处理从C中的函数返回的char数组

I am using a library called tinyXML, which parses XML files. Many of its methods return a const char*.

我正在使用一个名为tinyXML的库，它解析XML文件。它的许多方法都返回一个const char *。

After having read this question: how to return a char array from a function in C

读完这个问题之后：如何从C中的函数返回一个char数组

I now believe that every time a method returns a char* it is the caller's (my) responsibility to explicitly free it, because it is probably allocated dynamically on the heap. Am I right / wrong? What can I assume?

我现在相信每次方法返回一个char *时，调用者（我）有责任显式释放它，因为它可能是在堆上动态分配的。我是对还是错？我能假设什么？

(If I ever wrote a library I would much rather return std::string instead of char* arrays, because they are so much simpler for the user.)

（如果我写过一个库，我宁愿返回std :: string而不是char *数组，因为它们对于用户来说简单得多。）

2 个解决方案

#1

You can't assume anything and must check the documentation for the method you are calling to know if you must free the pointer or not. Sometimes a function returning a const char * might be returning a statically allocated string:

您不能假设任何事情，并且必须检查您要调用的方法的文档，如果您必须释放指针。有时返回const char *的函数可能会返回一个静态分配的字符串：

const char *getName(){
    return "SPQR3";
}

Or it might be a pointer that is someone else's responsibility to free. For example, strcpy's return value is the same as the pointer you pass to it as input.

或者它可能是一个指针，是别人有责任释放的。例如，strcpy的返回值与作为输入传递给它的指针相同。

#2

Read the library's documentation. Both usages (pointer to an internal buffer, or pointer to a dynamically-allocated buffer) are common, and aren't distinguishable from the function prototype alone.

阅读图书馆的文档。两种用法（指向内部缓冲区的指针或指向动态分配缓冲区的指针）都很常见，并且不能单独与函数原型区分开来。

#1