Slide 15.7: IDIV instruction (cont.)
Slide 16.1: Shift and rotate instructions
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Implementing Arithmetic Expressions

This program implements the following C++ expression in assembly language, using 32-bit signed integers: 
     x = ( y / z ) × ( x + y )
 x = ( y / z ) × ( x + y )  
 INCLUDE Irvine32.inc
 .data
           x  SDWORD  ?
           y  SDWORD  ?
           z  SDWORD  ?
           prompt1  BYTE  "x = ", 0
           prompt2  BYTE  "y = ", 0
           prompt3  BYTE  "z = ", 0
           prompt4  BYTE  "( y / z ) * ( x + y ) = ", 0
           prompt5  BYTE  "Divide by zero!", 0Dh, 0Ah, 0
           prompt6  BYTE  "Overflow!", 0Dh, 0Ah, 0

 .code
 main PROC
           ;; Read x.
           mov   edx, OFFSET prompt1
           call  WriteString
           call  ReadInt
           mov   x, eax

           ;; Read y.
           mov   edx, OFFSET prompt2
           call  WriteString
           call  ReadInt
           mov   y, eax

           ;; Read z.
           mov   edx, OFFSET prompt3
           call  WriteString
           call  ReadInt
           mov   z, eax

           ;; Calculate x = ( y / z ) × ( x + y ) .
           mov   eax, y
           cdq
           test  z, 0FFFFFFFFh
           jz    ZERO
           idiv  z
           mov   ebx, x
           add   ebx, y    

     

      
            

            
      		
      

    
    
     
     

      
     

    
           mov   x, eax

           ;; Print result.
           mov   edx, OFFSET prompt4
           call  WriteString
           mov   eax, x
           call  WriteInt
           call  Crlf
           exit
      
           ;; Divide by zero.
 ZERO:     mov   edx, OFFSET prompt5
           call  WriteString
           exit

           ;; Result is too large.
 OVERFLOW: mov   edx, OFFSET prompt6
           call  WriteString
           exit
 main ENDP
 END main