Thread: Reversing a String, void style

Good Evening.
I am trying to recreate the reverse String function. I realize there is a strrev() that does the same thing, but I wanted to create my own. After running into many problems, I started using the algorithm on programmingsimplified.com. Here is that algorithm

Code:

void ReverseString(char* String)  
{
  int i;
  int length = strlen(String);
  char* fromFront = String;
  char* fromBack = String;
  char temp;
  for (i=0;i<length-1;i++) fromBack++; //move pointer to end of string
  for (i=0;i<length/2;i++) {
      temp = *fromBack;
      *fromBack = *fromFront;
      *fromFront = temp;
      fromBack--;
      fromFront++;
  }
}

but when I run my main function to test it, it gives a seg fault when I am assigning *fromfront to *fromback.

Code:

  int main(int argc, char *argv[])
{
    char* hi="jewel";
    ReverseString(hi);
    printf ("%s", hi);
    return 0;
}

Please help

Just run(twice ) and it runned and did his job.Are you sure you post the whole code?

So this works?,
Cuz when I run it, it gives a seg fault.

1-You may use printf functions to see where your code crashes.
e.g

Code:

...
printf("here1\n");
here is the line that your project crashes
printf("here2\n");
...

output will be

Code:

here1

2-i would suggest to write your own code,not a copy paste one.
Hint- reverse the string character by character
post back -of course- if needed

Originally Posted by christop

Edit: std10093: yes, it's basic pointer arithmetic. you're getting a segfault because it's trying to write to a string constant.

I do not think i understand fully what the problem is.Can someone perhaps explain?

Originally Posted by std10093

I do not think i understand fully what the problem is.Can someone perhaps explain?

What it means is that u cannot rearrange the characters of a string when u declare it as a constant like char *str = "some_string". U can however rearrange the characters when u save the string as an array of characters eg char str[] = some_string

Code:

char *str = "jewel";

That declares str as a pointer. It initializes str to point to the place in memory where the string constant "jewel" is contained. Since "jewel" is used as a string constant, it exists in a read-only segment of memory. That means, that if your code tries to write to a read-only segment of memory, it violates the rules for that segment, hence the "segmentation violation" or "seg fault".

Code:

char arr[] = "jewel";

That declares arr as a char array of length 6 (5 for the letters in "jewel" + 1 for the null terminator). The array exists on the stack (if declared locally) or in global variable space (if declared globally), and thus is in a writable segment of memory, hence reversing it in place is allowed and doesn't cause a seg fault.

@std10093:
The reason this probably "worked" for you the first time is not a Linux/non-Linux issue. It is probably because you were using an old compiler (Turbo C or Dev-C++ perhaps?). Old compilers often used to place string constants in writable memory, but that meant the program could change the contents, and they weren't exactly constant any more. Try upgrading your other compiler.

Originally Posted by livin

What it means is that u cannot rearrange the characters of a string when u declare it as a constant like char *str = "some_string". U can however rearrange the characters when u save the string as an array of characters eg char str[] = some_string

You are right thank you

Originally Posted by anduril462

Code:

char *str = "jewel";

Tottaly agree. I had forgotten so thanks.

Originally Posted by anduril462

@std10093:
The reason this probably "worked" for you the first time is not a Linux/non-Linux issue. It is probably because you were using an old compiler (Turbo C or Dev-C++ perhaps?). Old compilers often used to place string constants in writable memory, but that meant the program could change the contents, and they weren't exactly constant any more. Try upgrading your other compiler.

I have netbeans the latest version,so the compiler may also be updated.