汇编学习(4), 整数，栈，浮点
本篇介绍
本篇介绍汇编的整数，浮点运算，还有栈。
整数运算
代码如下:
; icalc.asm
extern printf
section .data
number1 dq 128 ; the numbers to be used to
number2 dq 19 ; show the arithmetic
neg_num dq -12 ; to show sign extension
fmt db "The numbers are %ld and %ld",10,0
fmtint db "%s %ld",10,0
sumi db "The sum is",0
difi db "The difference is",0
inci db "Number 1 Incremented:",0
deci db "Number 1 Decremented:",0
sali db "Number 1 Shift left 2 (x4):",0
sari db "Number 1 Shift right 2 (/4):",0
sariex db "Number 1 Shift right 2 (/4) with "
db "sign extension:",0
multi db "The product is",0
divi db "The integer quotient is",0
remi db "The modulo is",0
section .bss
resulti resq 1
modulo resq 1
section .text
global main
main:
push rbp
mov rbp,rsp
; displaying the numbers
mov rdi, fmt
mov rsi, [number1]
mov rdx, [number2]
mov rax, 0
call printf
; adding---------------------------------------------------------------------
mov rax, [number1]
add rax, [number2] ; add number2 to rax
mov [resulti], rax ; move sum to result
; displaying the result
mov rdi, fmtint
mov rsi, sumi
mov rdx, [resulti]
mov rax, 0
call printf
; substracting---------------------------------------------------------------
mov rax, [number1]
sub rax, [number2] ; subtract number2 from rax
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, difi
mov rdx, [resulti]
mov rax, 0
call printf
; incrementing-----------------------------------------------------------------
mov rax, [number1]
inc rax ; increment rax with 1
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, inci
mov rdx, [resulti]
mov rax, 0
call printf
; decrementing-----------------------------------------------------------------
mov rax, [number1]
dec rax ; decrement rax with 1
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, deci
mov rdx, [resulti]
mov rax, 0
call printf
; shift arithmetic left------------------------------------------------------
mov rax, [number1]
sal rax, 2 ; multiply rax by 4
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, sali
mov rdx, [resulti]
mov rax, 0
call printf
; shift arithmetic right------------------------------------------------------
mov rax, [number1]
sar rax, 2 ; divide rax by 4
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, sari
mov rdx, [resulti]
mov rax, 0
call printf
; shift arithmetic right with sign extension ---------------------------------
mov rax, [neg_num]
sar rax, 2 ; divide rax by 4
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, sariex
mov rdx, [resulti]
mov rax, 0
call printf
; multiply-------------------------------------------------------------------
mov rax, [number1]
imul qword [number2] ; multiply rax with number2
mov [resulti], rax
; displaying the result
mov rdi, fmtint
mov rsi, multi
mov rdx, [resulti]
mov rax, 0
call printf
; divide---------------------------------------------------------------------
mov rax, [number1]
mov rdx, 0 ; rdx needs to be 0 before idiv
idiv qword [number2] ; divide rax by number2, modulo in rdx
mov [resulti], rax
mov [modulo], rdx ; rdx to modulo
; displaying the result
mov rdi, fmtint
mov rsi, divi
mov rdx, [resulti]
mov rax, 0
call printf
mov rdi, fmtint
mov rsi, remi
mov rdx, [modulo]
mov rax, 0
call printf
mov rsp,rbp
pop rbp
ret
结果如下:
The numbers are 128 and 19
The sum is 147
The difference is 109
Number 1 Incremented: 129
Number 1 Decremented: 127
Number 1 Shift left 2 (x4): 512
Number 1 Shift right 2 (/4): 32
Number 1 Shift right 2 (/4) with sign extension: -3
The product is 2432
The integer quotient is 6
The modulo is 14add，sub
可以用于有符号，也可以用于无符号，第二个操作数会加到第一个操作数上。对于有符号数，如果第一个操作数放不下结果，那么CF 标记会置位，对于无符号数，OF标记会置位，如果结果是0，ZF标记会置位，如果结果是负数，那么SF标记会置位。 sub 和add类似，无需重复。
sal，sar
算术左移，算术右移，在移动的 时候会考虑符号位，比如对于负数，在做sar的时候，就会在最高位补1。这儿会见证一次数字的神奇：
举一个例子， 比如一个8位的寄存器，需要表示 -4，那么结果如下：
0000 0100
=>
1111 1011
+ 1
1111 1100 （-4 的补码，也就是计算机中真正的存储形式）
算术右移1位
1111 1110 (-2)
也就是高位补一个符号位，就可以正确表示除以2的结果！imul, idiv
mul用于无符号数乘法，imul用于有符号数乘法，rax用来存放第一个参数，乘法会用到rdx，rax2个寄存器，rdx用来存放高64位，rax存放低64位。 对于除法，rax用来存放被除数，也用来存放结果的整数部分，rdx用来存放余数，因此使用idiv之前，需要将rdx清零。
栈
先看一个代码，反转字符串:
; stack.asm
extern printf
section .data
string db "ABCDE",0
stringLen equ $ - string-1 ; stringlength without 10 and 0
fmt1 db "The original string: %s",10,0
fmt2 db "The reversed string: %s",10,0
section .bss
section .text
global main
main:
push rbp
mov rbp,rsp
; Print the original string
mov rdi, fmt1
mov rsi, string
mov rax, 0
call printf
;push the string char per char on the stack
xor rax, rax
mov rbx, string ; address of string in rbx
mov rcx, stringLen ; length in rcx counter
mov r12, 0 ; use r12 as pointer
pushLoop:
; push char per char on the stack
mov al, byte [rbx+r12] ; move char into rax
push rax ;push rax on the stack
inc r12 ; increase char pointer with 1
loop pushLoop ; continue loop
;pop the string char per char from the stack
;this will reverse the original string
mov rbx, string ; address of string in rbx
mov rcx, stringLen ; length in rcx counter
mov r12, 0 ; use r12 as pointer
popLoop:
pop rax ; pop a char from the stack
mov byte [rbx+r12], al ;move the popped char into string
inc r12 ; increase char pointer with 1
loop popLoop ; continue loop
mov byte [rbx+r12],0 ; terminate string with 0
; Print the reversed string
mov rdi, fmt2
mov rsi, string
mov rax, 0
call printf
mov rsp,rbp
pop rbp
ret
结果：
The original string: ABCDE
The reversed string: EDCBA上面程序本质上就是按照栈的FILO规则，第一个loop先将字符串push到栈上，第二个loop再读出来，这样就把字符串反过来了。可以在第一个循环结束后打一个断点,显示rsp指向地址的内容
> i registers rsp
rsp 0x7fffffffe468 0x7fffffffe468
> x /5cg 0x7fffffffe468
0x7fffffffe468: 69 'E' 68 'D'
0x7fffffffe478: 67 'C' 66 'B'
0x7fffffffe488: 65 'A'浮点
首先看下浮点在计算机中的表示： 32位单精度 1 个符号位，8个指数位，23个分数位 64位双精度 1个符号位，11个指数位，52个分数位
对于指数位，还有一个偏移值，对于单精度，偏移值是127，双精度偏移值是1023。接下来看一个例子：
1101010.01011 (106 + 1/4 + 1/16 + 1/32 = 106 + 11/32)
=1.0101001011 x 2^6
由于是正数，符号位是0，指数位是 6 + 127 = 133，那么表示结果就是：
0 10000101 01010010110000000000000接下来看一点浮点的代码:
; fcalc.asm
extern printf
section .data
number1 dq 9.0
number2 dq 73.0
fmt db "The numbers are %f and %f",10,0
fmtfloat db "%s %f",10,0
f_sum db "The float sum of %f and %f is %f",10,0
f_dif db "The float difference of %f and %f is %f",10,0
f_mul db "The float product of %f and %f is %f",10,0
f_div db "The float division of %f by %f is %f",10,0
f_sqrt db "The float squareroot of %f is %f",10,0
section .bss
section .text
global main
main:
push rbp ; necessary because of printf with float
mov rbp,rsp
; print the numbers
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rdi,fmt
mov rax,2 ; two floats
call printf
; sum
movsd xmm2, [number1] ; double precision float into xmm
addsd xmm2, [number2] ; add into to xmm
; print the result
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rdi,f_sum
mov rax,3 ; one float
call printf
; difference
movsd xmm2, [number1] ; double precision float into xmm
subsd xmm2, [number2] ; subtract from xmm
; print the result
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rdi,f_dif
mov rax,3 ; three floats
call printf
; multiplication
movsd xmm2, [number1] ; double precision float into xmm
mulsd xmm2, [number2] ; multiply with xmm
; print the result
mov rdi,f_mul
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rax,3 ; three floats
call printf
; division
movsd xmm2, [number1] ; double precision float into xmm
divsd xmm2, [number2] ; dividdatae xmm0
; print the result
mov rdi,f_div
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rax,1 ; one float
call printf
; squareroot
sqrtsd xmm1, [number1] ; squareroot double precision float in xmm
; print the result
mov rdi,f_sqrt
movsd xmm0, [number1]
mov rax,2 ; one float
call printf
; exit
mov rsp,rbp
pop rbp ; undo the push at the beginning
ret
mov rdi,f_sum
mov rax,3 ; one float
call printf
; difference
movsd xmm2, [number1] ; double precision float into xmm
subsd xmm2, [number2] ; subtract from xmm
; print the result
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rdi,f_dif
mov rax,3 ; three floats
call printf
; multiplication
movsd xmm2, [number1] ; double precision float into xmm
mulsd xmm2, [number2] ; multiply with xmm
; print the result
mov rdi,f_mul
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rax,3 ; three floats
call printf
; division
movsd xmm2, [number1] ; double precision float into xmm
divsd xmm2, [number2] ; divide xmm0
; print the result
mov rdi,f_div
movsd xmm0, [number1]
movsd xmm1, [number2]
mov rax,1 ; one float
call printf
; squareroot
sqrtsd xmm1, [number1] ; squareroot double precision float in xmm
; print the result
mov rdi,f_sqrt
movsd xmm0, [number1]
mov rax,2 ; one float
call printf
; exit
mov rsp,rbp
pop rbp ; undo the push at the beginning
ret
结果如下：
The numbers are 9.000000 and 73.000000
The float sum of 9.000000 and 73.000000 is 82.000000
The float difference of 9.000000 and 73.000000 is -64.000000
The float product of 9.000000 and 73.000000 is 657.000000
The float division of 9.000000 by 73.000000 is 0.123288
The float squareroot of 9.000000 is 3.000000对于浮点，参数传递用的就是xmm0，xmm1,等，这时候调用printf时指定的rax值就作用出来了，会告诉系统从哪里获取参数。 movsd 用于双精度，movss用于单精度。
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原始发表：2022-12-03，如有侵权请联系 cloudcommunity@tencent.com 删除
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