not able to understand the role of getchar and putchar here - c

Question

#include <stdio.h>
#include <stdlib.h>
int main()
{
int c;
c = getchar();
while (c != EOF) {
putchar(c);
}
return 0;
}
when I compile and give input ABC and then press enter, the never ending loop starts like AAAAAAAAA....
And now look at this code below
#include <stdio.h>
#include <stdlib.h>
int main()
{
int c;
c = getchar();
while (c != EOF) {
putchar(c);
c = getchar (); // added this single line
}
return 0;
}
In this program, when I input ABC, the output is ABC.
Can anyone please explain why it is not showing just a single A as output?

Answer 1

Look at the below code you mentioned
int main(void){
int c;
c = getchar();
while (c != EOF) {
putchar(c);
}
return 0;
}
When c = getchar(); executes & if you provided input as ABC at runtime & press ENTER(\n), that time c holds first character A.
Next come to loop, your condition is c!=EOF i.e A!=EOF which always true & it will print A infinitely because you are not asking second time input so c holds A.
correct version of above code is
int main(void){
int c;
while ( (c = getchar())!=EOF) { /* to stop press ctrl+d */
putchar(c);
}
return 0;
}
case 2 :- Now looks at second code
int main(void){
int c;
c = getchar();
while (c != EOF) { /*condition is true */
putchar(c);
c = getchar ();/*After printing ABC, it will wait for second input like DEF, unlike case-1 */
}
return 0;
}
Can anyone please explain why it is not showing just a single A as output ? Why it should prints only A, it prints whatever input you given like ABC & so on. Just note that getchar() works with buffered input i.e when you press ENTER getchar() will read upto that & when there is nothing left to read getchar() returns EOF.