问题
I'm having a problem adding to a product of a 16 bit multiplication. I want to multiply a year such as 2015, by 365 to do so I
mov dx, 0 ; to clear the register
mov ax, cx ; cx holds the year such as 2015
mov dx, 365 ; to use as multiplier
mul dx ; multiply dx by ax into dx:ax
After checking the registers, I am getting the correct solution but is there a way so that I can store this product into a single register. I want to add separate values to the product and so I would like to move this product into a single 32 bit register. Thanks for the help!
回答1:
The usual method is to use a 32 bit multiply to start with. It's especially easy if your factor is a constant:
movzx ecx, cx ; zero extend to 32 bits
; you can omit if it's already 32 bits
; use movsx for signed
imul ecx, ecx, 365 ; 32 bit multiply, ecx = ecx * 365
You can of course also combine 16 bit registers, but that's not recommended. Here it is anyway:
shl edx, 16 ; move top 16 bits into place
mov dx, ax ; move bottom 16 bits into place
(There are other possibilities too, obviously.)
回答2:
mov dx, 0 ; to clear the register
mov ax, cx ; cx holds the year such as 2015
mov dx, 365 ; to use as multiplier
mul dx ; multiply dx by ax into dx:ax
You can start by simplifying this code (You don't need to clear any register before doing the multiplication):
mov ax, 365
mul cx ; result in dx:ax
Next to answer your title question and have the result DX:AX moved into a 32-bit register like EAX you could write:
push dx
push ax
pop eax
来源：https://stackoverflow.com/questions/33181034/moving-from-dxax-register-to-single-32-bit-register