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Shafik Yaghmour

Retro C++ quiz #57

Given the following in C++:

#include <cstdint>
#include <limits>

int main() {
int64_t s=32;
int64_t x=std::numeric_limits<int32_t>::max();

1 << s; // A
1 + x; // B
}

Without checking, assuming LP64, which invokes undefined behavior:

A. Both
B. None
C. A
D. B

#Cplusplus
#Cpppolls

A

36.4%

B

27.3%

C

31.8%

D

4.5%

Poll ended at Mar 17 at 4:52pm.

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Shafik Yaghmour Mar 17

The answer is C.

Usually we expect the operands of binary operators to undergo the usual arithmetic conversions but this does not happen for shift operators, only integral promotions: [expr.shift]p1 http://eel.is/c++draft/expr.shift#1.sentence-2

The type of the result is the promoted left operand which in this case would be int: [conv.prom] http://eel.is/c++draft/conv.prom

So we have undefined behavior since the right operand is greater than the bit-width of int in this case: [expr.shift]p1 http://eel.is/c++draft/expr.shift#1.sentence-4

For additive operators the usual arithmetic conversions are performed: [expr.add]p1 http://eel.is/c++draft/expr.add#1

The left operand which is an int is converted to int64_t since they are both signed and int64_t has a greater rank: [expr.arith.conv]p1.5.2 http://eel.is/c++draft/expr.arith.conv#1.5.2

There is no overflow and hence no undefined behavior.