Why is this C code causing a segmentation fault?

Question

I am trying to write code to reverse a string in place (I'm just trying to get better at C programming and pointer manipulation), but I cannot figure out why I am getting a segmentation fault:

int main() {
    char* s = "teststring";
    reverse(s);

    return 0;
}

reverse(char* s) {
    int i,j;
    char temp;

    for (i=0,j=(strlen(s)-1); i<j; i++,j--) {
        temp = *(s+i);     //line 1
        *(s+i) = *(s+j);   //line 2
        *(s+j) = temp;     //line 3
    }
}

It's lines 2 and 3 that are causing the segmentation fault. I understand that there may be better ways to do this, but I am interested in finding out what specifically in my code is causing the segmentation fault.

Update: I have included the calling function as requested.

Answer 1

You code could be segfaulting for a number of reasons. Here are the ones that come to mind

1. s is NULL
2. s points to a const string which is held in read only memory
3. s is not NULL terminated

I think #2 is the most likely. Can you show us the call site of reverse?

EDIT

Based on your sample #2 is definitely the answer. A string literal in C/C++ is not modifiable. The proper type is actually const char* and not char*. What you need to do is pass a modifiable string into that buffer.

Quick example:

char* pStr = strdup("foobar");
reverse(pStr);
free(pStr);

Answer 2

There's no way to say from just that code. Most likely, you are passing in a pointer that points to invalid memory, non-modifiable memory or some other kind of memory that just can't be processed the way you process it here.

How do you call your function?

Added: You are passing in a pointer to a string literal. String literals are non-modifiable. You can't reverse a string literal.

Pass in a pointer to a modifiable string instead

char s[] = "teststring";
reverse(s);

This has been explained to death here already. "teststring" is a string literal. The string literal itself is a non-modifiable object. In practice compilers might (and will) put it in read-only memory. When you initialize a pointer like that

char *s = "teststring";

the pointer points directly at the beginning of the string literal. Any attempts to modify what s is pointing to are deemed to fail in general case. You can read it, but you can't write into it. For this reason it is highly recommended to point to string literals with pointer-to-const variables only

const char *s = "teststring";

But when you declare your s as

char s[] = "teststring";

you get a completely independent array s located in ordinary modifiable memory, which is just initialized with string literal. This means that that independent modifiable array s will get its initial value copied from the string literal. After that your s array and the string literal continue to exist as completely independent objects. The literal is still non-modifiable, while your s array is modifiable.

Basically, the latter declaration is functionally equivalent to

char s[11];
strcpy(s, "teststring");

Answer 3

Are you testing this something like this?

int main() {
    char * str = "foobar";
    reverse(str);
    printf("%s\n", str);
}

This makes str a string literal and you probably won't be able to edit it (segfaults for me). If you define char * str = strdup(foobar) it should work fine (does for me).

Answer 4

Which compiler and debugger are you using? Using gcc and gdb, I would compile the code with -g flag and then run it in gdb. When it segfaults, I would just do a backtrace (bt command in gdb) and see which is the offending line causing the problem. Additionally, I would just run the code step by step, while "watching" the pointer values in gdb and know where exactly is the problem.

Good luck.

Answer 5

I think strlen can not work since s is not NULL terminated. So the behaviour of your for iteration is not the one you expect. Since the result of strlen will be superior than s length you will write in memory where you should not be.

In addition s points to a constant strings hold by a read only memory. You can not modify it. Try to init s by using the gets function as it is done in the strlen example

Answer 6

See Question 8.5 in the C FAQ list.

See also Back to Basics by Joel.

Answer 7

Your declaration is completely wrong:

char* s = "teststring";

"teststring" is stored in the code segment, which is read-only, like code. And, s is a pointer to "teststring", at the same time, you're trying to change the value of a read-only memory range. Thus, segmentation fault.

But with:

char s[] = "teststring";

s is initialized with "teststring", which of course is in the code segment, but there is an additional copy operation going on, to the stack in this case.

Answer 8

Please, read the standard C, there you will find surprises like this in the C language that you are not expecting to see...