相当于python代码的C ++：'function（* list）'

Question

In python i can do:

def foo(a,b,c):
    print a+b+c

l=[5,2,10]
foo(*l)

Can i do the same in c++?

Answer 1

I believe the closest equivalent is to use a tuple with the (experimental) apply function:

#include <tuple>
#include <iostream>
#include <experimental/tuple>

using std::cout;
using std::make_tuple;
using std::experimental::apply;

int main()
{
  auto foo = [](int a,int b,int c)
  {
    cout << a+b+c << "\n";
  };

  auto l = make_tuple(5,2,10);

  apply(foo,l);
}

Answer 2

The closest you can get to is creating a custom type with member variables for each argument, and overload the function to accept either individual values or the custom type. The second overload then calls the first.

Example:

#include <iostream>

void foo(int a, int b, int c)
{
    std::cout << a + b + c << "\n";
}

struct foo_args { int a; int b; int c; };

void foo(foo_args const& args)
{
    foo(args.a, args.b, args.c);
}

int main()
{
    foo_args const l { 5, 2, 10 };
    foo(l);
}

As you can see, it's not really worth the trouble, though. Different language, different idioms.

Answer 3

It's a little different in C++, but I believe you can still do what you want there.

#include <iostream>
#include <vector>

void foo(int a, int b, int c) {
    std::cout << a + b + c << std::endl;
}
void foo(const int (&arr)[3]) { //foo overloaded for arrays of size 3
    foo(arr[0], arr[1], arr[2]); //Calls the original foo
}
void foo(const std::vector<int>& v) { //foo overloaded for vectors
    if(v.size() >= 3) //Vector should at least have 3 elements
        foo(v[0], v[1], v[2]); //Calls the original foo
}

int main(void) {

    foo(10, 20, 30);
    int arr[3] {40, 50, 60};
    foo(arr);
    std::vector<int> v {70, 80, 90};
    foo(v);

    getchar();
    return 0;
}

Answer 4

不幸的是，这对于C ++是不可能的。