How to Replace Leading or Trailing Blank Characters with "X" - c

Question

Looking for a more efficient way to replace leading and trailing empty spaces (' ') and appending an 'X' to the front for each empty space.. It seems to work ok for trailing spaces but I'd like to know if there's a better / simpler way of going about this that I am missing.
Example:
Passed in string: '12345 '
Desired result 'XXXXX12345'
Removed 5 empty spaces and append 5 'X's to front.
Example:
Passed in string: ' 12345'
Desired result 'XX12345'
Remove 2 empty spaces and append 2 'X's to front.
void fixStr(char* str)
{
int i = 0;
int length = strlen(str);
char strCopy[10];
strcpy(strCpy, str);
for(i = 0; i < length; i++)
{
if(strCopy[i] == ' ')
{
strCopy[i] = '\0';
str[i] = '\0';
break;
}
}
for(i = 0; i < length - i + 2; i++)
{
str[i] = 'X';
str[i + 1] = '\0';
}
strcat(str, strCopy);
}

Answer 1

One way to achieve this is to find out the leading non-space position & trailing non-space position of the string, and then move the content in-between (leading nonspace, trailing nonspace) this to end of the string, then set all the empty space at the beginning to 'x'
This way you can get the expected output (function below)
void fixStr(char* str)
{
int i = 0;
int length = strlen(str);
int leadindex = length;
int tailindex = 0;
// First find the leading nonspace position
for(i = 0; i < length; i++)
{
if(str[i] != ' ')
{
leadindex = i;
break;
}
}
// if not found nonspace then no change
if( leadindex == length )
{
// all spaces, so no change required;
return;
}
// Find the trailing nonspace position
for(i = length - 1; i >= 0 ; i--)
{
if(str[i] != ' ')
{
tailindex = i;
break;
}
}
// move the buffer (in place) to exclude trailing spaces
memmove(str + (length - tailindex -1),str,(tailindex +1) );
// set the 'x' to all empty spaces at leading ( you may use for loop to set this)
memset(str, 'X', length - (tailindex - leadindex + 1) );
}

Answer 2

To solve a problem the engineer's way:
Define the needs.
Know your tools.
Use the tools as simple as possible, as accurate as necessary to make up a solution.
In your case:
Needs:
find the number of trailing spaces
move content of string to the end
set beginning to 'X's
Tools:
to measure, iterate, compare and count
to move a block of memory
to initialise a block of memory
Example for a solution:
#include <string.h> /* for strlen(), memmove () and memset() */
void fix_str(char * s)
{
if ((NULL != s) && ('\0' != *s)) /* Ignore NULL and empty string! */
{
/* Store length and initialise counter: */
size_t l = strlen(s), i = l;
/* Count space(s): */
for (; (0 != i) && (' ' == s[i-1]); --i); /* This for loop does not need a "body". */
/* Calculate the complement: */
size_t c = l - i;
/* Move content to the end overwriting any trailing space(s) counted before hand: */
memmove(s+c, s, i); /* Note that using memmove() instead of memmcpy() is essential
here as the source and destination memory overlap! */
/* Initialise the new "free" characters at the beginning to 'X's:*/
memset(s, 'X', c);
}
}

Answer 3

I didn't fix your code but you could use sprintf in combination with isspace, something along the lines of this. Also, remember to make a space for the '\0 at the end of your string. Use this idea and it should help you:
#include <ctype.h>
#include <stdio.h>
int main()
{
char buf[11];
char *s = "Hello";
int i;
sprintf(buf, "%10s", s); /* right justifies in a column of 10 in buf */
for(i = 0; i < 10; i++) {
if(isspace(buf[i])) /* replace the spaces with an x (or whatever) */
buf[i] = 'x';
}
printf("%s\n", buf);
return 0;
}