c - Why does my switch statement print a case and default? -

#include <stdio.h>  int main(void) {     char ch;     int end=0;     printf("\npick letter through f. (f ends program)");         {         scanf("%c", &ch);          switch (ch) {     case 'a':         printf("a. another: ");         break;     case 'b':         printf("b. another: ");         break;     case 'c':         printf("c: ");         break;     case 'd':         printf("d. another: ");         break;     case 'e':         printf("e. another:  ");         break;     case 'f':         printf("f. goodbye. ");         end=1;         break;     default:         printf("that wasn't through f. ");         break;         }     } while (end == 0);     return 0; } 

so if enter say:

a. another: wasn't through f. 

if enter g say:

that wasn't through f. wasn't through f.  

if enter f expected

f. goodbye. 

and program terminates.

any tips on how fix this? i've tried looking while , answers not in c or you're forgetting break; statement. i'm new c, maybe it's obvious i'm not noticing, thought might due while loop? time

scanf("%c", &ch) reads 1 character @ time. if you're typing letter , hitting enter, you're providing 2 characters: letter, , newline character (u+000a, '\n').

if want ignore newline characters, 1 option explicitly check them:

case '\n':     break;