Code:
#include<stdlib.h>
#include<string.h>
#include<stdarg.h>
char * konkatenacija(int n, ...);
int main()
{
char *s;
s = konkatenacija(3,
"First", " ", "last");
printf("%s", s);
free(s);
return 0;
}
char * konkatenacija(intn, ...)
{
char *result = 0, *next;
inti;
va_listargs;
va_start(args, n);
for (i = 0; i < n; i++)
{
intlen = result ? strlen(result) : 0;
next = va_arg(args, char *);
char *tmp = (char *)malloc(len +
strlen(next) + 1);
strcpy(tmp, result ? result : "");
strcat(tmp, next);
free(result);
result = tmp;
}
va_end(args);
return result;
}
What this code does it actually takes 3 strings and sends it to the function with variable number of arguments along with the number of strings used, i know that there are simple ways to do it, but i have to use function with variable arguments number, the problem is here:
Code:
intlen = result ? strlen(result) : 0;
next = va_arg(args, char *);
char *tmp = (char *)malloc(len +
strlen(next) + 1);
strcpy(tmp, result ? result : "");
strcat(tmp, next);
free(result);
result = tmp;
Why i can't use only two char* variables, one to take the next optional argument and other to contiue to like this:
Code:
char * konkatenacija(int n, ...)
{
char *result = 0, *next;
int i;
va_list args;
va_start(args, n);
for (i = 0; i < n; i++)
{
next=va_arg(args,char *);
strcat(result,next);
}
va_end(args);
return result;
}
instead of dynamic allocating the memory and making temporary variable, when i try to do it with function looking like this it just does nothing, is it because i cant use strcat on the variable that has value 0 even though it is char*, or is it something else?
