The output of a certain part of my for loop

Question

#include <stdio.h>
int num, i, k, a[5];
int main() {
    a[0]=2;
    a[1]=11;
    a[2]=12;
    a[3]=16;
    a[4]=28;
    num=a[1+3]
    i=4;
    while(num>0){
        a[i]=num%4;
        num=num/3;
        printf("%d ",num);
        i--;
    }
    printf("8\n");
    for(k=0;k<5;k++){
        printf("%c ",65+a[k]);
    }
    printf("\n);
}

The output of this program is:

9 3 1 0 8

CBDBA

I understand completely how the output for the first line is but am rather confused about the 2nd part.

for(k=0;k<5;k++){
    printf("%c ",65+a[k]);

This bit here confused me as the loop the first time from my understanding should go k=0 then print %c which comes from 65+a[k] which k is currently 0 so 65+a[0]. From the earlier part of the where its setting we see a[0]=2 and 65+2 is 67 which is the character "C". which is correct on the output but if I follow this same logic for the 2nd loop 65+a[k] where k=1 so 65+a[1] and a[1] is 11 and 65+11 is 76 that would equal the character "K" but that's wrong as it should be the character "B".

I feel that this line of code is where im missing something:

a[i]=num%4

but it doesn't actually set a number so still confused.

Any help is appreciated

Answer 1

a[i]=num%4 does set the number. This is how:

In your loop:

 while(num>0){
    a[i]=num%4;
    num=num/3;
    printf("%d ",num);
    i--;
}

num varies as in the first line of output.

a[i]=num%4;

actually sets the values in the array as follows:

Initially, i=4 and num=28 . Therefore,

a[i]=num%4; sets a[4] as 28%4=0 . Therefore, your last character is A+0=A .

Then i=3 , and num=9 . Therefore,

a[i]=num%4; sets a[3] as 9%4=1 . Therefore, your second last character is A+1=B .

Then i=2 , and num=3 . Therefore,

a[i]=num%4; sets a[2] as 3%4=3 . Therefore, your third last character is A+3=D .

Then i=1 , and num=1 . Therefore,

a[i]=num%4; sets a[1] as 1%4=1 . Therefore, your fourth last character is A+1=B .

Then i=0 , and num=0 . Therefore,

We do not enter the loop. a[0]=C , its initial value.

Hence we get: CBDBA