using atof function in x86 NASM

Question

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

Answer 1

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