I understand that standards before C++20 do not specifiy the representation of signed integers. Therefore, when left shifting a negative number, the result is undefined as it depends on the underlying representation. However, why when it comes to right shift, the behavior becomes implementation-defined instead of undefined? Doesn't the result still depend on the underlying representation?

I understand that standards before C++20 do not specifiy the representation of signed integers. Therefore, when left shifting a negative number, the result is undefined as it depends on the underlying representation. However, why when it comes to right shift, the behavior becomes implementation-defined instead of undefined? Doesn't the result still depend on the underlying representation?

• c++

• 1 that it is "implementation defined" does not imply that it would not depend on the underlying representation. It only means that the implementation must define and document the behavior (in accordance with the underlying representation) – 463035818_is_not_an_ai Commented 2 days ago
• 1 I presume... compiler vendors prefer the behavior to be implementation defined because having a C++ defined behavior can thwart platform specific optimizations. C++20 did not want to incur making compiler vendors and their customers irate by making an adversely performance impacting language defined behavior. Another area that continues to be UB is signed integer wrap-around, which could be well-defined now that integers are 2s complement, but that would also thwart existing optimizations. – Eljay Commented 2 days ago
• 1 related/dupe: stackoverflow/questions/3784996/… – NathanOliver Commented 2 days ago
• @user18676624 I think that the problem is that it is unclear whether the result value will be negative or positive. Neither can be said apriori. – Vlad from Moscow Commented 2 days ago

1 Answer 1

In C++ for signed integers when on a left shift occurs and a sign bit 1 was shifted out, the behaviour is undefined.

0b000xyz << 3 == 0bxyz
0b001xyz << 3 is undefined
0b1xyz   << 1 is undefined

For shift right the sign bit was repeatedly shifted in, the so called artithmetic bit shift. Except for -1 this is a division by 2 (-4 to -2 to -1, 4 to 2 to 1).

For the left shift one has for positive numbers this is a multiplication by 2 (1 to 2 to 4). But for negative numbers theoretically do -1 to -2 to -4 and it would behave (multiplication by 2), but if 0b10xyz << 3 happens there is a sign overflow and the choice 0b0xyz or 0b1xyz. In general a desirable behaviour cannot be found.

Hence a single test on the sign bit and then claiming "undefined" is a minimal checkable case.