16位MUL组件

Question

On x86 16-bit assembly, if there is a 1 in ax and 0x10 ( 16 ) in cl , the following code doesn't put 0x1000 in cl :

mul cl  ; ax = 0x  10
mul cl  ; ax = 0x 100
mul cl  ; ax = 0x0000 (not 0x1000)

Why doesn't this code work as expected?

Answer 1

Your source is 8 bits ( cl ) so the multiplication is performed over al only.

Use mul cx instead.

Explanation: If ax = 100h , then al = 00h . Since the result of mul cl is placed in ax , what you do is basically to replace the content of ax with 00h*cl , which is 00h .

Opcode MUL

CPU: i8086+ Type of Instruction: User

Affected FLags: CF, OF, AF, PF, SF, ZF

Instruction: MUL src

Note: Unsigned multiply of the accumulator by the source. If "src" is a byte value, then AL is used as the other multiplicand and the result is placed in AX. If "src" is a word value, then AX is multiplied by "src" and DX:AX receives the result. If "src" is a double word value, then EAX is multiplied by "src" and EDX:EAX receives the result. The 386+ uses an early out algorythm which makes multiplying any size value in EAX as fast in the 8 or 16 bit registers.

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Clocks (i486): MUL reg8 13-18 MUL reg16 13-26 MUL reg32 13-42 MUL mem8 13-18 MUL mem16 13-26 MUL mem32 13-42