C ++重载函数作为模板参数

Question

the simplified version of my code is here

int foo(int x)
{
  return x;
}

int foo(int x, int y)
{
  return x+y;
}

template<typename unary_func>
int bar(int k, unary_func f)
{
  return f(k);
}

int main()
{
  bar(3, foo);
  return 0;
}

Is there a way to tell the compiler what I want to pass as argument is the first `foo'?

Answer 1

You can give an explicit template argument:

bar<int(int)>(3, foo);

or cast the ambiguous function name to a type from which the template argument can be deduced:

bar(3, static_cast<int(*)(int)>(foo));

or wrap it in another function (or function object) to remove the ambiguity

bar(3, [](int x){return foo(x);});

Answer 2

I handle this problem with the following macro:

#define LIFT(fname) \
[] (auto&&... args) -> decltype (auto) \
{ \
    return fname (std::forward <decltype (args)> (args)...); \
}

Given your definitions of foo and bar , you can say

int main()
{
    bar(3, LIFT(foo));
    return 0;
}

and the correct overload will be selected. This uses some features of C++14, namely generic lambdas and decltype(auto) . If you're using C++11, you can get more or less the same effect with a little more work:

#define DECLARE_LIFTABLE(NAME) \
struct NAME##_lifter \
{ \
    template <typename... Args> \
    auto operator () (Args&&... args) -> decltype (NAME (std::forward <Args> (args)...)) \
    { \
        return NAME (std::forward <decltype (args)> (args)...); \
    } \
}
#define LIFT(NAME) (NAME##_lifter {})

DECLARE_LIFTABLE(foo);
int main()
{
    bar(3, LIFT(foo));
    return 0;
}

If you're using C++98, you're basically stuck with a cast to the appropriate function pointer type.

Answer 3

No, you can't, because you are calling the function always with only one argument, you need a type with only one argument. Instead, you can use template by value (no typename or class)

One argument:

int foo(int x)
{
    return x;
}

int foo(int x, int y)
{
    return x+y;
}

typedef int (*foo_fcn)(int);

template<foo_fcn unary_func>
int bar(int k)
{
    return unary_func(k);
}

int main()
{
    bar<foo>(3);
    return 0;
}

Two arguments:

int foo(int x)
{
    return x;
}

int foo(int x, int y)
{
    return x+y;
}

typedef int (*foo_fcn)(int, int);

template<foo_fcn unary_func>
int bar(int k)
{
    return unary_func(k, k);
}

int main()
{
    bar<foo>(3);
    return 0;
}

Both:

int foo(int x) // first foo
{
    return x;
}

int foo(int x, int y) // second foo
{
    return x+y;
}

typedef int (*foo_fcn)(int);
typedef int (*foo_fcn_2)(int, int);

template<foo_fcn unary_func>
int bar(int k)
{
    return unary_func(k);
}

template<foo_fcn_2 unary_func>
int bar(int a, int b)
{
    return unary_func(a, b);
}


int main()
{
    bar<foo>(3,1); // compiler will choose first foo
    bar<foo>(4); // compiler will choose second foo
    return 0;
}

Answer 4

Yes:

bar(3, static_cast<int(*)(int)>(&foo));

or:

bar<int(*)(int)>(3, &foo);