C++ 在C下通过引用传递指针参数？
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Passing pointer argument by reference under C?
提问by Jichao
#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\n", retstr);
return 0;
}
gccwould not compile this file, but after adding #include <cstring>I could use g++ to compile this source file.
gcc不会编译这个文件，但添加后#include <cstring>我可以使用 g++ 来编译这个源文件。
The problem is: does the C programming language support passing pointer argument by reference? If not, why?
问题是：C 编程语言是否支持通过引用传递指针参数？如果不是，为什么？
Thanks.
谢谢。
回答by Kirill V. Lyadvinsky
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.
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来代替引用。引用仅在 C++ 语言中可用。
回答by Bojan Resnik
References are a feature of C++, while C supports only pointers. To have your function modify the value of the given pointer, pass pointer to the pointer:
引用是 C++ 的一个特性，而 C 只支持指针。要让您的函数修改给定指针的值，请将指针传递给该指针：
void getstr(char ** retstr)
{
char *tmp = (char *)malloc(25);
strcpy(tmp, "hello,world");
*retstr = tmp;
}
int main(void)
{
char *retstr;
getstr(&retstr);
printf("%s\n", retstr);
// Don't forget to free the malloc'd memory
free(retstr);
return 0;
}
回答by PP.
Try this:
尝试这个：
void
getstr(char **retstr)
{
char *tmp = (char *)malloc(25);
strcpy(tmp, "hello,world");
*retstr = tmp;
}
int
main(void)
{
char *retstr;
getstr(&retstr);
printf("%s\n", retstr);
return 0;
}
回答by Sinan ünür
This should be a comment but it is too long for a comment box, so I am making it CW.
这应该是一条评论，但对于评论框来说太长了，所以我将其设为 CW。
The code you provided can be better written as:
您提供的代码可以更好地编写为：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void
getstr(char **retstr)
{
*retstr = malloc(25);
if ( *retstr ) {
strcpy(*retstr, "hello,world");
}
return;
}
int
main(void)
{
char *retstr;
getstr(&retstr);
if ( retstr ) {
printf("%s\n", retstr);
}
return 0;
}
回答by Dmytro Sirenko
There is an interesting trick in libgmp which emulates references:
typedef mpz_t __mpz_struct[1];
libgmp 中有一个模拟引用的有趣技巧：
typedef mpz_t __mpz_struct[1];
and then you can write like this:
然后你可以这样写：
mpz_t n;
mpz_init(n);
...
mpz_clear(n);
I would not recommend to use this method, because it may be incomprehensible for others, it still does not protect from being a NULL: mpz_init((void *)NULL), and it is as much verbose as its pointer-to-pointer counterpart.
我不建议使用这种方法，因为其他人可能无法理解它，它仍然不能防止成为 NULL: mpz_init((void *)NULL)，并且它与其指针到指针对应物一样冗长。
回答by Ashish
C lang does not have reference variables but its part of C++ lang.
C lang 没有引用变量，但它是 C++ lang 的一部分。
The reason of introducing reference is to avoid dangling pointers and pre-checking for pointers nullity.
引入引用的原因是为了避免悬空指针和预检查指针无效。
You can consider reference as constant pointeri.e. const pointer can only point to data it has been initialized to point.
您可以将引用视为常量指针，即常量指针只能指向它已初始化指向的数据。
