Question

The code on pages 20-21 of the Text book asks you to input a number, and it will then sum the numbers from 1 to that number. The Prompt asks you to input a number, not necessarily an integer. The program will abort if a floating point number is entered.

Your project, is to “fix” the program and allow for a floating point number to be entered. The program will NOT run, so your task is to convert the floating point number to an integer.

If a floating point number is entered, truncate the number and use that integer to run the program. Also, you must inform the user that they entered a floating point and it was truncated, output to the user, the floating point they entered and the integer you used.

Note: You must only use Shift, Rotate to manipulate the bits, no conversion instructions.

The following are some pointers as to what needs to be done.

1. The program reads an integer, that must be changed to read a floating point.
2. You will need to move that number into a floating point register and then that number must be copied into an integer register.
3. You will need to extract the exponent from the integer register and stored in another register.
4. You will need to insert the Implied bit. I would suggest, zero out the exponent part by shifting left 9 then shifting right 9. Then add 8388608 (2^23) to the number.
5. You will need to extract the fractional portion of the of the number. You will need the exponent to determine the shift. You only need to test to see that this is NOT EQUAL to 0 (if it is we have an integer)
6. Extract the Integer. You may want to “Rotate” the bits to the left.

You may want to use the following Assembler instructions in your code:

srl, add, sll, rol, sub, srlv and sllv

Please need an answer in Assembly language not in C/C++

Need to modify code below for floating point

# A program to find the sum of the integers from 1 to N, Where N is a value, which is read from the keyboard
# vo is: N;
# t0 is sum:

         .data
Prompt: .asciiz    "\n Please Input a value for N= "
Result: .asciiz    "The sum of the integers from 1 to N is"
Bye:    .asciiz    "\n **** Adios Amigos -Plz Have a Good Day****"
         .globl     main
         .text
main:
         li         $v0, 4        #system call for print
         la         $a0, Prompt   #load adress of prompt to $a0
         syscall                  #plz print the prompt message
         li         $v0, 5       #call code to read integer
         syscall                  # value of N into $v0
         blez       $v0, End      #to end if v0 <=0
         li         $t0, 0        #clear register $t0 to 0
Loop:
         add        $t0, $t0, $v0 #sum of integers in t0
         addi       $v0, $v0, -1 #summ all integers in reverse orders
         bnez       $v0, Loop     #loop if $v0 is !=0
         li         $v0, 4        #call code to print string
         la         $a0, Result   #load adress of message into $a0
         syscall                  #plz print the string
       
         li         $v0, 1        # call code for the print integer
         move       $a0, $t0      #move value to be printed to $a0
         syscall                  #print sum of integers
         b          main          #Plz branch to main
       
End:
         li         $v0, 4        #Call code for print string
         la         $v0, Bye      #load adress of msg. into $a0
         syscall                  #print string
         li         $v0, 10       #terminate program and run
         syscall                  #return control to syste
# t0 is sum:

         .data
Prompt: .asciiz    "\n Please Input a value for N= "
Result: .asciiz    "The sum of the integers from 1 to N is"
Bye:    .asciiz    "\n **** Adios Amigos -Plz Have a Good Day****"
         .globl     main
         .text
main:
         li         $v0, 4        #system call for print
         la         $a0, Prompt   #load adress of prompt to $a0
         syscall                  #plz print the prompt message
         li         $v0, 5       #call code to read integer
         syscall                  # value of N into $v0
         blez       $v0, End      #to end if v0 <=0
         li         $t0, 0        #clear register $t0 to 0
Loop:
         add        $t0, $t0, $v0 #sum of integers in t0
         addi       $v0, $v0, -1 #summ all integers in reverse orders
         bnez       $v0, Loop     #loop if $v0 is !=0
         li         $v0, 4        #call code to print string
         la         $a0, Result   #load adress of message into $a0
         syscall                  #plz print the string
       
         li         $v0, 1        # call code for the print integer
         move       $a0, $t0      #move value to be printed to $a0
         syscall                  #print sum of integers
         b          main          #Plz branch to main
       
End:
         li         $v0, 4        #Call code for print string
         la         $v0, Bye      #load adress of msg. into $a0
         syscall                  #print string
         li         $v0, 10       #terminate program and run
         syscall                  #return control to syste

Answer 1

Solution:-

int ftoi(float flt)

{

   int i;

   _asm

   {

         mov eax,flt;   //Loaded mem to acc

        rcl eax,1;     //left shift acc to removethe sign

        mov ebx,eax;   //save the acc

       mov edx,8388608; //clear reg edx

       and eax,edx;         //and acc to retrieve the exponent

       shr eax,24;   

       sub eax,7fh;       //subtract 7fh to get actual power

       mov edx,eax;     //save acc val power

       mov eax,ebx;     //retrieve from ebx

      rcl eax,8;          //trim the 8 left bits

      mov ebx,eax;   //store

      mov ecx,1fh;    //subtrat 17 h

     sub ecx,edx;

     mov edx,00000000h;

     cmp ecx,0;

     je loop2;

     shr eax,1;

     or eax,B0000000h;

loop1:

     shr eax,1;     //shift(total bits - power bits)

     sub ecx,1;

     add edx,1;

     cmp ecx,0;

     ja   loop1;

loop2:

     mov i,eax;

//check sign +/-

sign:

      mov eax,flt;

      and eax,80000000h;

      cmp eax,80000000h;

      je      putsign;

}

return i;

putsign:

return -i;

}

   }

}