将构造函数std :: string从char * [英] Move constructor for std::string from char*

问题描述

我有一个函数 f 返回一个 char * 。函数文档说：

I have a function f returning a char*. The function documentation says:

The user must delete returned string

我想从它构造一个 std :: string 。要做的小事是：

I want to construct a std::string from it. The trivial things to do is:

char* cstring = f();
std::string s(cstring);
delete cstring;

是否可以使用C ++功能做得更好？我想写如下

Is it possibile to do it better using C++ features? I would like to write something like

std::string(cstring)

避免泄露。

推荐答案

std :: string 将创建一个空终止字符串参数的副本，并管理该副本。没有办法让它拥有你传递给它的字符串的所有权。所以你正在做的是正确的，唯一的改进我建议是检查 nullptr ，假设是一个有效的返回值 f （）。这是必要的，因为使用 char const * 的 std :: string 构造函数要求参数指向有效的数组，而不是 nullptr 。

std::string will make a copy of the null terminated string argument and manage that copy. There's no way to have it take ownership of a string you pass to it. So what you're doing is correct, the only improvement I'd suggest is a check for nullptr, assuming that is a valid return value for f(). This is necessary because the std::string constructor taking a char const * requires that the argument point to a valid array, and not be nullptr.

char* cstring = f();
std::string s(cstring ? cstring : "");
delete[] cstring;   // You most likely want delete[] and not delete

现在，如果你不需要所有 std :: string 的接口，或者如果避免副本很重要，那么可以使用 unique_ptr

Now, if you don't need all of std::string's interface, or if avoiding the copy is important, then you can use a unique_ptr to manage the string instead.

std::unique_ptr<char[]> s{f()}; // will call delete[] automatically

您可以访问托管 char * 通过 s.get（），字符串将 delete code> s

超出范围。

You can get access to the managed char * via s.get() and the string will be deleted when s goes out of scope.

即使你使用第一个选项，我建议存储返回值在传递给 std :: string

Even if you go with the first option, I'd suggest storing the return value of f() in a unique_ptr before passing it to the std::string constructor. That way if the construction throws, the returned string will still be deleted.

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