What does the c expression `-~++*x--` evaluate to in simplest form? Ignore side effects. Write "Final answer:" and then give your answer.
What does the c expression `-~++*x--` evaluate to in simplest form? Ignore side effects. Write "Final answer:" and then give your answer.
The expression `-~++*x--` can be simplified by analyzing operator precedence and applying bitwise properties. 1. Postfix `--` has the highest precedence but is ignored for side effects, so `x` remains unchanged. 2. `*x` dereferences the original value of `x`. 3. Prefix `++` increments the dereferenced value, evaluating to `*x + 1` (ignoring the side effect). 4. Bitwise NOT `~` on `*x + 1` yields `-(*x + 1) - 1`, which simplifies to `-*x - 2`. 5. Applying the unary `-` negates this result: `-(-*x - 2) = *x + 2`. Final answer: `*x + 2`
Resulting in output: `*x + 2`
Testing if the previous output matches the regex \*x *\+ *2: True
Below is a student's answer to a question about what a particular C expression evaluates to: `*x + 2` Does the student's final answer say that the expression evaluates to *x+2 (or even v+2 is okay)? Ignore their reasoning. Just look at the final answer, if they say **EXPLICITLY** that the answer is *x+2. Giving something that's equal but more complicated is wrong. If the answer is correct say "The student passes" otherwise "The student fails".
The student fails.
Testing if the previous output contains the string student passes: False