C++ Test - Does sign bit 1 always mean negative?
c++, number-formatting
Solution
I would say it is undefined (and not implementation-defined) for a different reason.
From 5.8:3
The value of `E1 << E2` is E1 left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 × 2E2 , reduced modulo one more than the maximum value representable in the result type. Otherwise, if E1 has a signed type and non-negative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.
Problem
I was just taking a C++ test and I got the following question wrong: Q: What is the output of the following program? ``` #include <iostream> #include <stdint.h> using namespace std; int main() { int a = 0; for (int8_t i = 1; i > 0; i <<= 1) a++; cout << a; return 0; } ``` There were the following answers to choose from - 8 - 7 - Undefined Behavior - Compile Error The "correct" answer was 7. If there was "Implementation-Defined Behavior" in the answers, I would choose that, so I chose Undefined Behavior which was sort of the closest. I understand that in sign-and-magnitute, 1's complement, and 2's complement the answer will be 7. But doesn't the C++ standard theoretically allow any other number representations? For example, sign and magnitude, but 0 means negative? Am I correct in that the real correct answer to this question should be Implementation-Defined Behavior, and if not, could you please explain why the answer is 7 regardless of the implementation? I read the comments to the question and it appears that initially the type of `a` was char, which apparently had raised a lot of complaints about whether `char` is signed or not, so the testsetter changed it to int8_t. As a bonus question, is `<stdint.h>` part of C++? O_O