Integer to Boolean strange syntax

Question:

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.

转载于:https://www.cnblogs.com/vigorz/p/10499204.html