Slide 11.5: AND instruction

Slide 11.4: Boolean instructions
Slide 11.6: OR instruction
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AND Instruction (Cont.)

The operands can be 8, 16, or 32 bits, and they must be the same size. For each matching bit in the two operands, the output bit is 1 when both of the input bits are 1; otherwise, it is 0. The table on the right describes the boolean expression X AND Y.

`X`	`Y`	`X AND Y`
0	0	0
0	1	0
1	0	0
1	1	1

Code I: If we compare the ASCII codes of capital A and lowercase a, it becomes clear that only bit 5 is different:
```
    0 1 1 0 0 0 0 1 = 61h ('a')
    0 1 0 0 0 0 0 1 = 41h ('A')
```
The rest of the alphabetic characters have the same relationship. If we AND any character with 11011111 binary, all bits are unchanged except for bit 5, which is cleared.

Code II: To convert an integer digit into an ASCII digit, you must set bits 4 and 5. If, for example, AL=05h, you can OR it with 30h to convert it into the ASCII code for the digit 5 (35h).

Code I: Converting Lowercase
Letters into Uppercase Letters

Code II: Converting an Integer
Digit into an ASCII Digit

 INCLUDE Irvine32.inc
 .data
 buffer  BYTE  64 DUP(0)
 .code
 main PROC
     mov   edx, OFFSET buffer
     mov   ecx, SIZEOF buffer
     call  ReadString
     mov   ecx, eax
     mov   esi, OFFSET buffer

 L1: cmp   BYTE PTR [esi], 'a'
     jl    L2
     cmp   BYTE PTR [esi], 'z'
     jg    L2
     
      
      
      
      

    
    
    


    
    

      
     
    
 L2: inc   esi
     loop  L1

     mov   edx, OFFSET buffer
     call  WriteString
     exit
 main ENDP
 END main

 INCLUDE Irvine32.inc

 .data
 prompt1 BYTE "Enter a digit: ", 0
 prompt2 BYTE "The ASCII digit is ", 0

 .code
 main PROC
   call  Clrscr
   mov   edx, OFFSET prompt1
   call  WriteString
   call  ReadInt
     
      
   
      
      

    
    
    


    
    

      
     
    
   mov   edx, OFFSET prompt2
   call  WriteString
   call  WriteChar
   exit
 main ENDP
 END main