Sunday 28 May 2017

c++ - Integer to Boolean strange syntax





I'm less than a year into C++ development (focused on other languages prior to this) and I'm looking at a guy's code who's been doing this for two decades. I've never seen this syntax before and hopefully someone can be of some help.



bool b; // There exists a Boolean variable.
int i; // There exists an integer variable.

sscanf(value, "%d", &i); // The int is assigned from a scan.
b = (i != 0); // I have never seen this syntax before.


I get that the boolean is being assigned from the int that was just scanned, but I don't get the (* != 0) aspects of what's going on. Could someone explain why this person who knows the language much better than I is doing syntax like this?



Answer



Have a read here:
http://en.cppreference.com/w/cpp/language/operator_comparison



The result of operator != is a bool. So the person is saying "compare the value in i with 0". If 'i' is not equal to 0, then the '!=' returns true.



So in effect the value in b is "true if 'i' is anything but zero"



EDIT: In response to the OP's comment on this, yes you could have a similar situation if you used any other operator which returns bool. Of course when used with an int type, the != means negative numbers evaluate to true. If > 0 were used then both 0 and negative numbers would evaluate to false.


No comments:

Post a Comment

c++ - Does curly brackets matter for empty constructor?

Those brackets declare an empty, inline constructor. In that case, with them, the constructor does exist, it merely does nothing more than t...