从 C 中的字符串中删除空格

Question

从 C 中的字符串中删除空格的最简单和最有效的方法是什么？

Answer 1

Easiest and most efficient don't usually go together…

Here's a possible solution for in-place removal:

void remove_spaces(char* s) {
    char* d = s;
    do {
        while (*d == ' ') {
            ++d;
        }
    } while (*s++ = *d++);
}

Answer 2

Here's a very compact, but entirely correct version:

do while(isspace(*s)) s++; while(*d++ = *s++);

---

And here, just for my amusement, are code-golfed versions that aren't entirely correct, and get commenters upset.

If you can risk some undefined behavior, and never have empty strings, you can get rid of the body:

while(*(d+=!isspace(*s++)) = *s);

Heck, if by space you mean just space character:

while(*(d+=*s++!=' ')=*s);

Don't use that in production :)

Answer 3

As we can see from the answers posted, this is surprisingly not a trivial task. When faced with a task like this, it would seem that many programmers choose to throw common sense out the window, in order to produce the most obscure snippet they possibly can come up with.

Things to consider:

• You will want to make a copy of the string, with spaces removed. Modifying the passed string is bad practice, it may be a string literal. Also, there are sometimes benefits of treating strings as immutable objects .
• You cannot assume that the source string is not empty. It may contain nothing but a single null termination character.
• The destination buffer can contain any uninitialized garbage when the function is called. Checking it for null termination doesn't make any sense.
• Source code documentation should state that the destination buffer needs to be large enough to contain the trimmed string. Easiest way to do so is to make it as large as the untrimmed string.
• The destination buffer needs to hold a null terminated string with no spaces when the function is done.
• Consider if you wish to remove all white space characters or just spaces ' ' .
• C programming isn't a competition over who can squeeze in as many operators on a single line as possible. It is rather the opposite, a good C program contains readable code (always the single-most important quality) without sacrificing program efficiency (somewhat important).
• For this reason, you get no bonus points for hiding the insertion of null termination of the destination string, by letting it be part of the copying code. Instead, make the null termination insertion explicit, to show that you haven't just managed to get it right by accident.

What I would do:

void remove_spaces (char* restrict str_trimmed, const char* restrict str_untrimmed)
{
  while (*str_untrimmed != '\0')
  {
    if(!isspace(*str_untrimmed))
    {
      *str_trimmed = *str_untrimmed;
      str_trimmed++;
    }
    str_untrimmed++;
  }
  *str_trimmed = '\0';
}

In this code, the source string "str_untrimmed" is left untouched, which is guaranteed by using proper const correctness. It does not crash if the source string contains nothing but a null termination. It always null terminates the destination string.

Memory allocation is left to the caller. The algorithm should only focus on doing its intended work. It removes all white spaces.

There are no subtle tricks in the code. It does not try to squeeze in as many operators as possible on a single line. It will make a very poor candidate for the IOCCC . Yet it will yield pretty much the same machine code as the more obscure one-liner versions.

When copying something, you can however optimize a bit by declaring both pointers as restrict , which is a contract between the programmer and the compiler, where the programmer guarantees that the destination and source are not the same address. This allows more efficient optimization, since the compiler can then copy straight from source to destination without temporary memory in between.

Answer 4

In C, you can replace some strings in-place, for example a string returned by strdup():

char *str = strdup(" a b c ");

char *write = str, *read = str;
do {
   if (*read != ' ')
       *write++ = *read;
} while (*read++);

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

Other strings are read-only, for example those declared in-code. You'd have to copy those to a newly allocated area of memory and fill the copy by skipping the spaces:

char *oldstr = " a b c ";

char *newstr = malloc(strlen(oldstr)+1);
char *np = newstr, *op = oldstr;
do {
   if (*op != ' ')
       *np++ = *op;
} while (*op++);

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

You can see why people invented other languages ;)

Answer 5

if you are still interested, this function removes spaces from the beginning of the string, and I just had it working in my code:

void removeSpaces(char *str1)  
{
    char *str2; 
    str2=str1;  
    while (*str2==' ') str2++;  
    if (str2!=str1) memmove(str1,str2,strlen(str2)+1);  
}

Answer 6

#include <ctype>

char * remove_spaces(char * source, char * target)
{
     while(*source++ && *target)
     {
        if (!isspace(*source)) 
             *target++ = *source;
     }
     return target;
}

Notes;

• This doesn't handle Unicode.

Answer 7

The easiest and most efficient way to remove spaces from a string is to simply remove the spaces from the string literal. For example, use your editor to 'find and replace' "hello world" with "helloworld" , and presto!

Okay, I know that's not what you meant. Not all strings come from string literals, right? Supposing this string you want spaces removed from doesn't come from a string literal, we need to consider the source and destination of your string ... We need to consider your entire algorithm, what actual problem you're trying to solve, in order to suggest the simplest and most optimal methods.

Perhaps your string comes from a file (eg stdin ) and is bound to be written to another file (eg stdout ). If that's the case, I would question why it ever needs to become a string in the first place. Just treat it as though it's a stream of characters, discarding the spaces as you come across them...

#include <stdio.h>

int main(void) {
    for (;;) {
        int c = getchar();
        if (c == EOF) { break;    }
        if (c == ' ') { continue; }
        putchar(c);
    }
}

By eliminating the need for storage of a string, not only does the entire program become much, much shorter, but theoretically also much more efficient.

Answer 8

#include<stdio.h>
#include<string.h>
main()
{
  int i=0,n;
  int j=0;
  char str[]="        Nar ayan singh              ";
  char *ptr,*ptr1;
  printf("sizeof str:%ld\n",strlen(str));
  while(str[i]==' ')
   {
     memcpy (str,str+1,strlen(str)+1);
   }
  printf("sizeof str:%ld\n",strlen(str));
  n=strlen(str);
  while(str[n]==' ' || str[n]=='\0')
    n--;
  str[n+1]='\0';
  printf("str:%s ",str);
  printf("sizeof str:%ld\n",strlen(str));
}

Answer 9

Code taken from zString library

/* search for character 's' */
int zstring_search_chr(char *token,char s){
        if (!token || s=='\0')
        return 0;

    for (;*token; token++)
        if (*token == s)
            return 1;

    return 0;
}

char *zstring_remove_chr(char *str,const char *bad) {
    char *src = str , *dst = str;

    /* validate input */
    if (!(str && bad))
        return NULL;

    while(*src)
        if(zstring_search_chr(bad,*src))
            src++;
        else
            *dst++ = *src++;  /* assign first, then incement */

    *dst='\0';
    return str;
}

Code example

  Exmaple Usage
      char s[]="this is a trial string to test the function.";
      char *d=" .";
      printf("%s\n",zstring_remove_chr(s,d));

  Example Output
      thisisatrialstringtotestthefunction

Have a llok at the zString code, you may find it useful https://github.com/fnoyanisi/zString

Answer 10

That's the easiest I could think of (TESTED) and it works!!

char message[50];
fgets(message, 50, stdin);
for( i = 0, j = 0; i < strlen(message); i++){
        message[i-j] = message[i];
        if(message[i] == ' ')
            j++;
}
message[i] = '\0';

Answer 11

Here is the simplest thing i could think of. Note that this program uses second command line argument (argv[1]) as a line to delete whitespaces from.

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

/*The function itself with debug printing to help you trace through it.*/

char* trim(const char* str)
{
    char* res = malloc(sizeof(str) + 1);
    char* copy = malloc(sizeof(str) + 1);
    copy = strncpy(copy, str, strlen(str) + 1);
    int index = 0;

    for (int i = 0; i < strlen(copy) + 1; i++) {
        if (copy[i] != ' ')
        {
            res[index] = copy[i];
            index++;
        }
        printf("End of iteration %d\n", i);
        printf("Here is the initial line: %s\n", copy);
        printf("Here is the resulting line: %s\n", res);
        printf("\n");
    }
    return res;
}

int main(int argc, char* argv[])
{
    //trim function test

    const char* line = argv[1];
    printf("Here is the line: %s\n", line);

    char* res = malloc(sizeof(line) + 1);
    res = trim(line);

    printf("\nAnd here is the formatted line: %s\n", res);

    return 0;
}

Answer 12

/* Function to remove all spaces from a given string.
   https://www.geeksforgeeks.org/remove-spaces-from-a-given-string/
*/
void remove_spaces(char *str)
{
    int count = 0;
    for (int i = 0; str[i]; i++)
        if (str[i] != ' ')
            str[count++] = str[i];
    str[count] = '\0';
}

Answer 13

This is implemented in micro controller and it works, it should avoid all problems and it is not a smart way of doing it, but it will work :)

void REMOVE_SYMBOL(char* string, uint8_t symbol)
{
  uint32_t size = LENGHT(string); // simple string length function, made my own, since original does not work with string of size 1
  uint32_t i = 0;
  uint32_t k = 0;
  uint32_t loop_protection = size*size; // never goes into loop that is unbrakable
  while(i<size)
  {
    if(string[i]==symbol)
    {
      k = i;
      while(k<size)
      {
        string[k]=string[k+1];
        k++;
      }
    }
    if(string[i]!=symbol)
    {
      i++;
    }
    loop_protection--;
    if(loop_protection==0)
    {
      i = size;
      break;
    }
  }
}

Answer 14

While this is not as concise as the other answers, it is very straightforward to understand for someone new to C, adapted from the Calculix source code.

char* remove_spaces(char * buff, int len)
{
    int i=-1,k=0;
    while(1){
        i++;
        if((buff[i]=='\0')||(buff[i]=='\n')||(buff[i]=='\r')||(i==len)) break;
        if((buff[i]==' ')||(buff[i]=='\t')) continue;
        buff[k]=buff[i];
        k++;
    }
    buff[k]='\0';
    return buff;
}

Answer 15

I assume the C string is in a fixed memory, so if you replace spaces you have to shift all characters.

The easiest seems to be to create new string and iterate over the original one and copy only non space characters.

Answer 16

I came across a variation to this question where you need to reduce multiply spaces into one space "represent" the spaces.

This is my solution:

char str[] = "Put Your string Here.....";

int copyFrom = 0, copyTo = 0;

printf("Start String %s\n", str);

while (str[copyTo] != 0) {
    if (str[copyFrom] == ' ') {
        str[copyTo] = str[copyFrom];
        copyFrom++;
        copyTo++;

        while ((str[copyFrom] == ' ') && (str[copyFrom] !='\0')) {
            copyFrom++;
        }
    }

    str[copyTo] = str[copyFrom];

    if (str[copyTo] != '\0') {
        copyFrom++;
        copyTo++;
    }
}

printf("Final String %s\n", str);

Hope it helps :-)