How to run through a loop multiple times in C? - c

Question

Ok i modified my code but cannot get it to break when the user inputs 0. I tried 0, '0', and "0" and neither break the loop.
#include <stdio.h>
#include<conio.h>
int main(){
int word;
int countword = 0;
int countpunct = 0;
do{
printf("\nEnter the String: ");
while ((word = getchar()) != EOF && word != '\n'){
if (word == ' ' || word == '.' || word == '?' || word == '!' || word == '(' || word == ')' || word == '*' || word == '&'){
countword++;
}
if (word == '.' || word == '?' || word == '!' || word == '(' || word == ')' || word == '*' || word == '&'){
countpunct++;
}
}
printf("\nThe number of words is %d.", countword);
printf("\nThe number of punctuation marks is %d.", countpunct);
} while (word!= 0);
}

Answer 1

Your inner loops break when word is either EOF or \n. Since you never modify it when you get to the end of the outer loop, the condition will always be true.
Going back to your pre-edit code, all you really need is to change scanf("%c", word); to scanf("%c", &word);, although you should use a separate char variable for that, since the %c format specifier expected a pointer to char. So your code should look like this:
#include <stdio.h>
#include <stdlib.h>
int main(){
int word;
char cont;
for (;;){
int countword = 0;
int countpunct = 0;
printf("\nEnter the String: ");
while ((word = getchar()) != EOF && word != '\n'){
if (word == ' ' || word == '.' || word == '?' || word == '!' || word == '(' || word == ')' || word == '*' || word == '&'){
countword++;
}
if (word == '.' || word == '?' || word == '!' || word == '(' || word == ')' || word == '*' || word == '&'){
countpunct++;
}
}
printf("\nThe number of words is %d.", countword);
printf("\nThe number of punctuation marks is %d.", countpunct);
printf("\nContinue? Y/N?");
scanf("%c", &cont);
if (cont!= 'y' && cont!= 'Y'){
return 0;
}
}
}
Note also that countword and countpunct are moved inside of the outer loop. That way, they're initialized to 0 for each new set of words.