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What is the easiest and most efficient way to remove spaces from a string in C?
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++);
}
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 :)
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.
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 ;)
#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.
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);
}
#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));
}
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 (e.g. stdin) and is bound to be written to another file (e.g. 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.
/* 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';
}
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
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';
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;
}
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;
}
}
}
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;
}
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.
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 :-)
I am writing a program that takes a user's comment. Specifically one that has input outside of /* and */ and also inside. I have written my loop to find the char "/" in my array and I am unsure how to remove it and everything in between it until it appears again. For example if my input was "comment /* this is my comment */" I need to remove the /* */ and contents between. So my output would just be "comment". If there is no "/* and */" it doesn't remove anything. I know I need a loop but how would I write a loop that removes chars in the array until the next "/" appears and removes it as well?
My code is as follow:
#include <stdio.h>
#include <string.h>
void remove_comment(char *s1, char *s2){
for(; *s1 != '\0'; s1++){ //loops through array until null value
if(*s1 == '/'){ //if array has '/' stored
//clear array elements till next '/' and removes it as well
}
else{
return; //do nothing to array
}
strcpy(s2,s1); //copies new modified string to s2 for later use
}
int main(){
char s1[101]; //declares arrays up to 100 in length with room for null character
char s2[101];
printf("Enter a comment: "); //enter a comment
fgets(s1, 100, stdin); // saves comment to array
remove_comment(s1,s2); //calls function
printf("%s", s2); //prints my modified array
return 0;
}
Your code seems to build on a loop exploring the chars of the string. I propose you therefore the following solution:
void remove_comment(char *s1, char *s2)
{
for(int in_comment=0; *s1 ; s1++){ //loops through array until null value
if(!in_comment && *s1 == '/' && s1[1]=='*') { //if array has '/' follewed by '*' stored
in_comment=1; // we enter a comment area
s1++;
}
else if (in_comment) { // if we are in a comment area
if (*s1=='*' && s1[1]=='/') { // we only look for end of comment
in_comment = 0;
s1++;
}
}
else *s2++=*s1; // if we're not in comment, in all other cases we just copy current char
}
*s2='\0'; // don't forget to end the string.
}
It uses a in_comment variable to tell if we are currently exploring chars in the comment (and looking for an end of comment) or not (and looking eventually for a start of comment).
It uses *s1 and s1[1] to access the current and the next char.
It leaves the original string unchanged.
Online demo
I know I need a loop
Could use a loop, or use standard library functions. Consider char *strstr(const char *s1, const char *s2); as a candidate part of a solution.
The strstr function locates the first occurrence in the string pointed to by s1 of the sequence of characters (excluding the terminating null character) in the string pointed to
by s2.
The strstr function returns a pointer to the located string, or a null pointer if the string is not found.
Some untested code to give you an idea.
void remove_comment(const char *src, char *dest){
char *start = strstr(str, "/*"); // Look for beginning
if (start) { // Find the beginning?
char *end = strstr(start + 2, "*/"); // Now search 2 past
if (end) { // Find the end?
memcpy(dest, src, start - src);
strcpy(&dest[start - src], end+2);
return;
}
}
strcpy(dest, src);
}
If you want to avoid library funcitons, I'll pass along a hint
// if(*s1 == '/'){
if(s1[0] == '/' && s1[1] == '*') {
Of course this is insufficient for finding /* */ comments in C code like:
puts("/* this is not a C comment, but a string literal */");
int a = '/*', b = '*/';
// start of comment /* xyz */
As you can see below I have created a little program to concatenate 2 strings using C, as you may imagine this code doesn't work, I have already corrected it myself by using Array notation instead of pointers, and it works just fine, however I'm still not sure why is it that my code fails being almost a replica of my corrected code.
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
void concatena(char *str1, char *str2){
char *strAux;
int mover;
mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
*(strAux) = '\0';
if(str1 == '\0')
*strAux = '\0';
else
while(str1 != '\0'){
*(strAux+mover++)=*(str1++);
}
if(str2 == '\0')
*strAux = '\0';
else
while(str2 != '\0'){
*(strAux+mover++)=*(str2++);
}
strAux='\0';
str1=strAux;
printf("%s", str1);
free(strAux);
}
I´m still a C beginner (And yes, I'm aware that there are libraries like string.h, I'm asking this for academic reasons) and I have been told that char pointers and arrays are the same thing, something that confuses the heck out of me.
Any help is greatly appreciated.
The first problem I see is with this section:
if(str2 == '\0')
*strAux = '\0';
Just before this code, you've filled up strAux with the string from str1.
Then, if str2 is empty, you suddenly put a null-terminator at the beginning of strAux, eliminating all the work you've done so far!
I think what you intend is:
if(*str2 == '\0')
*(strAux+mover) = '\0';
Its the same thing again after your loop for str2, you have the code:
strAux='\0';
Again, this puts a null-terminator at the start of strAux, effectively ending the newly created string before it even gets started.
Here's how I'd re-write your code:
void concatena(char *str1, char *str2){
char *strAux;
int mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+1)); // Changed to +1, NOT +2
*(strAux) = '\0'; // Start the string as (empty)
while(*str1 != '\0'){ // Copy the first string over.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // Copy the second string over.
*(strAux+mover++)=*(str2++);
}
*(strAux+mover)='\0'; // End the new, combined string.
printf("%s", strAux); // Show the results.
free(strAux);
}
Accepting the same constraints, here is how I would (re)write your code. Unfortunately there is a specification shortcoming: should the concatenation occur to the first string passed? Or should a new string be created? Here are both methods:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
char *concatena (const char *str1, const char *str2)
{
char *op, *newStr = (char*)malloc (strlen (str1) + strlen (str2) + 1);
if (!newStr)
{
fprintf (stderr, "concatena: error allocating\n");
return;
}
op = newStr; // set up output pointer
while (str1 && *str1) // copy first string
*op++ = *str1++;
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
return newStr;
}
void concatenb (char *str1, const char *str2)
{
char *op;
if (!str1)
{
fprintf (stderr, "concatenb: NULL string 1\n");
return;
}
op = &str1 [strlen (str1)]; // set output pointer at trailing NUL
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
}
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
2 is not required, just 1 is sufficient for the termination character.
*(strAux) = '\0';
This should be happening only at the end of all your computation. Not in between the concatenation i.e.,
while(*str1 != '\0'){ // This loops copies the first string
// ^ Notice that you need to dereference to check for the termination character.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // This loop copies the second string
*(strAux+mover++)=*(str2++);
}
// Finally adding termination character
*(strAux+mover) = '\0'; // since with mover you are keeping track of locations.
The amount of errors in your code is disheartening. You should probably pick up a good C book and start over.
First off, there's a library function that you can use to concatenate strings:
const unsigned int len = strlen(str1) + strlen(str2) + 1;
char * dst = malloc(len);
strncat(dst, str1, len);
strncat(dst, str2, len);
Now, if you insist on doing it manually, you have to get pointers and dereferencing right:
char * d = dst;
while (*str1 != 0) *dst++ = *str1++;
while (*str2 != 0) *dst++ = *str2++;
*dst = 0;
// d now points to the beginning of the concatenated string
The two loops check if the current character in the input string is nonzero, and if so, then they copy that character to the current character in the output string, and then both input and output pointer are advanced. (This is all done in one wash by use of the postfix ++ operator.) Finally, the last character is set to zero to create a new null-terminator.
In the process we modified all three pointers dst, str1 and str2. The latter two came in as input function arguments by copy, so that's fine. For returning the concatenated string we made a copy of dst before the loop, which we can return in the end.
Please explain to me the working of strtok() function. The manual says it breaks the string into tokens. I am unable to understand from the manual what it actually does.
I added watches on str and *pch to check its working when the first while loop occurred, the contents of str were only "this". How did the output shown below printed on the screen?
/* strtok example */
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] ="- This, a sample string.";
char * pch;
printf ("Splitting string \"%s\" into tokens:\n",str);
pch = strtok (str," ,.-");
while (pch != NULL)
{
printf ("%s\n",pch);
pch = strtok (NULL, " ,.-");
}
return 0;
}
Output:
Splitting string "- This, a sample string." into tokens:
This
a
sample
string
the strtok runtime function works like this
the first time you call strtok you provide a string that you want to tokenize
char s[] = "this is a string";
in the above string space seems to be a good delimiter between words so lets use that:
char* p = strtok(s, " ");
what happens now is that 's' is searched until the space character is found, the first token is returned ('this') and p points to that token (string)
in order to get next token and to continue with the same string NULL is passed as first
argument since strtok maintains a static pointer to your previous passed string:
p = strtok(NULL," ");
p now points to 'is'
and so on until no more spaces can be found, then the last string is returned as the last token 'string'.
more conveniently you could write it like this instead to print out all tokens:
for (char *p = strtok(s," "); p != NULL; p = strtok(NULL, " "))
{
puts(p);
}
EDIT:
If you want to store the returned values from strtok you need to copy the token to another buffer e.g. strdup(p); since the original string (pointed to by the static pointer inside strtok) is modified between iterations in order to return the token.
strtok() divides the string into tokens. i.e. starting from any one of the delimiter to next one would be your one token. In your case, the starting token will be from "-" and end with next space " ". Then next token will start from " " and end with ",". Here you get "This" as output. Similarly the rest of the string gets split into tokens from space to space and finally ending the last token on "."
strtok maintains a static, internal reference pointing to the next available token in the string; if you pass it a NULL pointer, it will work from that internal reference.
This is the reason strtok isn't re-entrant; as soon as you pass it a new pointer, that old internal reference gets clobbered.
strtok doesn't change the parameter itself (str). It stores that pointer (in a local static variable). It can then change what that parameter points to in subsequent calls without having the parameter passed back. (And it can advance that pointer it has kept however it needs to perform its operations.)
From the POSIX strtok page:
This function uses static storage to keep track of the current string position between calls.
There is a thread-safe variant (strtok_r) that doesn't do this type of magic.
strtok will tokenize a string i.e. convert it into a series of substrings.
It does that by searching for delimiters that separate these tokens (or substrings). And you specify the delimiters. In your case, you want ' ' or ',' or '.' or '-' to be the delimiter.
The programming model to extract these tokens is that you hand strtok your main string and the set of delimiters. Then you call it repeatedly, and each time strtok will return the next token it finds. Till it reaches the end of the main string, when it returns a null. Another rule is that you pass the string in only the first time, and NULL for the subsequent times. This is a way to tell strtok if you are starting a new session of tokenizing with a new string, or you are retrieving tokens from a previous tokenizing session. Note that strtok remembers its state for the tokenizing session. And for this reason it is not reentrant or thread safe (you should be using strtok_r instead). Another thing to know is that it actually modifies the original string. It writes '\0' for teh delimiters that it finds.
One way to invoke strtok, succintly, is as follows:
char str[] = "this, is the string - I want to parse";
char delim[] = " ,-";
char* token;
for (token = strtok(str, delim); token; token = strtok(NULL, delim))
{
printf("token=%s\n", token);
}
Result:
this
is
the
string
I
want
to
parse
The first time you call it, you provide the string to tokenize to strtok. And then, to get the following tokens, you just give NULL to that function, as long as it returns a non NULL pointer.
The strtok function records the string you first provided when you call it. (Which is really dangerous for multi-thread applications)
strtok modifies its input string. It places null characters ('\0') in it so that it will return bits of the original string as tokens. In fact strtok does not allocate memory. You may understand it better if you draw the string as a sequence of boxes.
To understand how strtok() works, one first need to know what a static variable is. This link explains it quite well....
The key to the operation of strtok() is preserving the location of the last seperator between seccessive calls (that's why strtok() continues to parse the very original string that is passed to it when it is invoked with a null pointer in successive calls)..
Have a look at my own strtok() implementation, called zStrtok(), which has a sligtly different functionality than the one provided by strtok()
char *zStrtok(char *str, const char *delim) {
static char *static_str=0; /* var to store last address */
int index=0, strlength=0; /* integers for indexes */
int found = 0; /* check if delim is found */
/* delimiter cannot be NULL
* if no more char left, return NULL as well
*/
if (delim==0 || (str == 0 && static_str == 0))
return 0;
if (str == 0)
str = static_str;
/* get length of string */
while(str[strlength])
strlength++;
/* find the first occurance of delim */
for (index=0;index<strlength;index++)
if (str[index]==delim[0]) {
found=1;
break;
}
/* if delim is not contained in str, return str */
if (!found) {
static_str = 0;
return str;
}
/* check for consecutive delimiters
*if first char is delim, return delim
*/
if (str[0]==delim[0]) {
static_str = (str + 1);
return (char *)delim;
}
/* terminate the string
* this assignmetn requires char[], so str has to
* be char[] rather than *char
*/
str[index] = '\0';
/* save the rest of the string */
if ((str + index + 1)!=0)
static_str = (str + index + 1);
else
static_str = 0;
return str;
}
And here is an example usage
Example Usage
char str[] = "A,B,,,C";
printf("1 %s\n",zStrtok(s,","));
printf("2 %s\n",zStrtok(NULL,","));
printf("3 %s\n",zStrtok(NULL,","));
printf("4 %s\n",zStrtok(NULL,","));
printf("5 %s\n",zStrtok(NULL,","));
printf("6 %s\n",zStrtok(NULL,","));
Example Output
1 A
2 B
3 ,
4 ,
5 C
6 (null)
The code is from a string processing library I maintain on Github, called zString. Have a look at the code, or even contribute :)
https://github.com/fnoyanisi/zString
This is how i implemented strtok, Not that great but after working 2 hr on it finally got it worked. It does support multiple delimiters.
#include "stdafx.h"
#include <iostream>
using namespace std;
char* mystrtok(char str[],char filter[])
{
if(filter == NULL) {
return str;
}
static char *ptr = str;
static int flag = 0;
if(flag == 1) {
return NULL;
}
char* ptrReturn = ptr;
for(int j = 0; ptr != '\0'; j++) {
for(int i=0 ; filter[i] != '\0' ; i++) {
if(ptr[j] == '\0') {
flag = 1;
return ptrReturn;
}
if( ptr[j] == filter[i]) {
ptr[j] = '\0';
ptr+=j+1;
return ptrReturn;
}
}
}
return NULL;
}
int _tmain(int argc, _TCHAR* argv[])
{
char str[200] = "This,is my,string.test";
char *ppt = mystrtok(str,", .");
while(ppt != NULL ) {
cout<< ppt << endl;
ppt = mystrtok(NULL,", .");
}
return 0;
}
For those who are still having hard time understanding this strtok() function, take a look at this pythontutor example, it is a great tool to visualize your C (or C++, Python ...) code.
In case the link got broken, paste in:
#include <stdio.h>
#include <string.h>
int main()
{
char s[] = "Hello, my name is? Matthew! Hey.";
char* p;
for (char *p = strtok(s," ,?!."); p != NULL; p = strtok(NULL, " ,?!.")) {
puts(p);
}
return 0;
}
Credits go to Anders K.
Here is my implementation which uses hash table for the delimiter, which means it O(n) instead of O(n^2) (here is a link to the code):
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define DICT_LEN 256
int *create_delim_dict(char *delim)
{
int *d = (int*)malloc(sizeof(int)*DICT_LEN);
memset((void*)d, 0, sizeof(int)*DICT_LEN);
int i;
for(i=0; i< strlen(delim); i++) {
d[delim[i]] = 1;
}
return d;
}
char *my_strtok(char *str, char *delim)
{
static char *last, *to_free;
int *deli_dict = create_delim_dict(delim);
if(!deli_dict) {
/*this check if we allocate and fail the second time with entering this function */
if(to_free) {
free(to_free);
}
return NULL;
}
if(str) {
last = (char*)malloc(strlen(str)+1);
if(!last) {
free(deli_dict);
return NULL;
}
to_free = last;
strcpy(last, str);
}
while(deli_dict[*last] && *last != '\0') {
last++;
}
str = last;
if(*last == '\0') {
free(deli_dict);
free(to_free);
deli_dict = NULL;
to_free = NULL;
return NULL;
}
while (*last != '\0' && !deli_dict[*last]) {
last++;
}
*last = '\0';
last++;
free(deli_dict);
return str;
}
int main()
{
char * str = "- This, a sample string.";
char *del = " ,.-";
char *s = my_strtok(str, del);
while(s) {
printf("%s\n", s);
s = my_strtok(NULL, del);
}
return 0;
}
strtok() stores the pointer in static variable where did you last time left off , so on its 2nd call , when we pass the null , strtok() gets the pointer from the static variable .
If you provide the same string name , it again starts from beginning.
Moreover strtok() is destructive i.e. it make changes to the orignal string. so make sure you always have a copy of orignal one.
One more problem of using strtok() is that as it stores the address in static variables , in multithreaded programming calling strtok() more than once will cause an error. For this use strtok_r().
strtok replaces the characters in the second argument with a NULL and a NULL character is also the end of a string.
http://www.cplusplus.com/reference/clibrary/cstring/strtok/
you can scan the char array looking for the token if you found it just print new line else print the char.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
int len = strlen(s);
char delim =' ';
for(int i = 0; i < len; i++) {
if(s[i] == delim) {
printf("\n");
}
else {
printf("%c", s[i]);
}
}
free(s);
return 0;
}
So, this is a code snippet to help better understand this topic.
Printing Tokens
Task: Given a sentence, s, print each word of the sentence in a new line.
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
//logic to print the tokens of the sentence.
for (char *p = strtok(s," "); p != NULL; p = strtok(NULL, " "))
{
printf("%s\n",p);
}
Input: How is that
Result:
How
is
that
Explanation: So here, "strtok()" function is used and it's iterated using for loop to print the tokens in separate lines.
The function will take parameters as 'string' and 'break-point' and break the string at those break-points and form tokens. Now, those tokens are stored in 'p' and are used further for printing.
strtok is replacing delimiter with'\0' NULL character in given string
CODE
#include<iostream>
#include<cstring>
int main()
{
char s[]="30/4/2021";
std::cout<<(void*)s<<"\n"; // 0x70fdf0
char *p1=(char*)0x70fdf0;
std::cout<<p1<<"\n";
char *p2=strtok(s,"/");
std::cout<<(void*)p2<<"\n";
std::cout<<p2<<"\n";
char *p3=(char*)0x70fdf0;
std::cout<<p3<<"\n";
for(int i=0;i<=9;i++)
{
std::cout<<*p1;
p1++;
}
}
OUTPUT
0x70fdf0 // 1. address of string s
30/4/2021 // 2. print string s through ptr p1
0x70fdf0 // 3. this address is return by strtok to ptr p2
30 // 4. print string which pointed by p2
30 // 5. again assign address of string s to ptr p3 try to print string
30 4/2021 // 6. print characters of string s one by one using loop
Before tokenizing the string
I assigned address of string s to some ptr(p1) and try to print string through that ptr and whole string is printed.
after tokenized
strtok return the address of string s to ptr(p2) but when I try to print string through ptr it only print "30" it did not print whole string. so it's sure that strtok is not just returning adress but it is placing '\0' character where delimiter is present.
cross check
1.
again I assign the address of string s to some ptr (p3) and try to print string it prints "30" as while tokenizing the string is updated with '\0' at delimiter.
2.
see printing string s character by character via loop the 1st delimiter is replaced by '\0' so it is printing blank space rather than ''
What is the easiest and most efficient way to remove spaces from a string in C?
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++);
}
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 :)
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.
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 ;)
#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.
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);
}
#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));
}
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 (e.g. stdin) and is bound to be written to another file (e.g. 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.
/* 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';
}
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
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';
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;
}
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;
}
}
}
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;
}
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.
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 :-)