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.
Final answer: *x + 2 To explain briefly, consider the expression:   -~++*x-- First, the operators group as follows (ignoring side effects):   - (~ (++(*x--))) Working from the inside out: 1. x-- returns the current pointer x (ignoring the side effect of decrementing x). 2. *x-- dereferences that pointer. 3. The prefix ++ increments the value pointed to by x and yields the incremented value. 4. The bitwise NOT ~ is applied to that value. 5. Finally, the unary minus (-) negates the result. In two’s-complement arithmetic, the identity ~y = -y - 1 holds. Let y = ++*x (ignoring side effects), then:   -~y = -(-y - 1) = y + 1 Since y = ++*x is the same as *x + 1 when ignoring side effects, substituting back gives:   *y + 1 + 1 = *x + 2 Thus, the simplified form is *x + 2.
Resulting in output: *x + 2 to explain briefly, consider the expression:   -~++*x-- first, the operators group as follows (ignoring side effects):   - (~ (++(*x--))) working from the inside out: 1. x-- returns the current pointer x (ignoring the side effect of decrementing x). 2. *x-- dereferences that pointer. 3. the prefix ++ increments the value pointed to by x and yields the incremented value. 4. the bitwise not ~ is applied to that value. 5. finally, the unary minus (-) negates the result. in two’s-complement arithmetic, the identity ~y = -y - 1 holds. let y = ++*x (ignoring side effects), then:   -~y = -(-y - 1) = y + 1 since y = ++*x is the same as *x + 1 when ignoring side effects, substituting back gives:   *y + 1 + 1 = *x + 2 thus, the simplified form is *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 to explain briefly, consider the expression:   -~++*x-- first, the operators group as follows (ignoring side effects):   - (~ (++(*x--))) working from the inside out: 1. x-- returns the current pointer x (ignoring the side effect of decrementing x). 2. *x-- dereferences that pointer. 3. the prefix ++ increments the value pointed to by x and yields the incremented value. 4. the bitwise not ~ is applied to that value. 5. finally, the unary minus (-) negates the result. in two’s-complement arithmetic, the identity ~y = -y - 1 holds. let y = ++*x (ignoring side effects), then:   -~y = -(-y - 1) = y + 1 since y = ++*x is the same as *x + 1 when ignoring side effects, substituting back gives:   *y + 1 + 1 = *x + 2 thus, the simplified form is *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 passes.
Testing if the previous output contains the string student passes: True
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.
Final answer: ++*x + 1 To explain briefly: The expression parses as –~(++(*x--)). Ignoring sideâ€effects, we can write it as –~(++*x). Since the bitwise NOT operator satisfies ~n = –(n+1) for integers, we have   –~(++*x) = –[–((++*x)+1)] = (++*x) + 1.
Resulting in output: ++*x + 1 to explain briefly: the expression parses as –~(++(*x--)). ignoring sideâ€effects, we can write it as –~(++*x). since the bitwise not operator satisfies ~n = –(n+1) for integers, we have   –~(++*x) = –[–((++*x)+1)] = (++*x) + 1.
Testing if the previous output matches the regex \*x *\+ *2: False
Below is a student's answer to a question about what a particular C expression evaluates to: ++*x + 1 to explain briefly: the expression parses as –~(++(*x--)). ignoring sideâ€effects, we can write it as –~(++*x). since the bitwise not operator satisfies ~n = –(n+1) for integers, we have   –~(++*x) = –[–((++*x)+1)] = (++*x) + 1. 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