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Undefined vs. Implementation-Defined Behavior in Bitwise Shifts
The bitwise left shift operator (<<) raises questions regarding its behavior with negative left-hand operands. This article explores why this operation yields undefined behavior in C but only implementation-defined behavior in C .
C's Undefined Behavior
According to ISO C99, left-shifting a negative operand results in undefined behavior. This is because the standard dictates that the result should be the product of the left operand and 2E2, with the result modulo the range of the result type. However, when the left operand is negative, such a calculation may lead to undefined results in signed type contexts.
C 's Well-Defined Behavior
In contrast to C, C specifies that left-shifting an unsigned operand yields the expected mathematical multiplication. However, for signed types, C also defines undefined behavior if the result exceeds the representable range of the result type. Thus, in C , left-shifting negative operands remains undefined.
Reasons for the Divergence
The reason for C 's different approach likely stems from the fact that its negative behavior is already undefined due to the potential overflow. Extending the definition to cover all negative cases simplified the standard and clarified undefined behavior.
Implementation-Defined Right-Shift Behavior
Right-shifting a negative operand is implementation-defined in both C and C because it can involve a choice between sign extension and zero filling. Sign extension retains the original sign bit in the vacated bits, while zero filling replaces them with zeros. Different compilers and platforms may choose either behavior, hence the implementation-defined nature.
In Summary
In C, left-shifting a negative operand unconditionally results in undefined behavior. In C , such an operation is also undefined for signed types. Right-shifting a negative operand, on the other hand, is implementation-defined in both languages due to the choice between sign extension and zero filling.
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