character array reads 3 inputs instead of 5 in c?

Question

my code is given below

#include <stdio.h>
#include <ctype.h>

#define size 5

void main(){

    int i;
    char letter[size];
    for(i=0;i<size;i++)
        letter[i]=getchar();
    for(i=0;i<size;i++)
        putchar(toupper(letter[i]));
}

and the output is :

bitto@HP-ProBook-4430s:~$ gcc test.c
bitto@HP-ProBook-4430s:~$ ./a.out
a
s
d
A
S
D

why is this happening? it was to read 5 characters and convert them to upper case.

You need to read up a little on how terminals behave. See [this](http://stackoverflow.com/questions/421860/capture-characters-from-standard-input-without-waiting-for-enter-to-be-pressed) — JJF, Jun 30 '16 at 16:14
@EOF is certainly right. I expanded on your comment in an answer. — Eli Sadoff, Jun 30 '16 at 16:17

score 3 · Accepted Answer · answered Jun 30 '16 at 16:08

3

As EOF said, any input you have will be taking two chars as input because you are typing in A then Enter which C reads as a and \n so that is two chars, not one.

answered Jun 30 '16 at 16:08

Eli Sadoff

7,173
6
33
61

this explains it . typing in the characters without spaces in the same line solves the problem. – Jun 30 '16 at 16:18

character array reads 3 inputs instead of 5 in c?

1 Answers1