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Part of an exercise in my book requests "Write a function called bit_test that takes two arguments: an unsigned int anda bit number n. Have the function return 1 bit number n if it is on inside the
word, and 0 if it is off."
The if statement of the function that deals with this is:
if ( (word >> (31 - n)) & 0x1 )
return 1;
I don't understand what it is saying.
The argument shifts word to the right by (31-n) which is AND'ed with the hexidecimal 1. What does that mean?
I could answer that but since you were able to type all of that I wonder about your real question. Do you know what any of that means or does every operation need to be explained? I'd rather start by building on what you know.Quote:
The argument shifts word to the right by (31-n) which is AND'ed with the hexidecimal 1. What does that mean?
Oh. I actually didn't understand the question "Have the function return 1 bit number n if it is on inside the
word, and 0 if it is off." That is pasted verbatim, but is there a typo? I don't understand what it's saying.
OK so binary numbers, like all numbers, have a place value. 2^31, 2^30, 2^29, ... 2^0, and in each place value there is either a 0 where the place value is not included, or a 1, which it is, and that sums up the value of the number. What you were asked to do is test place value n if it is 1 or 0. The code you posted solved the problem. Did you solve the problem? Do you understand the solution at all? That is what I've asked you from the start.
if ( (word >> (31 - n)) & 0x1 )
So if the argument inside the if statement is 1, then 1 is returned.
The word is shifted over 31-n places to the right. If the value anded with 1 results in 1, it is on and returns 1. Is that correct?
Yes.