C return value using argument to function - c

Question

I'm trying to run function which will return two strings by passing a pointer to them:
#include <stdio.h>
void gen_str(char *str1, char *str2){
char *s1 = "abcd";
char *s2 = "defg";
str1= strdup(s1);
str2= strdup(s2);
printf("\n\r str1 %s str2 %s\n\r", str1, str2);
}
int main(void){
char *s1, *s2;
gen_str(s1, s2);
printf("\n\r s1 %s s2 %s\n\r", s1, s2);
return 0;
}
where output is:
str1 abcd str2 defg
s1 8, s2 8,
Can someone tell me, what I'm doing wrong? I thought that strdup() will alloc memory for my new strings, and return pointers filled with strings. But actual behavior is different. So I'm asking for help.

Answer 1

You need your function to modify the pointers s1 and s2, so you need to pass their addresses, so that the function can modify the originals, not a copy.
void gen_str(char **str1, char **str2){
char *s1 = "abcd";
char *s2 = "defg";
*str1= strdup(s1);
*str2= strdup(s2);
printf("\n\r str1 %s str2 %s\n\r", *str1, *str2);
}
int main(void){
char *s1, *s2;
gen_str(&s1, &s2);
printf("\n\r s1 %s s2 %s\n\r", s1, s2);
return 0;
}
Note also that you need to free these strings at some point, otherwise you have a memory leak.

Answer 2

You got your pointers confused. The basic type for a C string is already char*, and now you want a pointer to that, i.e. a char**:
void fill_me(char ** ps)
{
const char * t = "hello";
*ps = strdup(t);
}
/* elsewhere */
char * s;
fill_me(&s);
Moral: If you want a function to modify an argument passed by the caller, the caller needs to take the address-of (&) something at some point.

Answer 3

You're only changing the pointer inside the function. To change it outside the function, you'll need to pass the function the address of the two strings, like this:
#include <stdio.h>
void gen_str(char **str1, char **str2){
char *s1 = "abcd";
char *s2 = "defg";
*str1= strdup(s1);
*str2= strdup(s2);
printf("\n\r str1 %s str2 %s\n\r", *str1, *str2);
}
int main(void){
char *s1, *s2;
gen_str(&s1, &s2);
printf("\n\r s1 %s s2 %s\n\r", s1, s2);
return 0;
}

Answer 4

You can's use the value of a variable before you've initialized it!
char *s1, *s2;
gen_str(s1, s2);
The first line doesn't assign any particular values to s1 and s2. But then you pass them to gen_str. To have another function fill in the value of a variable, pass a pointer to that variable to the function, not its value.
#include <stdio.h>
#include <string.h>
void gen_str(char **str1, char **str2){
const char *s1 = "abcd";
const char *s2 = "defg";
*str1 = strdup(s1);
*str2 = strdup(s2);
printf("\n str1 %s str2 %s\n", *str1, *str2);
}
int main(void){
char *s1, *s2;
gen_str(&s1, &s2);
printf("\n s1 %s s2 %s\n", s1, s2);
free(s1);
free(s2);
return 0;
}