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\n", retstr);

 return 0;
}

gcc would 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,但是在添加#include 之后我可以使用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?如果没有,为什么?

Thanks.

6 个解决方案

#1

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。这是设计的。您可以使用指向C中的指针而不是引用。引用仅在C ++语言中可用。

#2

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。要让函数修改给定指针的值,请将指针传递给指针:

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;
}

#3

Try this:


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;
}

#4

This should be a comment but it is too long for a comment box, so I am making it CW.

这应该是一个评论,但它对于评论框来说太长了,所以我正在制作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;
}

#5

There is an interesting trick in libgmp which emulates references: typedef mpz_t __mpz_struct[1];

libgmp中有一个有趣的技巧,可以模拟引用:typedef mpz_t __mpz_struct [1];

and then you can write like this:

然后你可以像这样写:

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,因为它可能对其他人来说是不可理解的,它仍然不能防止成为NULL:mpz_init((void *)NULL),并且它与指针到指针的对应部分一样冗长。

#6

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指针无效的预检查。

You can consider reference as constant pointer i.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const指针只能指向已初始化为point的数据。

#1