Why is scanf not working as expected when writing to a string literal?

Question

include<stdio.h>
int main()
{
    //char b[10];
    char *a="goodone";
    //a=b;

    scanf("%s",a);//this scanf fails and thow segmentation fault.
    printf("%s",a);

}

Why is this not working? I tried a lot with this scanf but, when I reserve memory for my variable a*(by assigning a=b (commented)) it works fine. Otherwise it doesn't. I believe that char *a will allocate some memory for its string,("goodone")and return that memory location to its value. And printf working fine with this belief why scanf not? please save me from this....

I strongly recommend to read the [C FAQ entry](http://c-faq.com/aryptr/aryptr2.html) for this question. — Lundin, Feb 15 '12 at 10:56
This is definitely a good read: [What is the difference between char a\[\] = “string”; and char *p = “string”;](http://stackoverflow.com/questions/9460260/what-is-the-difference-between-char-a-string-and-char-p-string) — Alok Save, May 29 '12 at 09:32

score 2 · Accepted Answer · answered Feb 15 '12 at 10:10

2

This is because you are instructing scanf to write the data that it reads into the memory allocated for the const char* value, i.e. into read-only memory.

If you would like to make your string constant writable, change

char *a="goodone";

to

char a[]="goodone";

Note that this is not safe either: it may crash when the user enters more than seven characters. Add a limit to your format specifier in order to address that issue:

scanf("%7s",a);

P.S. The commented out a=b works fine because it is not modifying the string constant; rather, it modifies a pointer to character constant, which is allowed.

answered Feb 15 '12 at 10:10

Sergey Kalinichenko

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To emphasize the point, *never* use `scanf()` with a bare `%s` format. No matter how big a buffer you provide, you cannot prevent the user from entering enough characters to overflow it. It's as unsafe as `gets()` (which also should never be used). – Keith Thompson Feb 15 '12 at 10:19
@KeithThompson gets() is no longer standard C, since C11. – Lundin Feb 15 '12 at 10:58
@Lundin: Right -- but most or all implementations still provide it (perhaps with a warning), and sadly there are still plenty of books and tutorials that use it. – Keith Thompson Feb 15 '12 at 23:52

score 1 · Answer 2 · answered Feb 15 '12 at 10:15

char *a is just a pointer to a char. When you assign "goodone" to it, it points to that string literal (which is read-only), and scanf tries to overwrite that string in memory which causes the crash.

If you assign b to it, then you have a pointing to a writeable memory area of 10 chars (i.e., a string of maximum length 9 + the terminating NUL). So it works as long as scanf is not storing anything longer than that in there.

Likewise you could make a an array instead of a pointer (i.e., char a[] = "goodone";). Again you need to watch out not to store anything longer in there.

Why is scanf not working as expected when writing to a string literal?

2 Answers2