What happens when multiple scanf()s are included in a single for loop

Question

I have written a code for comparing the triplet problem. The problem statement is:

Input: 2 arrays of size 3

Task is to find their comparison points by comparing a[0] with b[0] , a[1] with b[1] , and a[2] with b[2] .

If a[i] > b[i] , then Alice is awarded point. If a[i] < b[i] then Bob is awarded the point. If a[i] == b[i] then neither person receives the point. Comparison points is the total points the person earned. Given a and b determine their respective comparison points.

The code I wrote is:

 #include<stdio.h>
 void main(){   
       int i, alice[3], bob[3];
       int a = 0;
       int b = 0;
                                                                                                                                          
       for(i=0; i<3; i++){
                 scanf("%d", &alice[i]);
       }
       for(i=0; i<3; i++){
                 scanf("%d", &bob[i]); 
       }
       for(i=0; i<3; i++){ 
                if(alice[i] > bob[i])  
                       a++; 
                else if (alice[i] < bob[i]) 
                       b++;
       }   
      printf("%d %d", a, b);
 } 

But when I put the two scanf()s in one line,

  for(i=0; i<3; i++){    
             scanf("%d", &alice[i]); 
             scanf("%d", &bob[i]); 
  }

The output is like 2 1 or 1 2 for all the inputs. Is it wrong to put two scanf() calls in a single for loop? I couldn't understand what is the reason behind this problem? Would someone kindly explain the reason?

Answer 1

The first version reads three values for alice and then three values for bob . The second version reads one value for alice , then one for bob , and repeats that 3 times.

If the entered numbers are 1, 2, 3, 4, 5, 6, then in the first example, alice gets 1, 2, 3 while bob gets 4, 5, 6; in the second example, alice gets 1, 3, 5 while bob gets 2, 4, 6. Quite different results.