char * /字符串连接不复制？ [英] char*/string concatenation without copying?

问题描述

我想在C或C ++中连接2个字符串，而不需要新的内存分配和复制。是否有可能？

I would like to concatenate 2 strings in C or C++ without new memory allocation and copying. Is it possible?

可能的C代码：

char* str1 = (char*)malloc(100);
char* str2 = (char*)malloc(50);
char* str3 = /* some code that concatenates these 2 strings
                without copying to occupy a continuous memory region */

然后，当我不再需要它们时，我只需：

Then, when I don't need them any more, I just do:

free(str1);
free(str2);

或者如果可能，我想在C ++中使用 std :: string 或者 char * ，但使用 new 和<$ c $ str3上的 delete （可能 void operator delete（void * ptr，std :: size_t sz） 。

Or if possible, I would like to achieve the same in C++, using std::string or maybe char*, but using new and delete (possibly void operator delete ( void* ptr, std::size_t sz ) operator (C++14) on the str3).

有很多关于字符串连接的问题，但我还没有找到一个问题。

There are a lot of questions about strings concatenation, but I haven't found one that asks the same.

推荐答案

不，不可能

在C中，malloc操作返回内存块彼此没有关系。但在C中，字符串必须是连续的字节数组。因此，没有办法在不复制的情况下扩展str1，更不用说连接。

In C, malloc operations return blocks of memory that have no relationship to each other. But in C, strings must be a continuous array of bytes. So there is no way to extend str1 without copying, let alone concatenate.

对于C ++，可能会感兴趣的是：查看此答案。

For C++, perhaps ropes may be of interest: See this answer.

绳索以不必连续的块分配。这支持O（1）级联。但是，访问器使其显示为单个字节字符串。我敢肯定，要将绳子转换回std :: string或C风格的字符串将需要一个副本，但这可能是最接近你想要的。

Ropes are allocated in chunks that do not have to be contiguous. This supports O(1) concatenation. However, the accessors make it appear as a single string of bytes. I'm certain that to convert ropes back to std::string or C style strings will take a copy however, but this is probably the closest to what you want.

此外，它可能是一个过早的优化，担心复制几个字符串的成本。除非您正在移动批次的资料，否则无关紧要

Also, it is probably a premature optimization to worry about the costs of copying a few strings around. Unless you are moving lots of data, it won't matter

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