c - The output of a certain part of my for loop -

#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 program is:

9 3 1 0 8

c b d b a

i understand how output first line rather confused 2nd part.

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

this bit here confused me loop first time understanding should go k=0 print %c comes 65+a[k] k 0 65+a[0]. earlier part of setting see a[0]=2 , 65+2 67 character "c". correct on output if follow same logic 2nd loop 65+a[k] k=1 65+a[1] , a[1] 11 , 65+11 76 equal character "k" that's wrong should character "b".

i feel line of code im missing something:

a[i]=num%4 

but doesn't set number still confused.

any appreciated

a[i]=num%4 set number. how:

in loop:

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

num varies in first line of output.

a[i]=num%4; 

actually sets values in array follows:

initially, i=4 , num=28. therefore,

a[i]=num%4; sets a[4] 28%4=0. therefore, last character a+0=a.

then i=3, , num=9. therefore,

a[i]=num%4; sets a[3] 9%4=1. therefore, second last character a+1=b.

then i=2, , num=3. therefore,

a[i]=num%4; sets a[2] 3%4=3. therefore, third last character a+3=d.

then i=1, , num=1. therefore,

a[i]=num%4; sets a[1] 1%4=1. therefore, fourth last character a+1=b.

then i=0, , num=0. therefore,

we not enter loop. a[0]=c, initial value.

hence get: c b d b a