从C ++中的void函数返回 [英] Returning From a Void Function in C++

问题描述

请考虑以下代码段：

void Foo()
{
  // ...
}

void Bar()
{
  return Foo();
}

在C ++中使用上述代码的合法理由常用方法：

What is a legitimate reason to use the above in C++ as opposed to the more common approach:

void Foo()
{
  // ...
}

void Bar()
{
  Foo();

  // no more expressions -- i.e., implicit return here
}

推荐答案

在你的示例中可能没有用，但在某些情况下，很难在模板代码中处理 void 我希望这个规则有时帮助。非常有用的示例：

Probably no use in your example, but there are some situations where it's difficult to deal with void in template code, and I expect this rule helps with that sometimes. Very contrived example:

#include <iostream>

template <typename T>
T retval() {
    return T();
}

template <>
void retval() {
    return;
}

template <>
int retval() {
    return 23;
}

template <typename T>
T do_something() {
    std::cout << "doing something\n";
}

template <typename T>
T do_something_and_return() {
    do_something<T>();
    return retval<T>();
}

int main() {
    std::cout << do_something_and_return<int>() << "\n";
    std::cout << do_something_and_return<void*>() << "\n";
    do_something_and_return<void>();
}

请注意，只有 main 必须处理在 void 情况下没有从 retval 返回的事实。中间函数 do_something_and_return 是通用的。

Note that only main has to cope with the fact that in the void case there's nothing to return from retval . The intermediate function do_something_and_return is generic.

当然这只会让你这么远 - 如果 do_something_and_return 想要，在正常情况下，在一个变量中存储 retval ，并在返回之前做一些事情，然后你仍然在麻烦 - 你必须专门化（或重载） do_something_and_return for void。

Of course this only gets you so far - if do_something_and_return wanted, in the normal case, to store retval in a variable and do something with it before returning, then you'd still be in trouble - you'd have to specialize (or overload) do_something_and_return for void.

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