Can someone help me for this. Please write this in X86 assembly language COMMENT

Question

Can someone help me for this. Please write this in X86 assembly language

COMMENT!
a. Create a constant for the number of minutes in an hour
b. Prompt the user for the starting time and read in the user input
c. Prompt the user for the ending time and read in the user input
   The user input has 2 digits for hour, 2 digits for minutes.
   For example: 0930 for 9:30am, or 2208 for 10:08pm
   The range of input hour is 0 - 23, and the range of input minute is 0 - 59
d. Calculate the time difference between the starting and ending times,
   make sure to use the constant of step a
e. Print the difference as number of hours and number of minutes
   Make sure there is a text explanation, such as:   2 hr 9 min
   The time difference can be negative if the starting time is later than the ending time

Except for text string variables, the program should have no other memory variables
Given the range of data used in the program, use the smallest data size (not DWORD)

Explanation / Answer

There are several directives that we use to define the constants.Some of them are listed below:-
1)EQU
2)%assign
3)%define

A]Seconds EQU (any default value)
This line creates the constants seconds and any default value can be provided for the variable defined above.

B] Prompt the user for the starting time.
This code have been taken by compiling the C code and after disassembling it from the online editor.

.Ltext0:
.section
.rodata

.LC0:
.string   "Enter the starting time"

movl   $.LC0, %edi
movl   $0, %eax
call   printf
leaq   -4(%rbp), %rdx
leaq   -8(%rbp), %rax
movq   %rax, %rsi
movl   $.LC1, %edi
movl   $0, %eax
call   __isoc99_scanf

C]
Prompt the user for the end time.
Similarly this also have been added by disassembling.

.Ltext0:
.section  
.rodata

.LC0:
.string   "Enter the ending time"
movl   $.LC0, %edi
movl   $0, %eax
call   printf
leaq   -4(%rbp), %rdx
leaq   -8(%rbp), %rax
movq   %rax, %rsi
movl   $.LC1, %edi
movl   $0, %eax
call   __isoc99_scanf

[D and I have consolidated the entire code into one,it have all the parts in it.To ensure the code is running I have done so.
The entire code which performs the explanation is given below:-

    .Ltext0:
        .section   .rodata
    .LC0:
        .string   "Enter the starting time"
  

  
  
    .LC1:
       .string   "%d"
    .LC2:
        .string   "Enter the ending time"

  
        .text
        .globl   main
    main:
    .LFB0:
        .cfi_startproc
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq   %rsp, %rbp
        .cfi_def_cfa_register 6
        subq   $16, %rsp
       movl   $.LC0, %edi
  
        movl   $0, %eax

        call   printf

        leaq   -8(%rbp), %rax
       movq   %rax, %rsi
       movl   $.LC1, %edi

       movl   $0, %eax

        call   __isoc99_scanf

        movl   -8(%rbp), %eax
        movl   %eax, %edi
        call   convert_format

movl   %eax, -8(%rbp)
        movl   $.LC2, %edi
  
        movl   $0, %eax
  
       call   printf

        leaq   -4(%rbp), %rax
        movq   %rax, %rsi
       movl   $.LC1, %edi

       movl   $0, %eax

       call   __isoc99_scanf

        movl   -4(%rbp), %eax
        movl   %eax, %edi
        call   convert_format

        movl   %eax, -4(%rbp)
        movl   -4(%rbp), %edx
        movl   -8(%rbp), %eax
        movl   %edx, %esi
        movl   %eax, %edi
        call   difft
  
        leave
        .cfi_def_cfa 7, 8
       ret
        .cfi_endproc
    .LFE0:
        .globl   convert_format
    convert_format:
    .LFB1:
        .cfi_startproc
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq   %rsp, %rbp
        .cfi_def_cfa_register 6
        movl   %edi, -4(%rbp)
        cmpl   $1299, -4(%rbp)
  
        jle   .L3
       movl   -4(%rbp), %eax
        subl   $1200, %eax

        jmp   .L4
    .L3:
       cmpl   $99, -4(%rbp)
       jg   .L5
        movl   -4(%rbp), %eax
        addl   $1200, %eax
  
        jmp   .L4
    .L5:
       movl   -4(%rbp), %eax
    .L4:
       popq   %rbp
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc
    .LFE1:
        .section   .rodata
    .LC3:
       .string   "Difference is "
  
    .LC4:
        .string   "The difference is %d"
  
        .text
        .globl   difft
    difft:
    .LFB2:
        .cfi_startproc
        pushq   %rbp
        .cfi_def_cfa_offset 16
        .cfi_offset 6, -16
        movq   %rsp, %rbp
        .cfi_def_cfa_register 6
        subq   $32, %rsp
        movl   %edi, -20(%rbp)
       movl   %esi, -24(%rbp)
        movl   $0, -4(%rbp)
  
        movl   -24(%rbp), %eax
        movl   -20(%rbp), %edx
        subl   %eax, %edx
        movl   %edx, %eax
        movl   %eax, -4(%rbp)
        cmpl   $0, -4(%rbp)
        jle   .L9
        movl   -4(%rbp), %eax
       movl   %eax, %esi
        movl   $.LC3, %edi

        movl   $0, %eax

        call   printf
  
        jmp   .L10
    .L9:
        movl   -4(%rbp), %eax
        negl   %eax
       movl   %eax, %esi
        movl   $.LC4, %edi
  
        movl   $0, %eax
  
        call   printf
  
    .L10:
        movl   -4(%rbp), %eax
        leave
        .cfi_def_cfa 7, 8
        ret
        .cfi_endproc
    .LFE2:
    .Letext0:

The code given above is the output after it has been diassembled into the Assembly language.   

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.