C Programming - Bitwise Operators - Discussion

Discussion Forum : Bitwise Operators - Point Out Correct Statements (Q.No. 4)
4.
Which of the following statements are correct about the program?
#include<stdio.h>

int main()
{
    unsigned int m[] = {0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80};
    unsigned char n, i;
    scanf("%d", &n);
    for(i=0; i<=7; i++)
    {
        if(n & m[i])
            printf("yes");
    }
    return 0;
}
It will put OFF all bits that are ON in the number n
It will test whether the individual bits of n are ON or OFF
It will put ON all bits that are OFF in the number n
It will report compilation errors in the if statement.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Quein said:   5 years ago
Please explain it clearly.

Rahul said:   7 years ago
It is clear now. Thanks @Abhishek.

Abhishek said:   7 years ago
Suppose n=2.

Binary representation as, 0000 0000 0000 0010.
Now when, i=0, m[0]=0x01(in hexa) hence in binary 0000 0000 0000 0001.
n & m gives 0 (as & gives 1 only when 1 is present in same position in both numbers).
i becomes 1, m[1]=0x02 hence in binary, 0000 0000 0000 0010
n&m gives 1 hence yes is printed.
i becomes 2, m[2]=0x04 hence in binary, 0000 0000 0000 0100
n&m gives 0 again.

Just like this, it will keep checking.
Remember when m=0x10 then in binary, 0000 0000 0001 0000.
and m=0x20 then in binary, 0000 0000 0010 0000.
As you can see m value converted to binary keeps shifting the value of 1 one position to the left till m=0x80 that is, 0000 0000 1000 0000.
Hope it helps.
(1)

Amit tiwari said:   7 years ago
I am not getting this, Someone please explain it.

Alex said:   7 years ago
It might make more sense modified this way to display a binary value. Run it, modify the value of n and observe.

/* Note: GCC Compiler (32 Bit Linux Platform). */

#include<stdio.h>
int main()
{
unsigned int m[] = {0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80};
unsigned char n, i;
n = 0x04;
for(i=7; i>0; i--)
{
if(n & m[i])
printf("1");
else
printf("0");
}
return 0;
}

Shweta kumari said:   7 years ago
Not getting this. Please explain me.

Manoj kumar said:   9 years ago
Anyone can explain this code in a right way?

Ketul said:   10 years ago
Here what is value of number considered?

Lucky said:   10 years ago
Can anybody explain that is it allowed to apply logical & op to an int var and char var?

Vaageesh said:   1 decade ago
Here i is declared as char and how it would be possible to take the integer value as i=0 and all that stuffs?


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