Is shifting the signed bit of a signed short undefined behaviour in C?

Question

I have heard that shifting into signed bit of an integer, i.e.

int test = INT_MAX;
test = (test<<1) + 1;

is undefined behaviour due to test being greater than INT_MAX. But will this behaviour be encountered in signed short variables, i.e.

short test1 = SHRT_MAX;
test1 = (test1<<1) + 1;

?

At the moment I have not come across any relevant documentation.

EDIT: I know that undefined behaviour will result in the case of integers, but not of short.

@Sander I wouldn't call it a duplicate, because shorts will be promoted to ints before shifting, which adds additional complexity here. — interjay
– interjay, Commented Dec 12, 2018 at 15:30
The problem behind the scenes is that the C shift operator is a 'logical shift', meaning that it doesn't care of sign handling. The promotion between integers follow well known rules that in the specific case, where a MAX_INT >= MAX_SHORT should always be true, should guarantee a correct representation of a short in an int. But when it come to the shift the behavior depends on machine and specific instructions coded by the compiler. I.e. a negative number can become positive right shifted, and the opposite for left shift. But what if CPU uses SAR/SAL arithmetic shifts instead? — Frankie_C
– Frankie_C, Commented Dec 12, 2018 at 16:33
Kind of related Why must a short be converted to an int before arithmetic operations in C and C++? — Shafik Yaghmour
– Shafik Yaghmour, Commented Dec 12, 2018 at 17:10

interjay · Accepted Answer · 2018-12-12 15:45:48Z

6

When a small integer type is used in most types of expressions, it is promoted to an int (if int can fit all values of the original type), and the expression is then performed on the int.

This gives two possibilites:

On a platform where int is the same size as short, the shift will be undefined behavior because it shifts into the int's sign bit. (More accurately, the reason is that the result cannot be represented as an int). See C11 6.5.7p4.
On a platform where int has more bits than short, the shift itself will be successful (though it can be undefined if you're shifting by more than 1). However, when you assign the result back into a short variable, the value will not be able to be represented as a short. The actual value that will be assigned is implementation-defined. See C11 6.3.1.3p3.

edited Dec 12, 2018 at 15:45

answered Dec 12, 2018 at 15:39

interjay

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