Segmentation fault reversing a string literal [duplicate]

Answer 1

If you do char s[6] = {'h','e','l','l','o','\0'}; you put 6 chars into an array on the stack. When you do char *s = "hello"; there's only a pointer on the stack and the memory that it points to may be read only. Writing to that memory causes undefined behavior.

Answer 2

With some versions of gcc you can allow modification of static strings with -fwritable-strings. Not that there is really any good excuse to do that.

Answer 3

You can try this:

void strrev(char *in, char *out, int len){
    int i;

    for(i = 0; i < len; i++){
        out[len - i - 1] = in[i];
    }
}

Note that it doesn't deal with the string terminator.

Answer 4

Your code attempts to modify a string literal which is not allowed in C or C++ If you change:

char *s = "hello"; 

to:

char s[] = "hello"; 

then you are modifying the contents of the array, into which the literal has been copied (equivalent to initialising the array with individual characters), which is OK.