How to get input using the left-shift-bitwise operator?

Question

#include<stdio.h>

long scan()
{
    long int t = 0;
    char c;
    c = getchar_unlocked();
    while (c < '0' || c>'9')
        c = getchar_unlocked();
    while (c >= '0' && c <= '9')
    {
        **t = (t << 3) + (t << 1) + c - '0'; **
            c = getchar_unlocked();
    }
    return(t);
}

int main()
{
    int t, a[110], n, i; long long int c, sum;
    t = scan();
    while (t--)
    {
        sum = 0;
        scanf("%d%lld", &n, &c);
        for (i = 0; i < n; i++)
        {
            a[i] = scan();
            sum += a[i];
        }
        if (sum <= c) 
            printf("Yes\n");
        else 
            printf("No\n");
    }
    return 0;
}

Please help me understand the line in '**'. I know that left shift operator n multiplies the number by 2^n. So t is first getting multiplied by 8, then by 2. But how is that helping us get the input?

The scan() function gets the input and returns it.

Answer 1

The line you indicated with the asterisks doesn't help get input, all it does it update the value of the variable t based on the current value of c .

It is the calls to getchar_unlocked() that get the next character from stdin and return it, and the program stores that value in the variable c .

The program expects that the characters will be numeric ASCII digits, so that the computed value of (c-'0') will be between 0 and 9 inclusive. For each of those value values, it does the equivalent of t = (t*8) + (t*2) + the_computed_value; -- I'm not sure what the intent is, but that's what it's doing.