Taking string input in char pointer

Question

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

int main(){
    char *s;
    printf("enter the string : ");
    scanf("%s", s);
    printf("you entered %s\n", s);
    return 0;
}

When I provide small inputs of length up to 17 characters (for example "aaaaaaaaaaaaaaaaa") the program works perfectly fine but on providing inputs of larger lengths, it gives me a runtime error saying "main.c has stopped working unexpectedly".

Is there some problem with my compiler (codeblocks) or my pc (windows 7)? Or is it somehow related to the input buffer of C?

It's called a buffer overflow... Don't use scanf() is you need to get large input values. — Kevin
– Kevin, Commented Feb 5, 2013 at 12:22
I don't think that the answers below mention that Kevin . Is it a buffer overflow ? — Nikunj Banka
– Nikunj Banka, Commented Feb 5, 2013 at 12:54
I don't think the answers below were there when I posted this comment, and yes it is a buffer overflow. Your input is larger than the boundaries of your buffer. — Kevin
– Kevin, Commented Feb 5, 2013 at 14:35
And to be clear: there's no buffer at all! (The pointer s doesn't point to a too-small buffer; it points nowhere at all.) — Steve Summit
– Steve Summit, Commented Feb 2, 2021 at 17:35

P.P · Accepted Answer · 2013-02-05 13:18:16Z

21

It's undefined behaviour as the pointer is uninitialized. There's no problem with your compiler but your code has problem :)

Make s point to valid memory before storing data in there.

To manage buffer overflow, you can specify the length in the format specifier:

scanf("%255s", s); // If s holds a memory of 256 bytes
// '255' should be modified as per the memory allocated.

GNU C supports an non-standard extension with which you don't have to allocate memory as allocation is done if %as is specified but a pointer to pointer should be passed:

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

int main() {
  char *s,*p;

  s = malloc(256);
  scanf("%255s", s); // Don't read more than 255 chars
  printf("%s", s);

  // No need to malloc `p` here
  scanf("%as", &p); // GNU C library supports this type of allocate and store.
  printf("%s", p);
  free(s);
  free(p); 
  return 0;
}

edited Feb 5, 2013 at 13:18

answered Feb 5, 2013 at 12:23

P.P

122k20 gold badges193 silver badges251 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

m0skit0 Over a year ago

+1 Best answer so far because of overflow protection. One question if you don't mind: why use malloc() instead of static allocation?

P.P Over a year ago

OP asked why used pointer and asked why it's failing. It's minimal example. Static array would work as well for the particular case.

Rami Jarrar · Accepted Answer · 2013-02-05 12:46:40Z

8

the char pointer is not initialized, you should dynamiclly allocate memory to it,

char *s = malloc(sizeof(char) * N);

where N is the maximum string size you can read, And its not safe to use scanf without specifying the maximum length for the input string, use it like this,

scanf("%Ns",s);

where N same as that for malloc.

edited Feb 5, 2013 at 12:46

answered Feb 5, 2013 at 12:23

Rami Jarrar

4,6537 gold badges39 silver badges54 bronze badges

3 Comments

Aniket Inge Over a year ago

thou shall not cast the return value from malloc()

Rami Jarrar Over a year ago

malloc return's void *, it should be cast ?

akp Over a year ago

@RamiJarrar malloc() returns void pointer so no need to type cast...its bad habbit as you are doing...refer stackoverflow.com/questions/605845/…

akp · Accepted Answer · 2013-02-05 12:23:34Z

1

You are not allocating any memory to the character array so first try to get memory by calling malloc() or calloc(). then try to use it.

s = malloc(sizeof(char) * YOUR_ARRAY_SIZE);
...do your work...
free(s);

answered Feb 5, 2013 at 12:23

akp

1,8334 gold badges20 silver badges26 bronze badges

1 Comment

m0skit0 Over a year ago

Why bother with malloc() and free()? Just reserve statically.

oleg_g · Accepted Answer · 2013-02-05 12:24:37Z

1

You need to allocate enough memory for buffer where your pointer will point to:

    s = malloc(sizeof(char) * BUF_LEN);

and then free this memory if you do not need it anymore:

    free(s);

answered Feb 5, 2013 at 12:24

oleg_g

5423 silver badges7 bronze badges

2 Comments

m0skit0 Over a year ago

Why bother with malloc() and free()? Just reserve statically.

oleg_g Over a year ago

Static reserve is better for this case, but question shows that TS is not familiar with pointers and memory allocation. Just helped him to understand this.

m0skit0 · Accepted Answer · 2013-02-05 12:25:09Z

1

You're not allocating memory for your string, and thus, you're trying to write in a non-authorized memory address. Here

char *s;

You're just declaring a pointer. You're not specifying how much memory to reserve for your string. You can statically declare this like:

char s[100];

which will reserve 100 characters. If you go beyond 100, it will still crash as you mentionned for the same reason again.

answered Feb 5, 2013 at 12:25

m0skit0

26k13 gold badges84 silver badges131 bronze badges

Comments

Aniket Inge · Accepted Answer · 2013-02-05 12:24:41Z

0

The problem is with your code .. you never allocate memory for the char *. Since, there is no memory allocated(with malloc()) big enough to hold the string, this becomes an undefined behavior..

You must allocate memory for s and then use scanf()(I prefer fgets())

answered Feb 5, 2013 at 12:24

Aniket Inge

25.9k5 gold badges54 silver badges80 bronze badges

1 Comment

m0skit0 Over a year ago

Why bother with malloc() and free()? Why not reserve statically?

Suraj Rao · Accepted Answer · 2019-07-15 13:31:28Z

0

#include"stdio.h"
#include"malloc.h"

int main(){

        char *str;

        str=(char*)malloc(sizeof(char)*30);

        printf("\nENTER THE STRING : ");
        fgets(str,30,stdin);

        printf("\nSTRING IS : %s",str);

        return 0;
}

edited Jul 15, 2019 at 13:31

Suraj Rao

29.7k11 gold badges96 silver badges104 bronze badges

answered Jul 15, 2019 at 13:28

ganesh borkar

32 bronze badges

Comments

Sai Ganesh Myneni · Accepted Answer · 2018-11-04 08:08:45Z

-1

The code in C to read a character pointer

#include<stdio.h>
 #include<stdlib.h>
 void main()
 {
    char* str1;//a character pointer is created 
    str1 = (char*)malloc(sizeof(char)*100);//allocating memory to pointer
    scanf("%[^\n]s",str1);//hence the memory is allocated now we can store the characters in allocated memory space
    printf("%s",str1);
    free(str1);//free the memory allocated to the pointer
 }

answered Nov 4, 2018 at 8:08

Sai Ganesh Myneni

11 bronze badge

Comments

cigien · Accepted Answer · 2022-01-11 12:49:22Z

-2

I was getting this problem. I tried this code below and it worked:

char *text; 
scanf("%s", *&text);

I dont know how it worked. I just felt like doing it.

edited Jan 11, 2022 at 12:49

cigien

61.2k11 gold badges86 silver badges124 bronze badges

answered Jan 11, 2022 at 11:41

Ziaur Rehman

1

1 Comment

MrSykkox Over a year ago

Try run the code and print the string.

Collectives™ on Stack Overflow

Taking string input in char pointer

9 Answers 9

2 Comments

3 Comments

1 Comment

2 Comments

Comments

1 Comment

Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

2 Comments

3 Comments

1 Comment

2 Comments

Comments

1 Comment

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related