在x86 NASM中使用atof函数 [英] using atof function in x86 NASM
问题描述
我有一些麻烦得到c函数atof（）在我的asm程序中工作。我试图从键盘上读取4个数字，最终打印出他们的平均值。然而，在我能做到这一点之前，我需要将数字转换为浮点数。我坚持成功让我的'总'变量工作。我已经尝试过在多个地方调用atof无济于事。
这是一个x86 NASM程序
pre > ; nasm -f elf -l prg2.lst prg2.asm
; gcc -o prg2 prg2.o
; ./prg2
SECTION .DATA
提示DB'输入测试分数'，13,10
fmt DB％s，0
fmtf DB％f，0
SECTION .bss
test1 resb 1000;保留变量名称到
test2 resb 1000;把东西放入
test3 resb 1000
test4 resb 1000
total resb 1000
SECTION .code
extern printf
extern scanf
extern atof
global main
main：
push ebp
mov ebp，esp
push prompt
call printf
add esp，4;提示用户
push test1; push test1变量
push fmt
call scanf
add esp，8; store test1变量
推送提示符
调用printf
add esp，4;提示用户
$ b $推测试2;推测试2变量
推送文件
调用scanf
add esp，8;存储test2变量
推送提示
调用printf
添加esp，4;提示用户
push test3; push test3变量
push fmt
call scanf
add esp，8; store test3变量
推送提示
call printf
add esp，4;提示用户
push test4; push test4变量
push fmt
call scanf
add esp，8; store test4变量
mov eax，[test1]
add eax，[test2]
add eax，[test3]
add eax，[test4]
call atof
mov [total]，eax
push total
call printf;不打印我想要的，
add esp，4;或者打印任何东西在所有
push test1;打印分数验证
call printf
add esp，4
push test2
call printf
add esp，4
push test3
call printf
add esp，4
push test4
call printf
add esp，4
mov esp，ebp
pop ebp
ret
编辑：修改后，我是一个可以使用这些代码块将输入的值转换为各自的数值。
mov eax，0;
add eax，[test1];把test1的值放入eax
mov [total]，eax
sub eax，'0'
add eax，[test2]
mov [total]，eax
sub eax，'0'
add eax，[test3]
mov [total]，eax
sub eax ，'0'
add eax，[test4];
mov [total]，eax
sub eax，'0'
push total
call printf
add esp，4
运行示例：
./ prg2b
输入一个测试分数。
1
输入一个测试分数。
1
输入一个测试分数。
1
输入一个测试分数。
1
41111
这个除了我的代码可以摆脱我的问题atof（）调用，但是只有当数字是一位数字，并且如果总数是< 10
，则它才是成功的如果任何人都可以提供关于如何正确使用atof，或如何正确地转换为使用scanf的程序中的浮点数，将不胜感激。我很新（阅读：2周的学习）到x86 asm。这是在UNIX系统的终端中编译的。您可以使用反引号在NASM中定义带有转义序列的C文本。例如。
提示DB`输入测试分数。不要忘记最后一个0
atof 需要堆栈中的内存地址，并将结果返回到FPU寄存器ST（0）。
SECTION .data
prompt DB你必须把每一个字符串转换成一个数字。 `输入一个测试分数\\\
`，0
fmt DB％s，0
fmtf DB`Sum：％f\\\
`，0
SECTION。 bss
test1 resb 1000
test2 resb 1000
test3 resb 1000
test4 resb 1000
double1 resq 1;预留Quadword = Double
double2 resq 1
double3 resq 1
double4 resq 1
sum resq 1
SECTION .code
extern printf， scanf，atof
全局主
主：
push ebp; Prolog
mov ebp，esp
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push test1
push fmt; ％s
call scanf
add esp，（2 * 4）; Pop 2 dword
push test1
call atof
fstp qword [double1]
add esp，（1 * 4）;流行1 dword
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push test2
push fmt; ％s
call scanf
add esp，（2 * 4）;流行2个单词
push test2
call atof
fstp qword [double2]
add esp，（1 * 4）;流行1 dword
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; pop 1 dword
push test3
push fmt; ％s
call scanf
add esp，（2 * 4）; Pop 2 dword
push test3
call atof
fstp qword [double3]
add esp，（1 * 4）;流行1 dword
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push test4
push fmt; ％s
call scanf
add esp，（2 * 4）; Pop 2 dword
push test4
call atof
fstp qword [double4]
add esp，（1 * 4）; Pop 1 dword
fld qword [double1]
fadd qword [double2]
fadd qword [double3]
fadd qword [double4]
fstp qword [ sum]
push dword [sum + 4];双推两步
push dword [sum + 0]
push fmtf; `result：％f\\\
`，0
call printf
add esp，（3 * 4）;流行音乐3 dwords
mov esp，ebp; Epilog
pop ebp
ret
您不需要 ATOF 。您可以让 scanf 将输入的字符串转换为格式字符串％lf。
SECTION .data
prompt DB`输入一个测试分数\\\
`，0
fmt DB％lf，0; scanf需要'lf'来存储双
fmtf DB`Sum：％f\\\
`，0; printf只需要'f'来打印一个double
SECTION .bss
double1 resq 1;预留Quadword = Double
double2 resq 1
double3 resq 1
double4 resq 1
sum resq 1
SECTION .code
extern printf， scanf，atof
全局主
主：
push ebp; Prolog
mov ebp，esp
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push double1
push fmt; ％lf
call scanf
add esp，（2 * 4）;流行2个单词
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push double2
push fmt; ％lf
call scanf
add esp，（2 * 4）;流行2个单词
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push double3
push fmt; ％lf
call scanf
add esp，（2 * 4）;流行2个单词
推送提示; `输入测试分数\ $`
call printf
add esp，（1 * 4）; Pop 1 dword
push double4
push fmt; ％lf
call scanf
add esp，（2 * 4）; Pop 2 dword
fld qword [double1]
fadd qword [double2]
fadd qword [double3]
fadd qword [double4]
fstp qword [ sum]
push dword [sum + 4];双推两步
push dword [sum + 0]
push fmtf; `result：％f\\\
`，0
call printf
add esp，（3 * 4）;流行音乐3 dwords
mov esp，ebp; Epilog
pop ebp
ret
I am having some trouble getting the c function atof() to work in my asm program. I'm trying to read in 4 numbers from the keyboard, and ultimately print their average. Before i can do that, however, i need to convert the numbers to floats. I'm stuck on successfully getting my 'total' variable to work. I have tried calling atof in multiple spots to no avail.
This is a x86 NASM program
; nasm -f elf -l prg2.lst prg2.asm
; gcc -o prg2 prg2.o
; ./prg2
SECTION .DATA
prompt DB 'enter a test score.', 13,10
fmt DB "%s",0
fmtf DB "%f",0
SECTION .bss
test1 resb 1000 ;reserves variable names to
test2 resb 1000 ;put stuff in
test3 resb 1000
test4 resb 1000
total resb 1000
SECTION .code
extern printf
extern scanf
extern atof
global main
main:
push ebp
mov ebp, esp
push prompt
call printf
add esp, 4 ;prompt user
push test1 ;push test1 variable
push fmt
call scanf
add esp, 8 ;store test1 variable
push prompt
call printf
add esp, 4 ;prompt user
push test2 ;push test2 variable
push fmt
call scanf
add esp, 8 ;store test2 variable
push prompt
call printf
add esp, 4 ;prompt user
push test3 ;push test3 variable
push fmt
call scanf
add esp, 8 ;store test3 variable
push prompt
call printf
add esp, 4 ;prompt user
push test4 ;push test4 variable
push fmt
call scanf
add esp, 8 ;store test4 variable
mov eax,[test1]
add eax,[test2]
add eax,[test3]
add eax,[test4]
call atof
mov [total], eax
push total
call printf ;not printing what i want,
add esp,4 ;or printing anything at all
push test1 ;printing scores for verification
call printf
add esp, 4
push test2
call printf
add esp, 4
push test3
call printf
add esp, 4
push test4
call printf
add esp, 4
mov esp, ebp
pop ebp
ret
EDIT: upon revision, i was able to turn the inputted values in their respective numeric values using these code blocks
mov eax, 0 ;
add eax,[test1] ;put test1 value in eax
mov [total], eax
sub eax, '0'
add eax,[test2]
mov [total], eax
sub eax,'0'
add eax,[test3]
mov [total], eax
sub eax,'0'
add eax,[test4] ;
mov [total], eax
sub eax,'0'
push total
call printf
add esp, 4
Sample run through:
./prg2b
enter a test score.
1
enter a test score.
1
enter a test score.
1
enter a test score.
1
41111
this addition to my code gets rid of my problem with the atof() call, but it is only successful if the numbers are one digit and if total is <10
If anyone could give a hint as to how to properly using atof, or how to properly convert to floating point numbers in a program that uses scanf, it would be greatly appreciated. I'm very new (read: 2 weeks of learning) to x86 asm. This is compiled in the terminal on a UNIX system
You can define a C literal with escape sequences in NASM by using backticks. E.g.
prompt DB `enter a test score.\n`, 0 ; Don't forget the last 0
atof needs a memory address on the stack and returns the result in register ST(0) of the FPU. You have to convert every single string to a number before you can calculate with it.
SECTION .data
prompt DB `Enter a test score\n`, 0
fmt DB " %s", 0
fmtf DB `Sum: %f\n`, 0
SECTION .bss
test1 resb 1000
test2 resb 1000
test3 resb 1000
test4 resb 1000
double1 resq 1 ; Reserve Quadword = Double
double2 resq 1
double3 resq 1
double4 resq 1
sum resq 1
SECTION .code
extern printf, scanf, atof
global main
main:
push ebp ; Prolog
mov ebp, esp
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push test1
push fmt ; " %s"
call scanf
add esp, (2*4) ; Pop 2 dwords
push test1
call atof
fstp qword [double1]
add esp, (1*4) ; Pop 1 dword
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push test2
push fmt ; " %s"
call scanf
add esp, (2*4) ; Pop 2 dwords
push test2
call atof
fstp qword [double2]
add esp, (1*4) ; Pop 1 dword
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push test3
push fmt ; " %s"
call scanf
add esp, (2*4) ; Pop 2 dwords
push test3
call atof
fstp qword [double3]
add esp, (1*4) ; Pop 1 dword
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push test4
push fmt ; " %s"
call scanf
add esp, (2*4) ; Pop 2 dwords
push test4
call atof
fstp qword [double4]
add esp, (1*4) ; Pop 1 dword
fld qword [double1]
fadd qword [double2]
fadd qword [double3]
fadd qword [double4]
fstp qword [sum]
push dword [sum + 4] ; Push a double in two steps
push dword [sum + 0]
push fmtf ; `result: %f\n`, 0
call printf
add esp, (3*4) ; Pop 3 dwords
mov esp, ebp ; Epilog
pop ebp
ret
You don't need atof. You can let scanf convert the inputted string with the format string " %lf".
SECTION .data
prompt DB `Enter a test score\n`, 0
fmt DB " %lf", 0 ; scanf needs 'lf' to store a double
fmtf DB `Sum: %f\n`, 0 ; printf needs only 'f' to print a double
SECTION .bss
double1 resq 1 ; Reserve Quadword = Double
double2 resq 1
double3 resq 1
double4 resq 1
sum resq 1
SECTION .code
extern printf, scanf, atof
global main
main:
push ebp ; Prolog
mov ebp, esp
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push double1
push fmt ; " %lf"
call scanf
add esp, (2*4) ; Pop 2 dwords
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push double2
push fmt ; " %lf"
call scanf
add esp, (2*4) ; Pop 2 dwords
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push double3
push fmt ; " %lf"
call scanf
add esp, (2*4) ; Pop 2 dwords
push prompt ; `enter a test score\n`
call printf
add esp, (1*4) ; Pop 1 dword
push double4
push fmt ; " %lf"
call scanf
add esp, (2*4) ; Pop 2 dwords
fld qword [double1]
fadd qword [double2]
fadd qword [double3]
fadd qword [double4]
fstp qword [sum]
push dword [sum + 4] ; Push a double in two steps
push dword [sum + 0]
push fmtf ; `result: %f\n`, 0
call printf
add esp, (3*4) ; Pop 3 dwords
mov esp, ebp ; Epilog
pop ebp
ret
这篇关于在x86 NASM中使用atof函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
