Why is Left Shifting Negative Numbers Undefined in C but Defined (with caveats) in C ?

Why is Left Shift Undefined on Negative Values in C but Well-Defined in C ?

The ISO C99 and C standard specify distinct behavior for left shift operations on negative values. In C, shifting a negative value invokes undefined behavior (UB) per the ISO C99 standard section 6.5.7/4. This is due to the definition only clearly stating the result for nonnegative values, leaving the behavior undefined for negative values.

However, the ISO C -03 standard (section 5.8/2) explicitly defines the behavior for signed types. If the shifted value can be represented within the result type, it is returned. Otherwise, undefined behavior is invoked.

Why is Undefined Behavior for Left Shift on Negatives?

The ISO C99 definition of the left shift operation doesn't specify what happens with the high-order bits of a negative value when shifted. This leaves it up to the implementation to decide what to do, which can result in different or incorrect results. To ensure consistent behavior across compilers, undefined behavior is the default.

Why is Implementation-Defined Behavior for Right Shift on Negatives?

In contrast to the left shift operation, the right shift operation doesn't have to deal with a truncation from the left. As a result, it's easier for implementations to define a consistent behavior for negative values. This behavior can vary depending on the compiler and system, but it is usually either arithmetic (similar to signed right shift in C ) or logical (the result is zero).

Conclusion

The reason for the different behavior of left and right shift operations on negative values in C and C is largely due to the ease of implementation. Right shift can be consistently defined for negative values without significant complexity, while left shift must either truncate bits or specify a non-portable result, which would lead to inconsistent behavior across compilers.

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