How do you reverse a string in place in C or C++?

How do you reverse a string in C or C++ without requiring a separate buffer to hold the reversed string?

Solution 1:

#include <algorithm>
std::reverse(str.begin(), str.end());

This is the simplest way in C++.

Solution 2:

Read Kernighan and Ritchie

#include <string.h>

void reverse(char s[])
{
    int length = strlen(s) ;
    int c, i, j;

    for (i = 0, j = length - 1; i < j; i++, j--)
    {
        c = s[i];
        s[i] = s[j];
        s[j] = c;
    }
}

Solution 3:

The standard algorithm is to use pointers to the start / end, and walk them inward until they meet or cross in the middle. Swap as you go.

Reverse ASCII string, i.e. a 0-terminated array where every character fits in 1 char. (Or other non-multibyte character sets).

void strrev(char *head)
{
  if (!head) return;
  char *tail = head;
  while(*tail) ++tail;    // find the 0 terminator, like head+strlen
  --tail;               // tail points to the last real char
                        // head still points to the first
  for( ; head < tail; ++head, --tail) {
      // walk pointers inwards until they meet or cross in the middle
      char h = *head, t = *tail;
      *head = t;           // swapping as we go
      *tail = h;
  }
}

// test program that reverses its args
#include <stdio.h>

int main(int argc, char **argv)
{
  do {
    printf("%s ",  argv[argc-1]);
    strrev(argv[argc-1]);
    printf("%s\n", argv[argc-1]);
  } while(--argc);

  return 0;
}

The same algorithm works for integer arrays with known length, just use tail = start + length - 1 instead of the end-finding loop.

(Editor's note: this answer originally used XOR-swap for this simple version, too. Fixed for the benefit of future readers of this popular question. XOR-swap is highly not recommended; hard to read and making your code compile less efficiently. You can see on the Godbolt compiler explorer how much more complicated the asm loop body is when xor-swap is compiled for x86-64 with gcc -O3.)

Ok, fine, let's fix the UTF-8 chars...

(This is XOR-swap thing. Take care to note that you must avoid swapping with self, because if *p and *q are the same location you'll zero it with a^a==0. XOR-swap depends on having two distinct locations, using them each as temporary storage.)

Editor's note: you can replace SWP with a safe inline function using a tmp variable.

#include <bits/types.h>
#include <stdio.h>

#define SWP(x,y) (x^=y, y^=x, x^=y)

void strrev(char *p)
{
  char *q = p;
  while(q && *q) ++q; /* find eos */
  for(--q; p < q; ++p, --q) SWP(*p, *q);
}

void strrev_utf8(char *p)
{
  char *q = p;
  strrev(p); /* call base case */

  /* Ok, now fix bass-ackwards UTF chars. */
  while(q && *q) ++q; /* find eos */
  while(p < --q)
    switch( (*q & 0xF0) >> 4 ) {
    case 0xF: /* U+010000-U+10FFFF: four bytes. */
      SWP(*(q-0), *(q-3));
      SWP(*(q-1), *(q-2));
      q -= 3;
      break;
    case 0xE: /* U+000800-U+00FFFF: three bytes. */
      SWP(*(q-0), *(q-2));
      q -= 2;
      break;
    case 0xC: /* fall-through */
    case 0xD: /* U+000080-U+0007FF: two bytes. */
      SWP(*(q-0), *(q-1));
      q--;
      break;
    }
}

int main(int argc, char **argv)
{
  do {
    printf("%s ",  argv[argc-1]);
    strrev_utf8(argv[argc-1]);
    printf("%s\n", argv[argc-1]);
  } while(--argc);

  return 0;
}

• Why, yes, if the input is borked, this will cheerfully swap outside the place.
• Useful link when vandalising in the UNICODE: http://www.macchiato.com/unicode/chart/
• Also, UTF-8 over 0x10000 is untested (as I don't seem to have any font for it, nor the patience to use a hexeditor)

Examples:

$ ./strrev Räksmörgås ░▒▓○◔◑◕●

░▒▓○◔◑◕● ●◕◑◔○▓▒░

Räksmörgås sågrömskäR

./strrev verrts/.