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

Answer 1

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 2

As some of the answers provided above, the string memory is read-only. However, some compilers provide an option to compile with writable strings. E.g. with gcc, 3.x versions supported -fwritable-strings but newer versions don't.

Answer 3

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 4

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 5

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 6

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)

See also Back to Basics by Joel.