)到 const char *
到<$的正式类型c $ c> bool 是一个标准转换序列,因为它仅使用内置类型。 编译器必须支持用户定义的转换序列,也就是说, std :: string
构造函数来自 const char *
。
重载的存在 void foo(long long int a)
是一条红色鲱鱼。
你可以在C ++ 11中通过将你的重载放到 bool
,并写入
#include< type_traits>
模板<
typename Y,
typename T = std :: enable_if_t< std :: is_same< Y,bool> {}>
>
void foo(Y)
{
std :: cout<< bool<<的std :: ENDL;
}
代替。然后编译器会在模板上支持 std :: string
作为 const char [N]
(因为这是一个的重载分辨率要求)。很好!
I have the following code:
#include <iostream>
#include <string>
void foo(bool a)
{
std::cout << "bool" << std::endl;
}
void foo(long long int a)
{
std::cout << "long long int" << std::endl;
}
void foo(const std::string& a)
{
std::cout << "string" << std::endl;
}
int main(int argc, char* args[])
{
foo("1");
return 0;
}
When executing I get this output:
bool
I would have expected as output:
string
Why does g++ 4.9 implicitly cast this string to bool?
Your compiler is interpreting the standard correctly. Yes, this is a tricky corner case that many interviewers ask so they appear smarter than they really are.
The route const char[2]
(the formal type of the literal "1"
) to const char*
to bool
is a standard conversion sequence, since it uses exclusively built-in types.
Your compiler must favour that to a user-defined conversion sequence, viz. the std::string
constructor from a const char*
.
The presence of the overload void foo(long long int a)
is a red herring.
You can rather elegantly work around this in C++11 by dropping your overload to bool
, and writing
#include <type_traits>
template <
typename Y,
typename T = std::enable_if_t<std::is_same<Y, bool>{}>
>
void foo(Y)
{
std::cout << "bool" << std::endl;
}
in its place. The compiler will then favour the std::string
for const char[N]
over the template (as that is one of the requirements of overload resolution). Nice!
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