Thread: strange behavior

  1. #1
    Registered User
    Join Date
    Oct 2005
    Hyderabad, India

    strange behavior

    Any reasons for the below mentioned behavior?? i use gcc .
    I think both should have run sucessfully.

    /* this gives segmentation fault */
    #include <stdio.h>
        char* a = "abcd";
        a[1] = '\0';
    /* this doesn't give segmentation fault */
    #include <stdio.h>
        char a[] = "abcd";
        a[1] = '\0';

  2. #2
    Gawking at stupidity
    Join Date
    Jul 2004
    Oregon, USA
    String literals are usually stored in read-only memory. When you do char *a = "abcd"; you're creating a pointer that points to a string that resides in read-only memory.

    When you do char a[] = "abcd"; you're creating an array that the string "abcd" is copied into. So when you do a[1] = '\0'; later you're modifying the array instead of the string in read-only memory.

    Trying to modify the read-only memory is what's giving you the segmentation fault.

    You can try giving gcc the -fwritable-strings option to make the pointer one work, but it's better to just avoid the practice altogether.
    itsme@itsme:~/C$ cat segfault.c
    #include <stdio.h>
    int main(void)
      char *a = "foo";
      a[1] = '\0';
      return 0;
    itsme@itsme:~/C$ gcc -Wall segfault.c -o segfault
    itsme@itsme:~/C$ ./segfault
    Segmentation fault (core dumped)
    itsme@itsme:~/C$ gcc -Wall -fwritable-strings segfault.c -o segfault
    cc1: note: -fwritable-strings is deprecated; see documentation for details
    itsme@itsme:~/C$ ./segfault
    You can see that the -fwritable-strings prevented the segmentation fault, but it's a deprecated feature that will probably go away eventually.
    Last edited by itsme86; 10-17-2005 at 12:10 PM.
    If you understand what you're doing, you're not learning anything.

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