I am working with a function that accept full name as input and print them out in order: first name, middle name, last name.
The problem is my while loop seem to go infinite and i haven't figured out how to fix it.
Here is my function:
void order(char ar[])
{
int i, count=0, a=1;
char* token;
char last[20], middle[20], first[20],c;
char s[]=" ";
for(i=0;i<strlen(ar)-1;i++)
if(ar[i]==' ') count++;
token=strtok(ar,s);
strcpy(last,token);
while(token!=NULL)
{
token=strtok(NULL,s);
if(a<count) strcpy(middle,token);
else strcpy(first,token);
a++;
}
printf("%s %s %s",first,middle,last);
}
When first is copied (i.e. all tokens identified), while loop will be entered once again(note that token is not NULL, but points to first in this case). Now strtok() will be called and it will return NULL. Code will go to the else part and try to strcpy(), which is causing the problem.
Check for NULL before strcpy(). I have done this like this and it works fine.
while(token!=NULL)
{
token=strtok(NULL,s);
if (token == NULL)
break;
if(a<count) strcpy(middle,token);
else strcpy(first,token);
a++;
}
This is more like how I'd present the code in the question — before I passed it through a compiler:
void order(char ar[])
{
int count = 0, a = 1;
char *token;
char last[20], middle[20], first[20];
char s[] = " ";
for (int i = 0; i < strlen(ar) - 1; i++)
{
if (ar[i] == ' ')
count++;
}
token = strtok(ar, s);
strcpy(last, token);
while (token != NULL)
{
token = strtok(NULL, s);
if (a < count)
strcpy(middle, token);
else
strcpy(first, token);
a++;
}
printf("%s %s %s", first, middle, last);
}
You can argue about brace placement — I use Allman but 1TBS is a widely used alternative style. I recommend choosing one of those two. I put braces around the body of a loop if it is more than one line; YMMV.
Use more white space around operators. Don't define unused variables (c vanished). I placed int i in the for loop. I would probably define one variable per line, and I probably wouldn't use the count or a.
The call to strlen() should not be in the condition of the loop. I'm not sure how you'd cope with my name — I don't have a middle name. I think you could avoid the while loop altogether. I'm not convinced you need the for loop either. Those are a separate discussion.
I'm not attempting to fix the algorithmic problem — this is not an answer to the question in the question (so it shouldn't be accepted), but is an answer to the question in the comment, and since code cannot be formatted in comments, I can't answer it sanely in a comment; hence, this 'not an answer' but 'intended to be helpful to an auxilliary question from the OP'.
Related
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 simple Shell for school assignment and stuck with a segmentation problem. Initially, my shell parses the user input to remove whitespaces and endofline character, and seperate the words inside the input line to store them in a char **args array. I can seperate the words and can print them without any problem, but when storing the words into a char **args array, and if argument number is greater than 1 and is odd, I get a segmentation error.
I know the problem is absurd, but I stuck with it. Please help me.
This is my parser code and the problem occurs in it:
char **parseInput(char *input){
int idx = 0;
char **parsed = NULL;
int parsed_idx = 0;
while(input[idx]){
if(input[idx] == '\n'){
break;
}
else if(input[idx] == ' '){
idx++;
}
else{
char *word = (char*) malloc(sizeof(char*));
int widx = 0; // Word index
word[widx] = input[idx];
idx++;
widx++;
while(input[idx] && input[idx] != '\n' && input[idx] != ' '){
word = (char*)realloc(word, (widx+1)*sizeof(char*));
word[widx] = input[idx];
idx++;
widx++;
}
word = (char*)realloc(word, (widx+1)*sizeof(char*));
word[widx] = '\0';
printf("Word[%d] --> %s\n", parsed_idx, word);
if(parsed == NULL){
parsed = (char**) malloc(sizeof(char**));
parsed[parsed_idx] = word;
parsed_idx++;
}else{
parsed = (char**) realloc(parsed, (parsed_idx+1)*sizeof(char**));
parsed[parsed_idx] = word;
parsed_idx++;
}
}
}
int i = 0;
while(parsed[i] != NULL){
printf("Parsed[%d] --> %s\n", i, parsed[i]);
i++;
}
return parsed;
}
In your code you have the loop
while(parsed[i] != NULL) { ... }
The problem is that the code never sets any elements of parsed to be a NULL pointer.
That means the loop will go out of bounds, and you will have undefined behavior.
You need to explicitly set the last element of parsed to be a NULL pointer after you parsed the input:
while(input[idx]){
// ...
}
parsed[parsed_idx] = NULL;
On another couple of notes:
Don't assign back to the same pointer you pass to realloc. If realloc fails it will return a NULL pointer, but not free the old memory. If you assign back to the pointer you will loose it and have a memory leak. You also need to be able to handle this case where realloc fails.
A loop like
int i = 0;
while (parsed[i] != NULL)
{
// ...
i++;
}
is almost exactly the same as
for (int i = 0; parsed[i] != NULL; i++)
{
// ...
}
Please use a for loop instead, it's usually easier to read and follow. Also for a for loop the "index" variable (i in your code) will be in a separate scope, and not available outside of the loop. Tighter scope for variables leads to less possible problems.
In C you shouldn't really cast the result of malloc (or realloc) (or really any function returning void *). If you forget to #include <stdlib.h> it could lead to hard to diagnose problems.
Also, a beginner might find the -pedantic switch helpful on your call to the compiler. That switch would have pointed up most of the other suggestions made here. I personally am also a fan of -Wall, though many find it annoying instead of helpful.
I have written code for parsing a string into words. Here is code. Can any one help here to fix the segmentation fault error during run time?
Calling fun :
int main()
{
int count = 0, i; // count to hold numbr of words in the string line.
char buf[MAX_LENTHS]; // buffer to hold the string
char *options[MAX_ORGS]; // options to hold the words that we got after parsing.
printf("enter string");
scanf("%s",buf);
count = parser(buf,options); // calling parser
for(i = 0; i < count; ++i)
printf("option %d is %s", i, options[i]);
return 0;
}
Called function:
int parser(char str[], char *orgs[])
{
char temp[1000];//(char *)malloc(strlen(str)*sizeof(char));
int list = 0;
strcpy(temp, str);
*orgs[list]=strtok(str, " \t ");
while(((*orgs[list++]=strtok(str," \t"))!=NULL)&&MAX_ORGS>list)
list++;
printf("count =%d",list);
return list;
}
Note : I'm trying to learn C these days, can any one help to get a good tutorial (pdf) or site to learn these strings with pointers, and sending string to functions as arguments?
You are using strtok wrong.
(It is generally best to not use strtok at all, for all its problems and pitfalls.)
If you must use it, the proper way to use strtok is to call it ONCE with the string you want to "tokenize",
then call it again and again with NULL as an indication to continue parsing the original string.
I also think you're using the orgs array wrong.
Change this assignment
*orgs[list++]=strtok(str, " \t ");
to this:
orgs[list++]=strtok(str, " \t ");
Because orgs is an array of character-pointers.
orgs[x] is a character-pointer, which matches the return-type of strtok
Instead, you are referring to *orgs[x], which is just a character.
So you are trying to do:
[character] = [character-pointer];
which will result in "very-bad-things™".
Finally, note that you are incrementing list twice each time through your loop.
So basically you're only filling in the even-elements, leaving the odd-elements of orgs uninitialized.
Only increment list once per loop.
Basically, you want this:
orgs[list++] = strtok(str, " \t ");
while(( (orgs[list++] = strtok(NULL," \t")) !=NULL) && MAX_ORGS > list)
/* do nothing */;
PS You allocate space for temp, and strcpy into it.
But then it looks like you never use it. Explain what temp is for, or remove it.
char buf[MAX_LENTHS];
You have not defined the array size, i. e. MAX_LENTHS should be defined like
#define MAX_LENTHS 25
And as Paul R says in his comment you also need to initialize your array of character pointers
char *options[MAX_ORGS];
with something .
int parser(char str[], char *orgs[]){
int list=0;
orgs[list]=strtok(str, " \t\n");
while(orgs[list]!=NULL && ++list < MAX_ORGS)
orgs[list]=strtok(NULL," \t\n");
printf("count = %d\n",list);
return list;
}
int main(){
int count=0,i;
char buf[MAX_LENTHS];
char *options[MAX_ORGS];
printf("enter string: ");
fgets(buf, sizeof(buf), stdin);//input include space character
count=parser(buf,options);
for(i=0;i<count;++i)
printf("option %d is %s\n",i,options[i]);
return 0;
}
I'm having a problem understanding how to read in a string to a structure member array. I have a structure called 'customer' and a member called 'char last_name[20]'. I prompt the user to enter in his last name and that last name is to be stored in the 'last_name[20]' variable. The condition is that I have to use a do...while loop.
Here's the code:
void get_customer_info(struct customer *p_customer_start, int customer_num)
{
struct customer *p_customer;
for (p_customer = p_customer_start; (p_customer - p_customer_start) <
customer_num; p_customer++)
{
printf("\nCustomer number %d: ", (p_customer - p_customer_start) + 1);
while (getchar() != NEW_LINE);
printf("\n Enter the customer's last name: ");
// *THIS PART IS THE PROBLEM*
do
{
p_customer->last_name = getchar();
p_customer->last_name++;
} while (*p_customer->last_name != NEW_LINE);
}
return;
}
Problem is, with that algorithm last_name[0] does not get checked, it moves to 'last_name[1]' before it gets checked for a new line. And yes, a do...while construct must be used (this is for a class).
I appreciate anyone's thoughts.
I think you may have much bigger problems there than you realise, with the attempted manipulation of an array address :-)
You can probably avoid all these problems with:
int i = 0;
do {
p_customer->last_name[i++] = getchar();
} while (p_customer->last_name[i-1] != NEW_LINE);
p_customer->last_name[i] = '\0';
Keep in mind that this is still open to buffer overflow problems (as was your original solution) since entering a name like "Pasquale Voldemort Fortescue del Mor" is going to blow out your 20-character array.
There are ways to fix that but it probably doesn't matter for classwork that much (it will in the real world but that comes with experience):
int i = 0;
do {
p_customer->last_name[i] = getchar();
if (i < sizeof (p_customer->last_name) - 1) // NEVER got to 20
i++;
} while (p_customer->last_name[i-1] != NEW_LINE);
p_customer->last_name[i] = '\0';
If you really want a pointer version of it, that's easy:
char *p = p_customer->last_name;
do {
*p = getchar();
if (p != &(p_customer->last_name[sizeof(p_customer->last_name) - 1])
p++;
} while (*(p-1) != NEW_LINE);
*p = '\0';
Ok, the solution with NO INDEXES would be;
char *pointer;
/* other code here */
pointer = p_customer->last_name
do {
*pointer = getchar();
pointer += sizeof( char );
} while ( *(pointer - sizeof( char ) )!= NEW_LINE );
if you want to be sure not to go out of array bound.......simply do not use do while =) (which I always reccomend, as you must read lots of line before understanding the loop condition (and in a multinested function this is't a real readibility problem
The code you posted won't compile. You mentioned that last_name is an array, but you can't assign to or increment arrays, so these two lines are invalid:
p_customer->last_name = getchar();
p_customer->last_name++;
Try doing something like this, which will get rid of this error as well as solve the problem of checking if the first character is a newline:
int i = 0;
char ch;
while ((ch = getchar()) != NEW_LINE)
p_customer->last_name[i++] = ch;
p_customer->last_name[i] = '\0'; // Don't forget the null terminator
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 :-)