Why is my c != 'o' || c != 'x' condition always true? [duplicate]

坚强是说给别人听的谎言 提交于 2019-11-26 02:15:41

It must be if(c!='o' && c!='x') instead of if(c!='o' || c!='x'). If you use the or operator the boolean expression will be always true.

T.J. Crowder

The condition (c != 'o' || c != 'x') can never be false. If c is 'o', then c != 'x' will be true and the OR condition is satisfied. If c is 'x', then c != 'o' will be true and the OR condition is satisfied.

You want && (AND), not || (OR):

while (c != 'o' && c != 'x') {
    // ...
}

"While c is NOT 'o' AND c is NOT `'x'..." (e.g., it's neither of them).

Or use De Morgan's law, covered here:

while (!(c == 'o' || c == 'x')) {
    // ...
}

"While it's NOT true that (c is 'o' or c is 'x')..."

Why is my c != 'o' || c != 'x' condition always true?

The expression combines two sub-expressions using the logical OR operator (||). The expression is true if at least one of the sub-expressions is true. In order to be false, both sub-expressions it connects must be false.

The sub-expressions are c != 'o' and c != 'x'.

The first sub-expression c != 'o' is false when c == 'o'. The same for the second one; it is false when c == 'x'.

Please note that they cannot be false on the same time because c cannot be 'o' and 'x' on the same time.

The condition should be if(!(c=='o' || c=='x')) or if(c!='o' && c!='x')

annu

even when you enter x or you enter o in that case if condition evaluates to true and hence game_start remains false.

it should be if(c!='o' && c!='x') or use a more straight forward way

if(c == 'o' || c == 'x')
   game_start=true;
else
    System.out.println("you can only enter o or x!");
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