如何在 C 或 C++ 中将字符串反转到位？

Question

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

Answer 1

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

This is the simplest way in C++.

Answer 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;
    }
}

Answer 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, ie 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/.

Answer 4

Reverse a string in place (visualization):

Answer 5

Non-evil C, assuming the common case where the string is a null-terminated char array:

#include <stddef.h>
#include <string.h>

/* PRE: str must be either NULL or a pointer to a 
 * (possibly empty) null-terminated string. */
void strrev(char *str) {
  char temp, *end_ptr;

  /* If str is NULL or empty, do nothing */
  if( str == NULL || !(*str) )
    return;

  end_ptr = str + strlen(str) - 1;

  /* Swap the chars */
  while( end_ptr > str ) {
    temp = *str;
    *str = *end_ptr;
    *end_ptr = temp;
    str++;
    end_ptr--;
  }
}

Answer 6

您可以使用C ++标准库中的std::reverse算法。

Answer 7

It's been a while and I don't remember which book taught me this algorithm, but I thought it was quite ingenious and simple to understand:

char input[] = "moc.wolfrevokcats";

int length = strlen(input);
int last_pos = length-1;
for(int i = 0; i < length/2; i++)
{
    char tmp = input[i];
    input[i] = input[last_pos - i];
    input[last_pos - i] = tmp;
}

printf("%s\n", input);

A visualization of this algorithm, courtesy of slashdottir :

Answer 8

Use the std::reverse method from STL :

std::reverse(str.begin(), str.end());

You will have to include the "algorithm" library, #include<algorithm> .

Answer 9

Note that the beauty of std::reverse is that it works with char * strings and std::wstring s just as well as std::string s

void strrev(char *str)
{
    if (str == NULL)
        return;
    std::reverse(str, str + strlen(str));
}

Answer 10

In the interest of completeness, it should be pointed out that there are representations of strings on various platforms in which the number of bytes per character varies depending on the character. Old-school programmers would refer to this as DBCS (Double Byte Character Set) . Modern programmers more commonly encounter this in UTF-8 (as well as UTF-16 and others). There are other such encodings as well.

In any of these variable-width encoding schemes, the simple algorithms posted here ( evil , non-evil or otherwise ) would not work correctly at all! In fact, they could even cause the string to become illegible or even an illegal string in that encoding scheme. See Juan Pablo Califano's answer for some good examples.

std::reverse() potentially would still work in this case, as long as your platform's implementation of the Standard C++ Library (in particular, string iterators) properly took this into account.

Answer 11

If you're looking for reversing NULL terminated buffers, most solutions posted here are OK. But, as Tim Farley already pointed out, these algorithms will work only if it's valid to assume that a string is semantically an array of bytes (ie single-byte strings), which is a wrong assumption, I think.

Take for example, the string "año" (year in Spanish).

The Unicode code points are 0x61, 0xf1, 0x6f.

Consider some of the most used encodings:

Latin1 / iso-8859-1 (single byte encoding, 1 character is 1 byte and vice versa):

Original:

0x61, 0xf1, 0x6f, 0x00

Reverse:

0x6f, 0xf1, 0x61, 0x00

The result is OK

UTF-8:

Original:

0x61, 0xc3, 0xb1, 0x6f, 0x00

Reverse:

0x6f, 0xb1, 0xc3, 0x61, 0x00

The result is gibberish and an illegal UTF-8 sequence

UTF-16 Big Endian:

Original:

0x00, 0x61, 0x00, 0xf1, 0x00, 0x6f, 0x00, 0x00

The first byte will be treated as a NUL-terminator. No reversing will take place.

UTF-16 Little Endian:

Original:

0x61, 0x00, 0xf1, 0x00, 0x6f, 0x00, 0x00, 0x00

The second byte will be treated as a NUL-terminator. The result will be 0x61, 0x00, a string containing the 'a' character.

Answer 12

Another C++ way (though I would probably use std::reverse() myself :) as being more expressive and faster)

str = std::string(str.rbegin(), str.rend());

The C way (more or less :) ) and please, be careful about XOR trick for swapping, compilers sometimes cannot optimize that.

In such case it is usually much slower.

char* reverse(char* s)
{
    char* beg = s, *end = s, tmp;
    while (*end) end++;
    while (end-- > beg)
    { 
        tmp  = *beg; 
        *beg++ = *end;  
        *end =  tmp;
    }
    return s;
} // fixed: check history for details, as those are interesting ones

Answer 13

#include <cstdio>
#include <cstdlib>
#include <string>

void strrev(char *str)
{
        if( str == NULL )
                return;

        char *end_ptr = &str[strlen(str) - 1];
        char temp;
        while( end_ptr > str )
        {
                temp = *str;
                *str++ = *end_ptr;
                *end_ptr-- = temp;
        }
}

int main(int argc, char *argv[])
{
        char buffer[32];

        strcpy(buffer, "testing");
        strrev(buffer);
        printf("%s\n", buffer);

        strcpy(buffer, "a");
        strrev(buffer);
        printf("%s\n", buffer);

        strcpy(buffer, "abc");
        strrev(buffer);
        printf("%s\n", buffer);

        strcpy(buffer, "");
        strrev(buffer);
        printf("%s\n", buffer);

        strrev(NULL);

        return 0;
}

This code produces this output:

gnitset
a
cba

Answer 14

如果您使用 GLib，它有两个函数， g_strreverse()和g_utf8_strreverse()

Answer 15

I like Evgeny's K&R answer. However, it is nice to see a version using pointers. Otherwise, it's essentially the same:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *reverse(char *str) {
    if( str == NULL || !(*str) ) return NULL;
    int i, j = strlen(str)-1;
    char *sallocd;
    sallocd = malloc(sizeof(char) * (j+1));
    for(i=0; j>=0; i++, j--) {
        *(sallocd+i) = *(str+j);
    }
    return sallocd;
}

int main(void) {
    char *s = "a man a plan a canal panama";
    char *sret = reverse(s);
    printf("%s\n", reverse(sret));
    free(sret);
    return 0;
}

Answer 16

Recursive function to reverse a string in place (no extra buffer, malloc).

Short, sexy code. Bad, bad stack usage.

#include <stdio.h>

/* Store the each value and move to next char going down
 * the stack. Assign value to start ptr and increment 
 * when coming back up the stack (return).
 * Neat code, horrible stack usage.
 *
 * val - value of current pointer.
 * s - start pointer
 * n - next char pointer in string.
 */
char *reverse_r(char val, char *s, char *n)
{
    if (*n)
        s = reverse_r(*n, s, n+1);
   *s = val;
   return s+1;
}

/*
 * expect the string to be passed as argv[1]
 */
int main(int argc, char *argv[])
{
    char *aString;

    if (argc < 2)
    {
        printf("Usage: RSIP <string>\n");
        return 0;
    }

    aString = argv[1];
    printf("String to reverse: %s\n", aString );

    reverse_r(*aString, aString, aString+1); 
    printf("Reversed String:   %s\n", aString );

    return 0;
}

Answer 17

如果您使用 ATL/MFC CString ，只需调用CString::MakeReverse() 。

Answer 18

Use std::reverse()

reverse(begin(str), end(str));

And that's it.

Answer 19

Share my code. As a C++ learner, as an option to use swap(), I am humbly asking for comments.

void reverse(char* str) {
    int length = strlen(str);
    char* str_head = str;
    char* str_tail = &str[length-1];
    while (str_head < str_tail) 
        swap(*str_head++, *str_tail--);
}

Answer 20

Yet another:

#include <stdio.h>
#include <strings.h>

int main(int argc, char **argv) {

  char *reverse = argv[argc-1];
  char *left = reverse;
  int length = strlen(reverse);
  char *right = reverse+length-1;
  char temp;

  while(right-left>=1){

    temp=*left;
    *left=*right;
    *right=temp;
    ++left;
    --right;

  }

  printf("%s\n", reverse);

}

Answer 21

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

unsigned char * utf8_reverse(const unsigned char *, int);
void assert_true(bool);

int main(void)
{
    unsigned char str[] = "mañana mañana";
    unsigned char *ret = utf8_reverse(str,  strlen((const char *) str) + 1);

    printf("%s\n", ret);
    assert_true(0 == strncmp((const char *) ret, "anãnam anañam", strlen("anãnam anañam") + 1));

    free(ret);

    return EXIT_SUCCESS;
}

unsigned char * utf8_reverse(const unsigned char *str, int size)
{
    unsigned char *ret = calloc(size, sizeof(unsigned char*));
    int ret_size = 0;
    int pos = size - 2;
    int char_size = 0;

    if (str ==  NULL) {
        fprintf(stderr, "failed to allocate memory.\n");
        exit(EXIT_FAILURE);
    }

    while (pos > -1) {

        if (str[pos] < 0x80) {
            char_size = 1;
        } else if (pos > 0 && str[pos - 1] > 0xC1 && str[pos - 1] < 0xE0) {
            char_size = 2;
        } else if (pos > 1 && str[pos - 2] > 0xDF && str[pos - 2] < 0xF0) {
            char_size = 3;
        } else if (pos > 2 && str[pos - 3] > 0xEF && str[pos - 3] < 0xF5) {
            char_size = 4;
        } else {
            char_size = 1;
        }

        pos -= char_size;
        memcpy(ret + ret_size, str + pos + 1, char_size);
        ret_size += char_size;
    }    

    ret[ret_size] = '\0';

    return ret;
}

void assert_true(bool boolean)
{
    puts(boolean == true ? "true" : "false");
}

Answer 22

With C++ lambda:

 auto reverse = [](std::string& s) -> std::string {
        size_t start = 0, end = s.length() -1;
        char temp;

        while (start < end) {
          temp = s[start];
          s[start++] = s[end];
          s[end--] = temp;
        } 

        return s;
   };

Answer 23

My answer would be similar to most of them, but please find my code here.

//Method signature to reverse string
string reverseString(string str);

int main(void){
    string str;
    getline(cin, str);
    str =  reverseString(str);
    cout << "The reveresed string is : " << str;
    return 0;
}

/// <summary>
///     Reverses the input string.
/// </summary>
/// <param name="str">
///    This is the input string which needs to be reversed.
/// </param>
/// <return datatype = string>
///     This method would return the reversed string
/// </return datatype>

string reverseString(string str){
    int length = str.size()-1;
    char temp;
    for( int i=0 ;i<(length/2);i++)
    {
        temp = str[i];
        str[i] = str[length-i];
        str[length-i] = temp;
    }
    return str;
}

Answer 24

I think there is another way to reverse the string. get the input from user and reverse it.

void Rev() {
    char ch;
    cin.get(ch);
    if(ch != '\n') {
        Rev();
        cout.put(ch);
    }
}

Answer 25

C++ multi-byte UTF-8 reverser

My thought is that you can never just swap ends, you must always move from beginning-to-end, move through the string and look for "how many bytes will this character require?" I attach the character starting at the original end position, and remove the character from the front of the string.

void StringReverser(std::string *original)
{
  int eos = original->length() - 1;
  while (eos > 0) {
    char c = (*original)[0];
    int characterBytes;
    switch( (c & 0xF0) >> 4 ) {
    case 0xC:
    case 0xD: /* U+000080-U+0007FF: two bytes. */
      characterBytes = 2;
      break;
    case 0xE: /* U+000800-U+00FFFF: three bytes. */
      characterBytes = 3;
      break;
    case 0xF: /* U+010000-U+10FFFF: four bytes. */
      characterBytes = 4;
      break;
    default:
      characterBytes = 1;
      break;
    }

    for (int i = 0; i < characterBytes; i++) {
      original->insert(eos+i, 1, (*original)[i]);
    }
    original->erase(0, characterBytes);
    eos -= characterBytes;
  }
}

Answer 26

void reverseString(vector<char>& s) {
        int l = s.size();
        char ch ;
        int i = 0 ;
        int j = l-1;
        while(i < j){
                s[i] = s[i]^s[j];
                s[j] = s[i]^s[j];
                s[i] = s[i]^s[j];
                i++;
                j--;
        }
        for(char c : s)
                cout <<c ;
        cout<< endl;
}

Answer 27

In C++ the reverse can be done in a function:

#include <algorithm>
#include <string>

void backwards(vector<string> &inputs_ref) {
    for (auto i = inputs_ref.begin(); i != inputs_ref.end(); ++i) {
        reverse(i->begin(), i->end());
    }
}

Answer 28

input string, return string, No other library required

std::string reverse_string(std::string &str)
{   
  const char*buf = str.c_str();
  char *start = const_cast<char*>(buf);
  char *end = start + strlen(buf) - 1;
  char t;

  while(start < end)
  {
      t = *start;
      *start = *end;
      *end = t;
      start ++;
      end --;
  }
  str = buf;
  return str;
}

std::string md1 = "abcdefghijklmnopqrstuvwxyz0123456789";
std::cout << reverse_string(md1) << std::endl;

//9876543210zyxwvutsrqponmlkjihgfedcba

Answer 29

If you don't need to store it, you can reduce the time spent like this:

void showReverse(char s[], int length)
{
    printf("Reversed String without storing is ");
    //could use another variable to test for length, keeping length whole.
    //assumes contiguous memory
    for (; length > 0; length--)
    {
        printf("%c", *(s+ length-1) );
    }
    printf("\n");
}

Answer 30

Here's my take on it in C. Did it for practice and tried to be as concise as possible! You enter a string via the command line, ie ./program_name "enter string here"

#include <stdio.h>
#include <string.h>

void reverse(int s,int e,int len,char t,char* arg) {
   for(;s<len/2;t=arg[s],arg[s++]=arg[e],arg[e--]=t);
}

int main(int argc,char* argv[]) {
  int s=0,len=strlen(argv[1]),e=len-1; char t,*arg=argv[1];
  reverse(s,e,len,t,arg);
  for(s=0,e=0;e<=len;arg[e]==' '||arg[e]=='\0'?reverse(s,e-1,e+s,t,arg),s=++e:e++);
  printf("%s\n",arg);
}

Answer 31

But I think the XOR swap algorithm is the best...

char str[]= {"I am doing reverse string"};
char* pStr = str;

for(int i = 0; i != ((int)strlen(str)-1)/2; i++)
{
    char b = *(pStr+i);
    *(pStr+i) = *(pStr+strlen(str)-1-i);
    *(pStr+strlen(str)-1-i) = b;
}

Answer 32

#include<stdio.h>
#include<conio.h>

int main()
{
    char *my_string = "THIS_IS_MY_STRING";
    char *rev_my_string = my_string;

    while (*++rev_my_string != '\0')
        ;

    while (rev_my_string-- != (my_string-1))
    {
        printf("%c", *rev_my_string);
    }

    getchar();
    return 0;
}

This is optimised code in the C language for reversing a string... And it is simple; just use a simple pointer to do the job...

Answer 33

Here is the cleanest, safest, easiest way to reverse a string in C++ (in my opinion):

#include <string>

void swap(std::string& str, int index1, int index2) {

    char temp = str[index1];
    str[index1] = str[index2];
    str[index2] = temp;

}

void reverse(std::string& str) {

    for (int i = 0; i < str.size() / 2; i++)
        swap(str, i, str.size() - i - 1);

}

An alternative is to use std::swap , but I like defining my own functions - it's an interesting exercise and you don't need to include anything extra.