[英]G++ problems with std::function
I have the following code: 我有以下代码:
#include <functional>
std::function<int,int> p;
int main()
{
return 0;
}
I am using MinGW g++ 4.8.1 which fails with 我正在使用MinGW g ++ 4.8.1,失败了
C:\main.cpp|4|error: wrong number of template arguments (2, should be 1)|
c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\functional|1866|error: provided for 'template<class _Signature> class std::function'|
C:\main.cpp|4|error: invalid type in declaration before ';' token|
is this a G++ bug or am I using std::function incorrectly 这是G ++错误还是我使用std :: function不正确
std::function<int(int)>
for function takes int and return int. std::function<int(int)>
接受int并返回int。 eg 例如
int foo(int);
std::function<void(int,int)>
for function takes two ints and no return value. std::function<void(int,int)>
带有两个整数,并且没有返回值。 eg 例如
void foo(int, int);
std::function
takes one template argument - the type of the callable object that it wraps around. std::function
接受一个模板参数-它包装的可调用对象的类型。 So, if you want to construct an std::function
which returns type Ret
and takes arguments of types Arg1, Arg2,..., ArgN
, you would write std::function<Ret(Arg1, Arg2,..., ArgN)>
. 所以,如果你想构建
std::function
返回类型Ret
和需要的类型参数Arg1, Arg2,..., ArgN
,你会写std::function<Ret(Arg1, Arg2,..., ArgN)>
。
(Note that the ellipses weren't meant to indicate parameter pack expansion - they are just being used in the regular mathematical sense.) (请注意,椭圆并不是要指示参数包的扩展,它们只是在常规数学意义上使用。)
As the compiler says, std::function takes one template argument. 正如编译器所说,std :: function接受一个模板参数。
Use the syntax returntype(argtype, ...) 使用语法returntype(argtype,...)
int foo(int a, int b) { return a+b; }
std::function<int(int,int)> p = foo;
int bar(int a) { return ++a; }
std::function<int(int)> q = bar;
void boo() { return; }
std::function<void()> r = boo;
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