是否有任何选择可以推迟可变参数模板中函数调用的求值

Question

Suppose a variadic template:

void f(T value, Args... args)

As one of the arguments a function is passed which returns a value.

Example: f(1, getName());

Is there a simple way to defer the evaluation of the function, so getName() is invoked inside f() and is not executed if not required.

And with a simple way I mean preferably without creating a complex calling syntax.

Answer 1

If you call f like you did in your example there is no way for f to even know that it was passed a function. Only the return value of getName is passed to f.

Like so:

auto x = getName();
f(1, x);

To only evaluate the function when needed, you can pass a function pointer:

f(1, getName);

I don't really understand the significance of variadic templates for this problem so I simplified it:

template <typename T, typename F>
void f(T a, F func)
{
    if (required)
        result = func();
}

---

If you want to pass arguments that should be used when the function is called, you can use the std::bind facilities or a lambda:

f(1, [&]{ return getName(arg1, arg2); });

Or even a case with arguments from outside and inside f:

template <typename T, typename F>
void f(T a, F func)
{
    if (required)
        result = func(3.0f);
}

f(1, [&](float otherArg){ return getName(arg1, arg2, otherArg); });