Hi all.
Why does the while loop evaluate to true, and the program steps into the while loop? Without the second condition, it evaluates to false. What am I missing?
Thanks in advance.
Code:#include <stdio.h> #include <stdlib.h> int main(void) { int x = 34; // ascii value of " while(x != '"' || x != '?') { printf("x != \""); exit(0); } return 0; }
It's true that x is not equal to '?'.
The condition
is true when one or two of the sub-conditions is true.Code:x!= '"' || x != '?'
when x is not a quote or a question mark, both sub-conditions are true and the whole condition is true
when x is a quote (but not a question mark), the first sub-condition is false and the 2nd one is true, making the whole condition true
when x is a question mark (but not a quote), the first sub-condition is true and the 2nd one is false, making the whole condition true
There is no way the whole condition can be false
Ahahaha! *facepalm*
Used || instead of &&.
In the while condition you have an OR logical operation.
This will result to true if at least one of the operants is true. This can be in this case the first , the second or both.
It will result to false, only if all operants are resulted to false.
But lets assume that x is the equal operator ( '=' )
So the first operant will be true, because ' = ' is different form ' " ' . Also the second will be true, likewise . As a result both are true, thus the condition results to true.Code:x != '"' || x != '?'
If now x is set to ' " ' , the first operant will be false but the second will be true again.Thus at least one operant of the OR logical operation is true, so at the end the condition will be true again.
Same situation if x is the question mark . First operant will be false and second true, but again the operation finally will result to true.
As a result your while condition has no meaning....It would be equivalent if i wrote
which in other words means that it will always be resulted to true.Code:while(1)
EDIT : Maybe i have to type faster next time
