~ operator in while not working as intended

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Sneha Chakravarthy
Sneha Chakravarthy 2021 年 4 月 5 日
I have this piece of code to make sure that the input is a whole number between 1 and 12
while (x~=1 || x~=2 || x~=3 || x~=4 || x~=5 || x~=6 || x~=7 || x~=8 || x~=9 || x~=10 || x~=11 || x~=12)
fprintf("\nOption entered is not present in the menu. Please enter a valid number");
x = input("\nEnter number corresponding to the conversion you want to perform: ");
end
However, the loop prints out the error message even the input is correct. Why is that so?
  2 件のコメント
Stephen23
Stephen23 2021 年 4 月 5 日
"Why is that so?"
Flawed logic. Can you name any value for which
x~=1 || x~=2
will return false?
Sneha Chakravarthy
Sneha Chakravarthy 2021 年 4 月 6 日
Oh dear, it is quite obvious isn't it. Thanks for the help

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回答 (2 件)

Mike
Mike 2021 年 4 月 5 日
In your code the or condition ( || ) ensures it always executes the error. You need to do a logical and i.e. change all || to &&.

Steven Lord
Steven Lord 2021 年 4 月 5 日
Let's take a simpler example.
check = @(x) (x ~= 1) | (x ~= 2)
check = function_handle with value:
@(x)(x~=1)|(x~=2)
Does this return true for 1.5?
check(1.5)
ans = logical
1
Does it return true for 3?
check(3)
ans = logical
1
Does it return true for 1?
check(1)
ans = logical
1
Whoops. Why? Well, 1 is equal to 1 so (x ~= 1) returns false. But 1 is not equal to 2 so (x ~= 2) returns true. And false or true returns true.
For something like this, you could use round (to make sure x is an integer value) and > and < to determine if it's in the bounds that you want. Alternately, if you want to limit x to being in a specific set of values use ismember.

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