Why does my C program not print output from an if statement?

Question

This program ought to output two variables which are in an if statement. However it does not but anything printed outside does.

#include <stdio.h>
#include <string.h>
char tracks[][80] = {
    "I left my heart in Harvard Med School",
    "Newark, Newark - a wonderful town",
    "Dancing with a Dork",
    "From here to maternity",
    "The girl from Iwo Jima",
};
void find_track(char search_for[])
{
    int i;
    for (i = 0; i < 5; i++) {
        if (strstr(tracks[i], search_for))
        printf("Track %i: '%s'\n", i, tracks[i]);
   }
}
int main()
{
    char search_for[80];
    printf("Search for: \n");
    fgets(search_for, 80, stdin);
    find_track(search_for);
    return 0;
}

I expected it to output to show the track number and the associated string but it does not.

Answer 1

The fgets function will read a line of text up to a newline, and it will read and store the newline if there is space for it. So if for example you enter "here" then search_for will contain "here\\n" . None of your strings contain a newline so you never find anything.

You'll need to strip the newline from the string that you read in:

search_for[strcspn(search_for,"\n")]=0; 

The strcspn function returns the number of characters from the start of the string which are not in the given list. So if your string contains a newline it will return the index of the newline, otherwise it contains the index of the terminating null byte.