在对 NASM 浮点运算的简单测试中，我开始收集数据以求解两个线性方程组。但是，应该采用 a1 * a4 的程序的输出太长而无法理解。我输入了个位数的整数作为我的数字，但仍然得到了一个遍历多个控制台行的结果。我究竟做错了什么？
我的代码：
extern printf
extern scanf
extern exit
global main
SECTION .data
message1: db "Enter the value of ",0
welcome: db "Welcome to the linear equation solver.",10,10,0
eq1:db " a1 * x + a2 * y = b1",10,0
eq2:db " a3 * x + a4 * y = b2",10,10,0
formats: db "%s",0
formatss: db "%s %s: ",0
formatf: db "%f",0
a1: db "a1",0
a2: db "a2",0
a3: db "a3",0
a4: db "a4",0
b1: db "b1",0
b2: db "b2",0
SECTION .bss
float1: resd 1
float2: resd 1
float3: resd 1
float4: resd 1
float5: resd 1
float6: resd 1
a1a4: resd 1
a2a3: resd 1
SECTION .text
main:
;Welcome to the linear equation solver
push welcome
push formats
call printf
add esp, 8
;show equation type
push eq1
push formats
call printf
add esp, 8
push eq2
push formats
call printf
add esp, 8
;get a1
push a1
push message1
push formatss
call printf
add esp, 12
push float1
push formatf
call scanf
add esp, 8
;get a2
push a2
push message1
push formatss
call printf
add esp, 12
push float2
push formatf
call scanf
add esp, 8
;get b1
push b1
push message1
push formatss
call printf
add esp, 12
push float5
push formatf
call scanf
add esp, 8
;get a3
push a3
push message1
push formatss
call printf
add esp, 12
push float3
push formatf
call scanf
add esp, 8
;get a4
push a4
push message1
push formatss
call printf
add esp, 12
push float4
push formatf
call scanf
add esp, 8
;get b2
push b2
push message1
push formatss
call printf
add esp, 12
push float6
push formatf
call scanf
add esp, 8
;Initiate floats
FINIT
;Bring a1 into st float register.
fld dword [float1] ;ST0
fmul dword [float4] ;ST0 *= float4
fstp dword [a1a4] ;Store result: a1 * a4, pop float stack
;print a1 * a4
fld dword [a1a4]
sub esp, 8
fstp qword [esp]
push formatf
call printf
add esp, 12
call exit