strange scanf behaviour

Hi all,

after alot of java i've decided to start learning some C, i've an exercise that i cannot complete because i've a strange (to me) behaviour, here's the code

Code:

---

#include <stdio.h>
#include <stdlib.h>

int main()
{
        char *strings [10];
        int i, n = 4, str_size = 255;
       
        printf ("Enter %d strings to sort:\n", n);
       
        for (i=0; i<n; i++) {
                strings[i] = (char *) malloc (str_size);
                scanf ("%s", &strings[i]);
        }
       
        printf("\nRESULT:\n", &strings[i]);
        for (i=0; i<n; i++) {
                printf("%s \n", &strings[i]);
        }

}

---

and here is the terminal output

Code:

---

Enter 4 strings to sort:
one two three four

RESULT:
one
two
threfour
four

---

What happen to the "three" number? in general all string with a length greater then 3 are concatenated with the following string ...

i've re-read the scanf documentation but really can't find where's the mistake

You dont need & while using %s with scanf and printf. %s takes pointer as its argument. Also no need to cast malloc, it returns void*. Also look at this.

Code:

---

printf("\nRESULT:\n", &strings[i]);

---

I suppose you shouldn't take an address of the char* that strings[i] is.

Code:

---

scanf ("%s", strings[i]);
printf("%s \n", strings[i]);

---

If you happen to use GCC, then turning on warnings can help with errors in format string.

Quote:

---

a.c:13: warning: char format, pointer arg (arg 2)
a.c:16: warning: too many arguments for format
a.c:18: warning: char format, pointer arg (arg 2)
a.c:21: warning: control reaches end of non-void function

---

(Actually the warning seems to be incorrectly worded: what you have is char* format and char** argument? Anyway, this shows there is something wrong and it is unlikely to work.)

thanks BEN10 and anon, the error was the use of the "&" operator...