word entry in an array of strings - c

Question

I'm trying to make a program that will read up to 20 words entered by the user and stored in an array of strings. The program will ask for additional words until 20 words have been entered or until the word 'done' has been entered. The idea is that these words will then be entered into a matrix to create a word search program. I'm stuck on scanning in the words entered by the user. I'm a new programmer so any words of advice is very beneficial.
#include <stdio.h>
int main(void)
{
char string[20][100];
printf("Enter up to 20 words to hide in the puzzle.\n");
printf("Enter the word 'done' after your last word if entering less than 20 words.\n");
scanf("%s\n",c)
printf("Entered words:\n");
return 0;
}

Answer 1

#include<stdio.h>
#include<string.h>
int main(void)
{
char words[20][100];
char temp[100]="\0";
int i=0;
int end=0; //0 false and 1 true
printf("Enter 20 words or enter done to exit.\n");
while(i <=19 && end==0)
{
strset(temp,'\0');// resets array temp to NULL's everytime
scanf(" %99[^\n]",temp); //this is scan set, to read a string without '\n'
printf("Given:%s\n\n",temp);
if(strcmpi(temp,"done")==0)//compares given input with "done".if "done" is entered. zero is returned
end=1;//when 0 is returned this end=1 will break the loop.
else//if input is not given "done" then copy temp array to words[i].
{
strcpy(words[i],temp);
i++;
}
}
}
Instead of using words[20][100] directly, I have used a temporary array named temp to initially store the input,because at the end i don't want to store "done" into words[20][100].Assuming that "done" is used only to end the input process and it is not the actual word to store.But you can change this program to your need.

Answer 2

This is a straight-forward but bug-fixed variant of the code provided by developer3466402 in his answer.
I've used a for loop instead of a while loop since that neatly summarizes the action in the while loop. I added n to record how many words were entered, leaving i as a loop control variable (yes, once upon a long time ago I wrote Fortran too).
#include <stdio.h>
#include <string.h>
int main(void)
{
char words[20][100];
int i = 0;
int n;
printf("Enter up to 20 words to hide in the puzzle.\n");
printf("Enter the word 'done' after your last word if entering less than 20 words.\n");
for (i = 0; i < 20; i++)
{
printf("Enter word %2d:\n", i+1);
if (scanf("%99s", words[i]) != 1 || strcmp(words[i], "done") == 0)
break;
}
n = i;
printf("%d words entered\n", n);
for (i = 0; i < n; i++)
printf("Word %2d = [%s]\n", i+1, words[i]);
return 0;
}
It worked OK for me entering 0, 1, many and 20 words. We can debate the newline after the prompt; it has pros and cons, but that's generally true. I chose to echo the data after the loop rather than within the loop. Be aware that with some programs, echoing in a loop can appear to work where echoing after the loop shows a problem.
Example run (with an early exit):
$ ./rw
Enter up to 20 words to hide in the puzzle.
Enter the word 'done' after your last word if entering less than 20 words.
Enter word 1:
aleph
Enter word 2:
null
Enter word 3:
absolute
Enter word 4:
twaddle and nonsense
Enter word 5:
Enter word 6:
Enter word 7:
elephants are done for
Enter word 8:
Enter word 9:
8 words entered
Word 1 = [aleph]
Word 2 = [null]
Word 3 = [absolute]
Word 4 = [twaddle]
Word 5 = [and]
Word 6 = [nonsense]
Word 7 = [elephants]
Word 8 = [are]
$

Answer 3

You can dynamically allocate memory to store words. If the user inputs more than 20 words, then you can use realloc function to allocate more memory.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
int wlimit = 20;
int wcount = 0;
int retval; // to save the return value of scanf
char wstring[100+1]; // +1 for the terminating null byte
char **wlist = malloc(wlimit * sizeof *wlist);
if(wlist == NULL) {
// handle it
printf("not enough memory to allocate\n");
return 1;
}
char *temp;
while(1) {
if(wcount >= wlimit) {
wlimit *= 2;
temp = wlist;
wlist = realloc(wlist, wlimit);
if(temp == NULL) {
printf("not enough memory to allocate\n");
wlist = temp;
}
}
retval = scanf("%100s", wstring);
// if the input string is done, then break out of the
// loop else keep taking input from the user
if(retval != 1 || strcmp(wstring, "done") == 0)
break;
// strdup function creates a new string which is a duplicate
// of the input string and returns a pointer to it which can
// be freed using free
wlist[wcount++] = strdup(wstring);
}
// do stuff with wlist
// after done, free the memory
for(int i = 0; i < wcount; i++)
free(wlist[i]);
free(wlist);
wlist = NULL;
// stuff
return 0;
}

Answer 4

I think you wish something like below:
#include<stdio.h>
int main(void)
{
char words[20][100];
int i = 0;
printf("Enter up to 20 words to hide in the puzzle.\n");
printf("Enter the word 'done' after your last word if entering less than 20 words.\n");
while (i < 20) {
printf("Entered words:\n");
if (scanf("%99s", words[i]) != 1 || strcmp(words[i], "done") == 0)
break;
printf("%s\n", words[i]);
i++;
}
return 0;
}