#include<stdio.h> #include<stdlib.h> int main() { char c='y'; int n=0; while(c=='y') { printf("this is loop number %d\n", n); n++; printf("do you wish to continue? (y/n)"); c=getchar(); } return 0; } The while loop is exiting after the initial iteration and the getchar is not taking the character input as expcted! what's wrong with the implementation here.
When the user presses Enter, two characters are generated: 'y' or 'n' and a newline character '\n'. The first call to getchar() gets the letter, the next call gets the newline. Your code must ignore all newlines.
do { c = getchar(); } while(c == '\n'); 
do; while; loop instead of a while; loop? It's just a question not a critic. You have an issue if getchar() return EOF. You should change your answer to fix this issue.c first and only then decide whether to keep it or not. With while, you must check the condition before executing the action. As for EOF, getchar() returns it only at the end of file, which happens only when the program is terminated by ^C or ^D, or at keyboard error.
getchar() is standard ?When you type "y" with enter on your terminal, you have entered two characters. 'y' and '\n' so getchar() return '\n' on the second call. Plus getchar() return a int not a char but you stock it in char c.
#include <stdio.h> #include <stdlib.h> int main() { int c = 'y'; int n = 0; while (c == 'y') { printf("this is loop number %d\n", n); n++; printf("do you wish to continue? (y/n)"); while ((c = getchar()) == '\n' && c != EOF); } return 0; } You can use scanf() because getchar() is very limited.
#include <stdio.h> #include <stdlib.h> #include <stdbool.h> int main(void) { int n = 0; while (true) { printf("this is loop number %d\n", n++); printf("do you wish to continue? (y/n)\n"); char c; if (scanf(" %c", &c) != 1 || c != 'y') { break; } } } Or:
#include <stdio.h> #include <stdlib.h> int main(void) { int n = 0; { char c; do { printf("this is loop number %d\n", n++); printf("do you wish to continue? (y/n)\n"); } while (scanf(" %c", &c) == 1 && c == 'y'); } } " %c"?'\n' must stop the loop and the first answer has the same behavior.
(scanf("%c\n", &c) is that on its first call and the user types <x>, <Enter> , the function does not return right away. The '\n' tells scanf() to consume all white space until some non-white space is found. With buffered input, that means another line of input like <y>, <Enter> scanf() sees the y, puts it back into stdin and returns 1 with x in c. So the user had to type 2 lines of input to read one character on the first call.
\n in scanf format string does not do what you think it does, please read the manual. The ideone example does not illustrate the problem because ideone does not have interactive input
getche()instead ofgetchar()scanf.cat the end?<conio.h>is, but that's Microsoft specific and won't work on another platform because it's not standardised