是否可以编写C ++模板/宏来检查两个函数是否具有相同的签名 [英] Is it possible to write c++ template/macros to check whether two functions have the same signatures
问题描述
是否可以编写c ++模板/宏来检查两个函数是否具有相同的签名(返回类型和参数列表)?
Is it possible to write c++ template/macros to check whether two functions have the same signatures (return type and arguments list) ?
这是一个简单的示例我想使用它:
Here's a simple example of how I want to use it:
int foo(const std::string& s) {...}
int bar(const std::string& s) {...}
if (SAME_SIGNATURES(foo, bar))
{
// do something useful... make Qt signal-slot connection for example...
}
else
{
// signatures mismatch.. report a problem or something...
}
那么有可能以某种方式还是只是一个空想?
So is it possible somehow or is it just a pipe dream ?
PS
实际上我对c ++ 2003标准很感兴趣。
P.S. Actually I'm interesting in c++ 2003 standard.
推荐答案
C ++ 11解决方案
无需自己编写任何模板。
C++11 Solution
No need to write any template yourself.
您可以将 decltype
与 std :: is_same
:
You can use decltype
along with std::is_same
:
if (std::is_same<decltype(foo),decltype(bar)>::value )
{
std::cout << "foo and bar has same signature" << std::endl;
}
此处 decltype
返回类型(在这种情况下为函数),以及 std :: is_same
比较两个类型,如果两者相同,则返回 true
,否则 false
。
Here decltype
returns the type of the expression which is function in this case, and std::is_same
compares the two types, and returns true
if both are same, else false
.
在C ++ 03中,您没有 decltype
,因此可以实现重载的函数模板,如下所示:
In C++03, you don't have decltype
, so you can implement overloaded function templates as:
template<typename T>
bool is_same(T,T) { return true; }
template<typename T, typename U>
bool is_same(T,U) { return false; }
现在,您可以将其用作:
Now you can use it as:
if (is_same(foo, bar))
{
std::cout << "foo and bar has same signature" << std::endl;
}
现在在这种情况下, is_same
是功能模板,而不是类模板。因此,它是在运行时而不是编译时进行评估的。因此,这将产生错误:
Now that in this case is_same
is a function template, not class template. So it is evaluated at runtime as opposed to compile-time. So this will give error:
int a[is_same(foo,bar) ? 10 : 20]; //error (in Standard C++03)
//the size must be known at compile-time!
但是,如果您需要在编译时知道它,则您需要做更多的工作并实现功能如:
However, if you need to know it at compile-time, then you've to work more, and implement the functionality as:
typedef char same[1];
typedef char different[2];
template<typename T>
same& is_same_helper(T,T); //no need to define it now!
template<typename T, typename U>
different& is_same_helper(T,U); //no definition needed!
#define is_same(x,y) (sizeof(is_same_helper(x,y)) == sizeof(same))
现在将其用作:
if (is_same(foo, bar))
{
std::cout << "foo and bar has same signature" << std::endl;
}
您也可以在编译时使用它。所以你可以这样写:
You can use it at compile-time also. so you can write it:
int a[is_same(foo,bar) ? 10 : 20]; //okay
希望有帮助。
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