从函数返回一个字符串 [英] returning a string from a function

问题描述

我想编写一个跨平台的函数（win32& linux），并返回datetime的字符串表示形式[hh：mm：ss dd-mm-yyyy]。

I wanted to write a function that'll be cross platform (win32 & linux), and return a string representation of the datetime [hh:mm:ss dd-mm-yyyy].

知道我只想使用返回的字符串作为临时流如下所示：

Knowing that I just want to use the returned string as a temporary in a stream fashion as below:

std::cout << DateTime() << std::endl;

我考虑用以下原型写一个函数

I considered writing a function with the following prototype

const char* DateTime();

如果返回字符数组，必须在完成后将其删除。但我只想要一个临时的，我不想担心取消分配字符串。

If you return a character array, you must delete it once you're done. But I just want a temporary, I don't want to have to worry about de-allocating the string.

所以我写了一个函数，只返回一个std :: string：

So I've written a function that just returns an std::string:

#include <ctime>
#include <string>
#include <sstream>

std::string DateTime()
{
    using namespace std;

    stringstream ss;
    string sValue;
    time_t t = time(0);
    struct tm * now = localtime(&t);

    ss << now->tm_hour << ":";
    ss << now->tm_min << ":";
    ss << now->tm_sec << " ";
    ss << now->tm_mday + 1 << " ";
    ss << now->tm_mon + 1 << " ";
    ss << now->tm_year + 1900;

    sValue = ss.str();

    return sValue;
}

$ b $ p我意识到我在DateTime中返回一个堆栈变量的副本。这是没有效率的，因为我们在DateTime堆栈上创建字符串，填充它，然后返回一个副本并销毁堆栈上的副本。

I realize that I'm returning a copy of the stack variable in DateTime. This is inefficient in that we create the string on the DateTime stack, populate it, then return a copy and destroy the copy on the stack.

有c ++ 11移动语义革命做了什么来解决这种低效率 - 我可以改进这个吗？

Has the c++11 move-semantics revolution done anything to resolve this inefficiency - can I improve upon this?

推荐答案

+11代码。在C ++ 98/03中，由于编译器优化，您的代码可能会 效率很高，但这些优化不能保证。在C ++ 11中，这些相同的优化可能仍然使您的返回免费，但是如果没有，您的字符串将被移动，而不是复制。

lapin, your code is fine C++11 code. In C++98/03 your code will probably be efficient due to compiler optimizations, but those optimizations aren't guaranteed. In C++11, those same optimizations will probably still make your return free, but just in case they don't, your string will be moved instead of copied.

所以返回值无罪！ ： - ）

So return by value guilt-free! :-)

次要：

最好的做法是在首次使用时声明您的值，而不是在块的顶部：

It is best practice to declare your values at the point of first use, instead of at the top of a block:

string sValue = ss.str();
return sValue;

或者甚至：

return ss.str();

但这只是一个小小的。您的代码效果很好 。

But this is just a minor nit. Your code is fine and efficient.

这篇关于从函数返回一个字符串的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！