Can't pass pointers to scanf("%s",ptr);

Question

So I commented out the scanf part (and just initialized it with my own string), why does it crash if I use scanf? I believe the actual arguments I've put in scanf(); are correct.

#include <stdio.h>
#include <stdlib.h>
int strendmilan(char *s,char *t)
{
    int scntr = 0,tcntr = 0;
    while(*(s+(scntr++)) != '\0')
        ;
    --scntr;
    while(*(t+(tcntr++)) != '\0')
        ;
    --tcntr;
    while(tcntr >= 0)
        if(*(s+scntr--) == *(t+tcntr--))
            ;
        else
            return 0;
    return 1;
}
int main()
{
    char *s,*t; 
    /*
    scanf("%s",s);
    scanf("%s",t);
    */
    s = "HAHAHACOOL";
    t = "COOL";
    if(strendmilan(s,t) == 0)
        printf("NOT");
    else
        printf("YES");
    getch();
}

Assign some memory to `s` before calling `scanf`. – cnicutar Sep 09 '12 at 17:11 — cnicutar, Sep 09 '12 at 17:11

score 7 · Accepted Answer · answered Sep 09 '12 at 17:12

7

The problem is that you are sending scanf() a pointer to a buffer (e.g., s) where it can store the information read, but you haven't allocated any buffer space pointed to by s (and same for t)

char *s,*t; 
scanf("%s",s);

So you can either allocate an array of char for s, or use malloc() to allocate some storage and assign the return value of that call to s.

answered Sep 09 '12 at 17:12

Levon

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But s = "HAHAHACOOL"; t = "COOL"; works even if i dont allocate space – jantristanmilan Sep 09 '12 at 17:13
@vincentbelkin that really doesn't change anything I posted. `scanf` wants a pointer to a location where it can store data. You create the pointer, and then immediately call `scanf` *without* having allocated any storage for `s` to point to (and for `scanf` to successfully use). – Levon Sep 09 '12 at 17:16
2

@vincentbelkin, `t="COOL"` implicitly allocates some memory to store the string literal, then assigns the location of that memory to the pointer variable. `scanf` expects the pointer variable to already point to allocated memory, into which it will place the characters read. – Dave Costa Sep 09 '12 at 17:17
I think i'm misunderstanding something, but doesn't s = "HAHA"; require me to use malloc(); to allocate storage space too? why does it work? – jantristanmilan Sep 09 '12 at 17:18
@vincentbelkin It's a string literal (think of it as a constant if that helps). Also see DaveCosta's comment. Also, even if this worked, you are doing this *after* your call to scanf ... – Levon Sep 09 '12 at 17:19

Can't pass pointers to scanf("%s",ptr);

1 Answers1