std :: unique_ptr用于需要免费的C函数 [英] std::unique_ptr for C functions that need free

问题描述

想一个返回必须 free d的C函数，例如POSIX的 strdup（） 。我想在C ++ 11中使用该函数，并避免任何泄漏的机会，这是正确的方法吗？

Think to a C function that return something that must be freed, for example the POSIX's strdup(). I want to use that function in C++11 and avoid any chance of leaks, is this a correct way?

#include <memory>
#include <iostream>
#include <string.h>

int main() {
    char const* t { "Hi stackoverflow!" };
    std::unique_ptr<char, void(*)(void*)>
        t_copy { strdup(t), std::free };

    std::cout << t_copy.get() << " <- this is the copy!" <<std::endl;
}

假设有意义，指针？例如对于POSIX的函数 open ，返回 int ？

Assuming it makes sense, it is possible to use a similar pattern with non-pointers? For example for the POSIX's function open that returns an int?

推荐答案

你有什么很可能在实践中工作，但不是严格正确。但是，正确的版本可以更可读：

What you have is extremely likely to work in practice, but not strictly correct. However, the correct version is arguably even more readable:

std::unique_ptr<char, decltype(std::free) *>
    t_copy { strdup(t), std::free };

原因是函数类型 std :: free 不能保证 void（void *）。它保证取一个 void * ，并返回 void ，但有两个函数类型匹配该规范：一个具有C键，一个具有C ++键。大多数编译器不会注意到，但为了正确，你应该避免对它做出假设。

The reason is that the function type of std::free is not guaranteed to be void(void*). It is guaranteed to take a void*, and to return void, but there are two function types that match that specification: one with C linkage, and one with C++ linkage. Most compilers pay no attention to that, but for correctness, you should avoid making assumptions about it.

假设有意义，要使用非指针类似的模式？

Assuming it makes sense, it is possible to use a similar pattern with non-pointers?

不能使用 unique_ptr 这是真正特定于指针。但是你可以创建自己的类，类似于 unique_ptr ，但没有对被包裹对象做出假设。

Not with unique_ptr, which is really specific to pointers. But you could create your own class, similar to unique_ptr, but without making assumptions about the object being wrapped.

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