c-fill a one dimensional array with for loop

Question

I can understand the first solution.But in the second solution i am confused about the way scanf , accept 4 values at the same time and apply them to the for loop.

//first solution
#include <stdio.h>

int main() {
    int pin[4],i;
    
    for(i=0; i<4; i++){
        printf("Give value: ");
        scanf("%d", &pin[i]);
    }
  
    return 0;
}

//second solution
#include <stdio.h>

int main() {
    int pin[4],i;
    
    printf("Give 4 values: ");
    
    for(i=0; i<4; i++){
        
        scanf("%d", &pin[i]);
    }
  
    return 0;
}

Answer 1

The difference is only that in the first example there is a printf that asks you for the input values at every iteration of the cycle while in the first example there is a printf (only one) before the cycle.

The operation that matters (the scanf) is exactly the same in the two examples.

Answer 2

The scanf commamd does not read four values at the same time.

This is an example of a loop : a piece of code that does the same thing multiple times.

for(i=0; i<4; i++){
    
    scanf("%d", &pin[i]);
}

What the for statement does is, first initialize i to zero, then check that i is less than four, then, execute the code within its curly braces. That's the scanf statement, which attempts to read an integer to the address specified by &pin[i] . Since i is zero, that means, the address of pin[0] , which is the first element of the array pin .

After the scanf executes (and whether or not it succeeds, which is something you may want to look into) the for statement increments the value of i , checks that it is still less than four, and executes the block of code between the braces again. This time, the scanf statement attempts to read to pin[1] .

The loop executes two more times, potentially storing integers in pin[2] and pin[3] before terminating just after incrementing i to 4.