Why is this string reversal 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:

#include <string.h>

void reverse(char *s);

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

    return 0;
}

void 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.

Segfault almost always means you're trying to dereference a null pointer. Run your code through GDB, find which line it's segfaulting on, and see which pointer is null (0x000000) there. — Matt Ball, Oct 23 '09 at 17:03
Also, why are you using two variables (i and j)? You can absolutely do this with counter. — Matt Ball, Oct 23 '09 at 17:04
Can you post the code that creates the char* and passes it to reverse()? — noctonura, Oct 23 '09 at 17:07
Asked and answered many times. E.g. http://stackoverflow.com/questions/1405594/writing-into-c-string — Sinan Ünür, Oct 23 '09 at 17:43
If you really are trying to improve, you ought to read the FAQ list: http://c-faq.com/charstring/strlitinit.html — Sinan Ünür, Oct 23 '09 at 18:07
@Carl Norum: If you lok carefully, `strlen` is not used in the loop condition in the above code sample. — AnT stands with Russia, Oct 23 '09 at 21:20

AnT stands with Russia · Accepted Answer · 2009-10-24T15:22:29.490

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");

This worked. Thanks. But can you please explain in more detail what the difference is between the two string initializations? I.e. why does the way I did create a string literal whereas the 'array' syntax creates a modifiable string? — james, Oct 23 '09 at 17:10
@james String literals go in read-only storage. But by doing "char []s = ..." you're declaring an array and initializing it to a literal, rather than getting a pointer to a literal. — asveikau, Oct 23 '09 at 17:15

JaredPar · Answer 2 · 2009-10-23T17:09:45.517

10

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

s is NULL
s points to a const string which is held in read only memory
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);

edited Oct 23 '09 at 17:09

answered Oct 23 '09 at 17:03

JaredPar

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The literal "teststring" is located in read only memory somewhere and you are not allowed to write to it. This used to be common practice, ie a way of allocating some memory, but most modern systems won't allow it. JaredPar's example works because strdup allocates memory which you then own. – Tim Allman Oct 23 '09 at 17:19

score 3 · Answer 3 · answered Oct 23 '09 at 17:08

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).

score 3 · Answer 4 · answered Oct 23 '09 at 18:37

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.

score 1 · Answer 5 · edited Jun 20 '20 at 09:12

See Question 1.32 in the C FAQ list:

What is the difference between these initializations?
char a[] = "string literal";
char *p  = "string literal";
My program crashes if I try to assign a new value to p[i].

Answer:

A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways:

As the initializer for an array of char, as in the declaration of char a[], it specifies the initial values of the characters in that array (and, if necessary, its size).

Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element.

Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching).

(emphasis mine)

Why is this string reversal C code causing a segmentation fault?

8 Answers8

Answer:

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