Well I'm looking for a function that reduce multiple space characters ' ' in a string.
For example for string s given :
s="hello__________world____!"
The function must return "hello_world_!"
In python we can do it via regexp simply as:
re.sub("\s+", " ", s);
A version that modifies the string in place, run it on a copy if the original must be preserved:
void compress_spaces(char *str)
{
char *dst = str;
for (; *str; ++str) {
*dst++ = *str;
if (isspace(*str)) {
do ++str;
while (isspace(*str));
--str;
}
}
*dst = 0;
}
++str. Is it the same as str++ in this situation?There is no such function in the C standard library. One must write a function to do so or use a third-party library.
The following function should do the trick. Use the source string as the destination pointer to perform the operation in place. Otherwise, ensure that the destination buffer is sufficiently sized.
void
simplifyWhitespace(char * dst, const char * src)
{
for (; *src; ++dst, ++src) {
*dst = *src;
if (isspace(*src))
while (isspace(*(src + 1)))
++src;
}
*dst = '\0';
}
void remove_more_than_one_space(char *dest, char *src)
{
int i, y;
assert(dest && src);
for(i=0, y=0; src[i] != '\0'; i++, y++) {
if(src[i] == ' ' && src[i+1] == ' ') {
/* let's skip this copy and reduce the y index*/
y--;
continue;
}
/* copy normally */
dest[y] = src[i];
}
dest[y] = '\0';
}
int main()
{
char src[] = "Hello World ! !! !";
char dest[strlen(src) + 1];
remove_more_than_one_space(dest, src);
printf("%s\n", dest);
}
I just made this, hope it helps.
strlen(src) then why does your code go char dest[strlen(src) + 1];? ;)
#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main()
{
char word[100];
gets(word);
//the word has more than a single space in between the words
int i=0,l,j;
l=strlen(word);
for (i=0;i<l;i++)
{
if(word[i]==' '&&word[i+1]==' ')
{
for(j=i+1;j<l;j++)
word[j]=word[j+1];
}
}
puts(word);
return 0;
}
This code is very simple and it worked like a charm for me. I don't know if this code will have some other problems I haven't come across, but for now this works.
I'm just learning C, so I'm using much more basic code. I'm reading the first chapter of "The C programming language", and I was trying to find the answer to a task set in there.
This is what I came up with:
#include <stdio.h>
int main()
{
/* Set two integers:
c is the character being assessed,
lastspace is 1 if the lastcharacter was a space*/
int c, lastspace;
lastspace = 0;
/* This while loop will exit if the character is EOF
The first "If block" is true if the character is not a space,
and just prints the character
It also tells us that the lastcharacter was not a space
The else block will run if the character is a space
Then the second IF block will run if the last character
was not also a space (and will print just one space) */
while((c = getchar()) != EOF){
if (c != ' '){
putchar(c);
lastspace = 0;
}
else {
if (lastspace != 1)
putchar(c);
lastspace = 1;
}
}
return 0;
}
Hope that helps! Also, I am well aware that this code is perhaps not optimised, but it should be simple for a beginner like me to understand!
Thanks, Phil
another way of doing this to print only the first occurrence of space until next character comes, here is my brute force solution.
#include<stdio.h>
typedef int bool;
#define True 1
#define False 0
int main()
{
int t;
bool flag = False;
while ((t = getchar()) != EOF)
if (t == ' ' && !flag)
{
putchar(' ');
flag = True; // flag is true for the first occurence of space
}
else if(t == ' '&& flag)
continue;
else
{
putchar(t);
flag = False;
}
return 0;
}
hope it helps.
