Does x86 instruction SHL/SHR actually rotate the bits?

Question

I was debugging a program. When the register di=0x01, cl=0x11, The instruction "shl di,cl" will actually make di=0x1.

Shouldn't the bits that slide off the end disappear?

Answer 1

(Disregarding the example in the question as it clearly has typos.)

The higher bits of the shift amount are ignored, unless you're running your program on an original 8086: See "IA-32 Architecture Compatibility" at the bottom of the page.

Answer 2

Full disclosure: I am the author of the following article.

Accessing and Modifying Upper Bits in x86 and x64 Registers

The aim of the article is to provide the reader with a detailed, example- and code-laden explanation of exactly how the shift and rotate instructions work in both x86 and x64 contexts. The examples therein are directly applicable to OP's inquiry.

Answer 3

SHL and SHR work using the following syntax dest, count

now di is a 16 bit register and CL is basically the count (3)

so you are right it should be zeroed out and you can prove this by using a online x86 emulator

http://carlosrafaelgn.com.br/asm86/

and enter the following asm instructions: mov di, 01h mov cl, 11h shl di, cl

so i dont know what would make it behave this way.