Thread: Really confused on how to attack this question

Hello everyone, im preparing for my first c exam and going through some practice questions.

Display the appropriate code to reset to 0 the middle two bits of variable x who type is unsigned char.

Im really confused on how to start. Im assuming that i need to convert the variable x into binary, and then use some-type of bit operator to shift through the code and replace the appropriate values. Any suggestions would be appreciated, thanks.

No, you mainly need to use bitwise and. No "conversion to binary" is needed, since you already have control over the bits with bitwise operations and assignment.

Originally Posted by Memloop

The most portable way to do this is probably going to involve the one's complement operator to set all bits to 1 first, and then use bit shifting to create a XOR mask.

Sounds good (i.e., use bitwise not on 0), but I still think that bitwise and is the one to use since the other bits could just be ANDed with 1.

Originally Posted by Memloop

I'm not sure if the number 0 is stored as 11...1 or 00...1 on systems that use one's complement instead of two's complement though.

This concerns an unsigned char, so we do not need to worry about signed integer representation

Originally Posted by Memloop

But you're going to have to hard code the bitmask, and that's going to be trouble if CHAR_BITS change. What are you going to do on machines that only have 16 bit bytes?

You do not have to hard code the bit mask, but after thinking more carefully on how I would actually do it, let me guess: when you say "use bit shifting to create a XOR mask", you actually mean to use bit shifting and bitwise xor to create the mask that will be used with bitwise and? That seems to be the simplest solution that I can come up with at this point, e.g.,

Code:

unsigned char mask = 3U << (CHAR_BIT / 2 - 1);
mask ^= ~0U;

Originally Posted by Memloop

I'm guessing 0 is always interpreted as +0 even if you are using signed numbers (a lot of existing code would be broken otherwise)?

That certainly appears to be the case. C99 states that:

Originally Posted by C99 Section 6.2.6.2 Paragraph 3

If the implementation supports negative zeros, they shall be generated only by:

• the &, |, ^, ~, <<, and >> operators with arguments that produce such a value;
• the +, -, *, /, and % operators where one argument is a negative zero and the result is zero;
• compound assignment operators based on the above cases.

It is unspecified whether these cases actually generate a negative zero or a normal zero, and whether a negative zero becomes a normal zero when stored in an object.