for loop, how to print integers/string two by two - c

Question

This drives me crazy because it seems so easy, but I cannot figure it out.
I have a for-loop and it prints out words what I have entered. This version of for-loop prints it out this way:
"car - wheel - bike - handlebar",
instead I want to print it this way:
"car - wheel
bike - handlebar"
for(int i=0;i<numberOfWords;i++)
printf("%s - ",words[i]);
printf("\n");
EDIT: It prints out from an array. I have a function that takes in words and stores in an array. Then I want to pair two words side by side.

Answer 1

for(int i=0;i<numberOfWords;i++) }
printf("%s",words[i]);
if(i%2) printf("\n");
else printf(" - ");
}
printf("\n");
Explanation. Prints every word. And if i is odd (so after printing words[1] and words[3]) print a newline. Otherwise, print a - since another word will be printed on that line.
If the total number of words might not be even (of multiple of whatever number of words per line you want, since this code can be adapted for other than 2), then a specific code should be written for the last word. For example, adding
if(i==numberOfWords-1) break after the first printf, so that last loop iteration does not print any separator - nor newline.

Answer 2

As other answers have pointed out, some test of 'parity' of the loop counter will select between outputting a "spacer" or a "newline" after each word is printed. Some of those methods are arcane, and some of those methods will not properly handle the case of an odd number of words.
If you are going to make the effort to code this algorithm, it should be as capable as possible. The following works for "pairs", but it seems more lyrical to show it working for "triplets" of sequential words in an array. (Omitting checks for negative numbers and empty strings. Those are not a few of my favourite things.)
#include <stdio.h>
#include <string.h>
void out( char *w[], size_t n, size_t per ) {
// output sequential words (array elements) several per line.
for( size_t r = 0; r < n; r += per ) {
char *prfx = ""; // prefix separator for words...
for( size_t c = r; c < r+per && c < n; c++ ) {
printf( "%s%s", prfx, w[c] );
prfx = " - "; // subsequent words prefixed by this string
}
putchar( '\n' );
}
putchar( '\n' );
}
int approved( char *w ) { // suppress adding some words to the array
char *stop[] = { "and", "up", "are", "of" }; // "stopwords"
for( int i = 0; i < sizeof stop/sizeof stop[0]; i++ )
if( strcmp( w, stop[i] ) == 0 ) return 0;
return 1;
}
int main() {
char meat[] =
"Raindrops on roses and whiskers on kittens "
"Bright copper kettles and warm woolen mittens "
"Brown paper packages tied up with strings "
"These are a few of my favourite things";
char *words[100]; // sufficient
size_t nWords = 0;
// Break apart the meat into individual (approved) words
for( char *p = meat; (p = strtok( p, " \n") ) != NULL; p = NULL )
if( approved( p ) )
words[ nWords++ ] = p;
// out( words, nWords, 2 ); // now, output pairs
out( words, nWords, 3 ); // or output triplets
// out( words, nWords, 4 ); // or output quadruplets
return 0;
}
Raindrops - on - roses
whiskers - on - kittens
Bright - copper - kettles
warm - woolen - mittens
Brown - paper - packages
tied - with - strings
These - a - few
my - favourite - things
Slow day... Instead of multiple calls to strcmp() to approve (or not) certain words, there's this: (presumes that any word is <= 30 characters long.)
int approved( char *w ) {
char *stop = " and up a of ", buf[1+30+1=1], *p = buf+1;
while( (*p = *w) != '\0' ) p++, w++;
buf[0] = p[0] = ' '; p[1] = '\0';
return strstr( stop, buf ) == NULL;
}

Answer 3

It seems you mean the following
for ( int i = 0; i < numberOfWords; i++ )
{
if ( i % 2 == 0 )
{
printf( "\"%s - ", words[i] );
}
else
{
printf( "%s\"\n", words[i] );
}
}
The output will be
"car - wheel"
"bike - handlebar"

Answer 4

If you have an even number of words:
for (int i = 0; i < numberOfWords; i++)
printf("%s%s", words[i], i % 2 ? "\n" : " - ");
If you have an even or odd number of words:
int i = 0;
while (i < numberOfWords)
{
printf("%s", words[i]);
printf(i++ % 2 || i == numberOfWords ? "\n" : " - ");
}

Answer 5

My minimum working example uses command line parameters (argc, argv) to provide the test case array of strings, but the main difference to the other examples I have seen here is that this iterates over that array in steps of two.
It prints all pairs of words, and then catches and handles the single leftover array item in case the array contains an odd number of items.
int main(const int argc, const char *const argv[]) {
for (int i=0; i<argc; i+=2) {
if ((i+1)<argc) {
printf("word pair: %s - %s\n", argv[i], argv[i+1]);
} else {
printf("leftover word: %s\n", argv[i]);
}
}
return 0;
}