Removing spaces from a string - c

Question

I am trying to copy from one string to another, but the second string should omit the space. I tried to approach this by saying, if a character in the original string is a space, do not copy it. Instead, copy the next character. However, the program deletes everything after the space. Any ideas?
char deleteSpaces(char phrase[256], int length){
int j, i=0;
char phrase2[length];
for(j=0;j<length;j++){
if(phrase[i]==' '){
phrase2[j]=phrase[i+1];
i++;
}
phrase2[j]=phrase[i];
}
return phrase2;
}

Answer 1

Here is a solution:
void deleteSpaces(char src[], char dst[]){
// src is supposed to be zero ended
// dst is supposed to be large enough to hold src
int s, d=0;
for (s=0; src[s] != 0; s++)
if (src[s] != ' ') {
dst[d] = src[s];
d++;
}
dst[d] = 0;
}

Answer 2

First, you're returning a static tab and this is wrong.
Secondly you don't increment 'i' if there is no space.
And to finish you will copy spaces if there is more than one space in a row. And you do not control if you reach the end of your source.
for(j = 0; j < length; j++)
{
while (src[i] == ' ' && i < length) i++;
if (i < length)
dest[j] = src[i++];
else
{
dest[j] = 0;
break;
}
}

Answer 3

My solution. Arguments and their order are chosen to match strncpy().
#include <ctype.h>
char *strip_whitespace(char *dest, const char *src, size_t n)
{
char *s, *d;
/*
* Copy 'src' to 'dest', omitting whitespace and making sure we don't
* overflow 'dest'.
*/
for(s=src, d=dest; *s && (d-dest)<n; s++) {
if( !isspace(*s) ) {
*d = *s;
d++;
}
}
/* Ensure that dest is NUL terminated in any event */
if( d-dest < n ) {
*d = '\0';
} else {
dest[n-1] = '\0';
}
return dest;
}

Answer 4

As pointed out already you need to allocate a new string and return a pointer to it. This code works:
char* strip (char* input)
{
int loop;
char *output = (char*) malloc (strlen(input));
char *dest = output;
if (output)
{
for (loop=0; loop<strlen(input); loop++)
if (input[loop] != ' ')
*dest++ = input[loop];
*dest = '\0';
}
return output;
}
int main (void)
{
char srcString[] = " this is a test with spaces at the end ";
char* dstString = strip (srcString);
printf ("source string = '%s'\n", srcString);
printf (" dest string = '%s'\n", dstString ? dstString : "malloc failed in strip");
free (dstString);
return 0;
}
Output:
source string = ' this is a test with spaces at the end '
dest string = 'thisisatestwithspacesattheend'
It takes the input string and allocates a destination string the same size which is safe, although wasteful of a few bytes.
The method is simple; only copy a character if it is not a space. After all the characters are copied, I write the terminator on the end and return the pointer.

Answer 5

****WHAT YOU WERE DOING** IS THAT if a character in the original string is a space, do not copy it. Instead, copy the next character. IS NOT A GOOD idea because
Case 1. When multiple spaces are present, it simply discard the first spaces and copies the second one...
Case 2: Your programme copies the string only after space is found,,it is simply skipping the first word of string which doesn't start with spaces.
case 3. You are only returning the character pointed by phrase2[0] as return type is char,and the scope of local variable is limited to only that function....
//the corrected programme
int deleteSpaces(char phrase[256],charphrase2[256])
{
int i,j;
i=j=0;
while(phrase[i]!=NULL){
if(phrase[i]!=' '){
phrase2[j]=phrase[i];
j++;
}
i++;
}//end while
phrase2[j]=phrase[i] //nulcharceter copied
return 0;
}//end deleteSpaces function

Answer 6

I use this:
char input[100], foreval[100];
for(i=0;i<strlen(input);i++){
if(isspace(input[i])){
continue;
}
strncat(foreval,&input[i],1);
}