Question

Sometimes I want to do something like this (with i and j being ints).

(if i==4 && j==9)
{
   ...
}

Where it’ll go through the brackets if i equals 4 and j equals 9. I’ve been using a single ampersand (&) instead of a double one and my code’s been compiling and running.

Is it doing the same thing as a double ampersand &&, and if not what has it been doing?

Edit: Oh and I’ve been doing the same thing with or, using ‘|’ instead of ‘||’

Answer 1

Presumably you mean if (i==4 && j==9).

Under the circumstances, changing this from && to & shouldn’t change much. The big thing that’ll change is that with &&, the j==9 would only be evaluated if the i==4 part was true, but with &, they’ll both be evaluated regardless.

When you have something like if (x != NULL && x->whatever ...) you want to ensure that the second part (that dereferences x) is only evaluated if x is not a null pointer. In your case, however, comparing what appear to be ints is unlikely to produce any problems.

It’s also possible to run into a problem when you’re dealing with something that may produce a value other than 1 to signal true. Again, it’s not a problem here because == will always produce either 0 or 1. If (for example) you were using isalpha, islower, etc., from <ctype.h>, they’re only required to produce 0 or non-zero values. If you combined those with a &, you’d get a bit-wise or which could produce 0 for two non-zero inputs (e.g., 1 & 2 == 0, but 1 && 2 == 1).

When you use bitwise and on the results from ==, you’re going to get 0 & 0 or 0 & 1 or 1 & 0 or 1 & 1. 1 & 1 will yield 1 (true). All the others will yield 0 (false) — just like && would have.