remove leading spaces from string in c - c

Question

I am reading in a string character by character then for each word found(separated by space) counting the length of each word, and finally printing all that information to the screen.
Sample run: trickier to master
n=8, s=[trickier]
n=2, s=[to]
n=6, s=[master]
n=0, s=[]
n=-1, s=[]
This is correct what I get is this:
n=0, s=[]
n=0, s=[]
n=8, s=[trickier]
n=2, s=[to]
n=0, s=[]
n=6, s=[master]
n=0, s=[]
n=-1, s=[]
The problem is the leading spaces in the string I have looked at a lot of examples of how to trim the leading spaces, but I couldn't get anything to work with my current source code.
Code:
#include "getword.h"
int getword(char *w) {
int iochar;
int index = 0;
int numberofchars = 0;
if (numberofchars == 0 && iochar == '\n') {
w[index] = '\0';
iochar = EOF;
return 0;
}
if (iochar == EOF && numberofchars == 0) {
w[index] = '\0';
return -1;
}
while ((iochar = getchar()) != EOF) {
if (iochar != ' ' && iochar != '\n') {
w[index] = iochar;
index++;
numberofchars++;
} else {
w[index] = '\0';
if (strcmp(w, "done") == 0) {
return -1;
} else {
return numberofchars;
}
}
} //after while loop
} // end of function
int main() {
int c;
char s[STORAGE];
for (;;) {
(void)printf("n=%d, s=[%s]\n", c = getword(s), s);
if (c == -1)
break;
}
}

Answer 1

The code is way too complicated, with some useless and bogus tests producing undefined behavior:
testing iochar in getword() before you even read it with getchar() makes no sense.
combining reading, testing and writing the words in a single printf() call is bogus too: you should instead read, then test, then output if not done.
Here is a simplified version:
#include <stdio.h>
#include <string.h>
#define STORAGE 50
// read a word into an array of size `size`.
// return the number of characters read.
int getword(char *w, size_t size) {
int c;
size_t i = 0;
while (i + 1 < size && (c = getchar()) != EOF) {
if (c == ' ' || c == '\t' || c == '\n') {
if (i == 0)
continue; // ignore leading spaces
else
break; // stop on white space following the word.
}
w[i++] = c;
}
w[i] = '\0';
return i;
}
int main() {
char s[STORAGE];
int len;
while ((len = getword(s, sizeof s)) != 0) {
if (!strcmp(s, "done"))
break;
printf("n=%d, s=[%s]\n", len, s);
}
return 0;
}

Answer 2

I tried to make the code produce desired output, but this is all I could do. There were some bugs which I thought and so I fixed them. See the comments in the code for more details. Hope that solves the problem.
int getword(char * w) {
int iochar = 0;
int index = 0;
int numberofchars = 0;
// I really don't know why those if conditions were required
// Thought they were useless so removed them
while ((iochar = getchar()) != EOF) {
if (iochar != ' ' && iochar != '\n') {
w[index++] = iochar; // slight change here
numberofchars++;
} else {
w[index] = '\0';
// I don't know what this condition is supposed to mean
// so I ignored it
if (strcmp(w, "done") == 0) {
return -1;
} else {
return numberofchars;
}
}
} //after while loop
// Since EOF is encountered, no more characters to read
// So terminate the string with '\0'
w[index] = '\0';
// Here after the loop you should check if some characters were read, but not
// handled. If there are any, return them because that's what you last read
// before EOF was encountered
return (numberofchars > 0 ? numberofchars : -1);
} // end of function
int main()
{
int c;
char s[STORAGE];
for (;;) {
// Put the if condition before printing because if -1 is returned
// it doesn't make sense to print the string at all
c = getword(s);
if (c == -1) break;
printf("n=%d, s=[%s]\n", c, s);
}
}
Here's where I tested this: http://ideone.com/bmfaA3