strcmp giving segmentation fault

Question

this is my code giving segmentation fault

#include <stdio.h>
#include <signal.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>

int main(void) {
    char *get;
    scanf("%s", get);
    int k = strcmp("sachin", get);
    printf("%d", k);
}

thanks for any help;

Is the `C++` tag here a mistake? And if not, then you have much bigger problems. — BoBTFish, Mar 20 '14 at 08:24

score 4 · Accepted Answer · answered Mar 20 '14 at 08:35

char *get;

The above statement defines get to be a pointer to a character. It can store the address of an object of type char, not a character itself. The problem is with both scanf and strcmp call. You need to define an array of characters to store the input string.

#include <stdio.h>
#include <string.h>

int main(void) {
    // assuming max string length 40
    // +1 for the terminating null byte added by scanf

    char get[40+1];

    // "%40s" means write at most 40 characters
    // into the buffer get and then add the null byte
    // at the end. This is to guard against buffer overrun by
    // scanf in case the input string is too large for get to store

    scanf("%40s", get);
    int k = strcmp("sachin", get);
    printf("%d", k);

    return 0;
}

score 2 · Answer 2 · answered Mar 20 '14 at 08:22

2

You need to allocate memory for the pointer get.

Or use a char array:

char   get[MAX_SIZE];
scanf("%s",get);

When you declare a pointer, it is not pointing to any memory address. You have to explicitly allocate memory for that pointer by malloc (c-style) or new (c++ style).

You will also need to manage the clean operation yourself by free / delete respectively.

answered Mar 20 '14 at 08:22

Abhishek Bansal

12,589
4
31
46

i do not understand why i have to allocate memory it is already allocated when i declare it; – Sachin Setiya Mar 20 '14 at 08:23
@SachinSetiya if you declare something memory is only allocated for the declared object itself - in this case the pointer. – vlad_tepesch Mar 20 '14 at 08:27
@SachinSetiya No. while declaration you are just specifying that the pointer can point to a char. It does not start pointing immediately. – Abhishek Bansal Mar 20 '14 at 08:27
thank you sir i under stand the way but is there any way to get variable length string which does not have maximum length – Sachin Setiya Mar 20 '14 at 08:37

vlad_tepesch · Answer 3 · 2014-03-20T08:35:37.743

0

of sure you get a segment fault. you told your computer to write user input to uninitialized pointer.

char   *get; // just a pointer that may wildly points anywhere
scanf("%s",get);

define get to be some char array

char get[SOME_LARGE_CONSTANTS_THAT_CLEARLY_IS_LARGER_THAN_YOUR_USER_INPUT];

edited Mar 20 '14 at 08:35

answered Mar 20 '14 at 08:22

vlad_tepesch

6,681
1
38
80

i do not want spoil space by using a fixed array so used char * – Sachin Setiya Mar 20 '14 at 08:35
@SachinSetiya that is not possible! your pointer does not point to any buffer! you have to allocate sufficient memory where the input routine may write to. – vlad_tepesch Mar 20 '14 at 08:37

strcmp giving segmentation fault

3 Answers3

Linked