Question

I have the following code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* user;
char* passwd;
int nr;

void test()
{
    int i=0;
    for(i=0;i<argc;i++)
    printf("Hello %s \n",user);
}

int main(int argc,char*argv[])
{
    int i;
    nr=argc;
    for (i=0; i<argc; i++)
    {
        user=strdup(argv[i]);

    }

    test();
return 0;
}

The result is the argv[argc] on all the positions. How can I fix this? I wwant to have that test() outside the loop.

**

EDIT

**
After the ANSWERS here this is my new code, which is not working. Can anyone say why?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char* user;


void test(int n)
{
    int i=0;
    for(i=0;i<n;i++)
    printf("%s \n",user[i]);
}
int main(int argc,char*argv[])
{
     user = (char*) malloc(argc*sizeof(char));
int i;
for (i=0;i<argc;i++)
{
user[i]=argv[i];
}
test(argc);
return 0;
}

Answer 1

Because you overwrite the pointers user and passwd in every iteration. Hence, you’ll only see the last string.

If you can tell your aim of the program, a better answer can be provided. Because I am not sure whether you want to read one user and passwd Or an array of users and passwds.

After you edit, I see you want to read an array of strings:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char** user;
// or char *user[100]; /* If you want a fix length array of pointers. Now, you dont have to malloc. /*
char* passwd;
int nr;

void test(int argc)
{
    int i=0;
    for(i=0;i<argc;i++)
    printf("Hello %s \n",user[i]);
}

int main(int argc,char*argv[])
{
    int i;
    nr=argc;
    user = malloc(argc*sizeof(char*));

    for (i=0; i<argc; i++)
    {
        user[i]=strdup(argv[i]);

    }
    test(argc);
return 0;
}