- Forum
- General C++ Programming
- !(1 && !(0 || 1))Help
!(1 && !(0 || 1))Help
This !(1 && !(0 || 1)) is said to be true, if so, why?
Can someone explain how this is interpreted?
Can someone explain how this is interpreted?
well, if I'm going through this right, here's the breakdown:
(0 || 1) comes out to be true, as the one is true and it's saying "either '0' or '1.'"
!(0 || 1) is then false, as the ! takes the opposite, making the statement "not true."
(1 && !(0 || 1)) is false, as it is saying "if both '1' and '!(0 || 1)'" and the latter is false, making the whole statement false.
Then, the ! in front of the whole thing makes it "not false," or true.
I THINK that's how it goes.
(0 || 1) comes out to be true, as the one is true and it's saying "either '0' or '1.'"
!(0 || 1) is then false, as the ! takes the opposite, making the statement "not true."
(1 && !(0 || 1)) is false, as it is saying "if both '1' and '!(0 || 1)'" and the latter is false, making the whole statement false.
Then, the ! in front of the whole thing makes it "not false," or true.
I THINK that's how it goes.
OK, I understand all except...
!(0 || 1) is then false, as the ! takes the opposite, making the statement "not true."
Are you saying that with the () is True however because of !() it means Not True?
If within () resulted as false would this have been the have been Not False?
Thanks BTW, I am a newbie to programming. Actually, just started to look at C++ and have
had no prior programming experience.
!(0 || 1) is then false, as the ! takes the opposite, making the statement "not true."
Are you saying that with the () is True however because of !() it means Not True?
If within () resulted as false would this have been the have been Not False?
Thanks BTW, I am a newbie to programming. Actually, just started to look at C++ and have
had no prior programming experience.
Topic archived. No new replies allowed.