在C下通过引用传递指针参数?

#include <stdio.h>
#include <stdlib.h>

void
getstr(char *&retstr)
{
 char *tmp = (char *)malloc(25);
 strcpy(tmp, "hello,world");
 retstr = tmp;
}

int
main(void)
{
 char *retstr;

 getstr(retstr);
 printf("%s
", retstr);

 return 0;
}

gcc 不会编译这个文件，但是在添加 #include <cstring> 之后我可以使用 g++ 来编译这个源文件.

gcc would not compile this file, but after adding #include <cstring> I could use g++ to compile this source file.

问题是:C语言是否支持通过引用传递指针参数?如果没有，为什么?

The problem is: does the C programming language support passing pointer argument by reference? If not, why?

谢谢.

推荐答案

不，C 不支持引用.这是设计使然.您可以在 C 中使用指向指针的指针代替引用.引用仅在 C++ 语言中可用.

No, C doesn't support references. It is by design. Instead of references you could use pointer to pointer in C. References are available only in C++ language.