#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.
The problem is: does the C programming language support passing pointer argument by reference? If not, why?
Thanks.
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.
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:
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;
}
malloc() returns. If you've included stdlib.h, it does absolutely nothing. If you haven't, it hides that fact.
malloc returns void* I find it good practice to cast it explicitly to the required type instead of relying on implicit conversions. If stdlib.h is not included, casting doesn't hide that fact - the compiler will warn you that malloc has not been declared.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;
}
malloc. It can hide a failure to #include <stdlib.h> (which you did forget in the code above). See c-faq.com/malloc/mallocnocast.html Also, don't forget to check the return value of malloc.
free but I guess some people miss the obvious!This should be a comment but it is too long for a comment box, so I am making it CW.
The code you provided can be better written as:
#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;
}
There is an interesting trick in libgmp which emulates references:
typedef mpz_t __mpz_struct[1];
and then you can write like this:
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.
C lang does not have reference variables but its part of C++ lang.
The reason of introducing reference is to avoid dangling pointers and pre-checking for pointers nullity.
You can consider reference as constant pointer i.e. const pointer can only point to data it has been initialized to point.
