按位运算符

Question

I'm new here but had some questions about my Computing 2 HW.

Given the main function:

void set_flag(int* flag_holder, int flag_position);
int check_flag(int flag_holder, int flag_position);

int main(int argc, char* argv[])
{
    int flag_holder = 0;
    int i;
    set_flag(&flag_holder, 3);
    set_flag(&flag_holder, 16);
    set_flag(&flag_holder, 31);
    for(i = 31; i >= 0; i--) {
        printf("%d", check_flag(flag_holder, i));
        if(i % 4 == 0)
            printf(" ");
    }
    printf("\n");
    return 0;
}

We're supposed to complete set_flag , and check_flag , and its supposed to display:

1000 0000 0000 0001 0000 0000 0000 1000

Now I'm really struggling to understand this. is flag_holder supposed to equal 0 the whole time? My professor said to use bitwise and no multiplication. But it looks like I could just implement an array of 32 '0's and set the 3rd, 16th, and 31st elements as 0s? I apologize in advance for the messy set up, and thanks to any who actually read this and help!

Answer 1

Let me tell you the similarity between multiplication and bitwise

left shift & right shift are used to set flags mostly

2 = 0010 

when you left shit 2 by 1, all the bits are shifted to left and with zero appended.

0010 << 1 
0100 

and the result will be

0100 = 4 

and further left shifting make 0100 to 1000 = 8

left shifting is similar to multiply by 2 and right shifting is similar to divide by 2

Setting flags

set_flag(&flag_holder, 3);

the function will more or less be like this

void set_flag( int *flag_holder, int shifter) {
*flag_holder = *flag_holder | 1 << shifter;
}

what this will do is that 3rd bit is set without affecting any other bits