将两个 4 位数字合并为一个 8 位数字

Question

所以问题是：
有2个字符。
我需要构建一个像这样构建的 8 位数字：
第一个数左4位，第二个数右4位

编辑：如果数字将放在 al 中，位应该是这样的：

al 位 0 到 3 = 第二个字符的最低 4 位。 al 位 4 到 7 = 第一个字符的最高 4 位。

我试过将数字右移 4 位以获得左 4 位。为了得到右边的 4 位，我试图将数字的左边 4 位变为 0。然后我把ax的左4位加起来，左移4次，再把左4位加起来。

    mov dl,[si]       ; the value of the character, it is inside of a char array
    shr dl,4
    add al,dl
    and dl,00001111b
    shl ax,4          ; ax value was 0
    inc si
    mov dl,[si]
    and dl,00001111b
    add al,dl
    shl ax,4

我认为这应该有效，但显然无效。

我该怎么做？

Answer 1

I need to build a 8-bit number that is built like this: the left 4 bits from the first number, the right 4 bits from the second number

不知道你是否想要这样的东西：

    mov ax,[si]     ;al = first character, ah = second character
    shl al,4        ;al bits 4 to 7 = lowest 4 bits of first character
    shr ax,4        ;al bits 0 to 3 = lowest 4 bits of first character, al bits 4 to 7 = lowest 4 bits of second character

..或类似这样的东西：

    mov ax,[si]     ;al = first character, ah = second character
    and ax,0xF00F   ;al bits 0 to 3 = lowest 4 bits of first character, ah bits 4 to 7 = highest 4 bits of second character
    or al,ah        ;al bits 0 to 3 = lowest 4 bits of first character, al bits 4 to 7 = highest 4 bits of second character

..或类似这样的东西：

    mov ax,[si]     ;al = first character, ah = second character
    and ax,0x0FF0   ;al bits 4 to 7 = highest 4 bits of first character, ah bits 0 to 3 = lowest 4 bits of second character
    or al,ah        ;al bits 0 to 3 = lowest 4 bits of second character, al bits 4 to 7 = highest 4 bits of first character

..或其他。

将两个 4 位数字合并为一个 8 位数字

Merge two 4-bit numbers into an 8-bit number

assembly

x86-16