Why does left shift operation invoke Undefined Behaviour when the left side operand has negative value?

Question

In C bitwise left shift operation invokes Undefined Behaviour when the left side operand has negative value.

Relevant quote from ISO C99 (6.5.7/4)

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

But in C++ the behaviour is well defined.

ISO C++-03 (5.8/2)

The value of E1 << E2 is E1 (interpreted as a bit pattern) left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 multiplied by the quantity 2 raised to the power E2, reduced modulo ULONG_MAX+1 if E1 has type unsigned long, UINT_MAX+1 otherwise. [Note: the constants ULONG_MAXand UINT_MAXare defined in the header ). ]

That means

int a = -1, b=2, c; c= a << b ; 

invokes Undefined Behaviour in C but the behaviour is well defined in C++.

What forced the ISO C++ committee to consider that behaviour well defined as opposed to the behaviour in C?

On the other hand the behaviour is implementation defined for bitwise right shift operation when the left operand is negative, right?

My question is why does left shift operation invoke Undefined Behaviour in C and why does right shift operator invoke just Implementation defined behaviour?

P.S : Please don't give answers like "It is undefined behaviour because the Standard says so". :P

Answer 1

The paragraph you copied is talking about unsigned types. The behavior is undefined in C++. From the last C++0x draft:

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 × 2^E2, 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 E1×2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

EDIT: got a look at C++98 paper. It just doesn't mention signed types at all. So it's still undefined behavior.

Right-shift negative is implementation defined, right. Why? In my opinion: It's easy to implementation-define because there is no truncation from the left issues. When you shift left you must say not only what's shifted from the right but also what happens with the rest of the bits e.g. with two's complement representation, which is another story.

Answer 2

In C bitwise left shift operation invokes Undefined Behaviour when the left side operand has negative value. [...] But in C++ the behaviour is well defined. [...] why [...]

The easy answer is: Becuase the standards say so.

A longer answer is: It has probably something to do with the fact that C and C++ both allow other representations for negative numbers besides 2's complement. Giving fewer guarantees on what's going to happen makes it possible to use the languages on other hardware including obscure and/or old machines.

For some reason, the C++ standardization committee felt like adding a little guarantee about how the bit representation changes. But since negative numbers still may be represented via 1's complement or sign+magnitude the resulting value possibilities still vary.

Assuming 16 bit ints, we'll have

 -1 = 1111111111111111  // 2's complement  -1 = 1111111111111110  // 1's complement  -1 = 1000000000000001  // sign+magnitude 

Shifted to the left by 3, we'll get

 -8 = 1111111111111000  // 2's complement -15 = 1111111111110000  // 1's complement   8 = 0000000000001000  // sign+magnitude 

What forced the ISO C++ committee to consider that behaviour well defined as opposed to the behaviour in C?

I guess they made this guarantee so that you can use << appropriately when you know what you're doing (ie when you're sure your machine uses 2's complement).

On the other hand the behaviour is implementation defined for bitwise right shift operation when the left operand is negative, right?

I'd have to check the standard. But you may be right. A right shift without sign extension on a 2's complement machine isn't particularly useful. So, the current state is definitely better than requiring vacated bits to be zero-filled because it leaves room for machines that do a sign extensions -- even though it is not guaranteed.