Editing Intel 8080 (section)

===Example code===
The following 8080/8085 [[assembly language|assembler]] source code is for a subroutine named {{code|memcpy}} that copies a block of data bytes of a given size from one location to another. The data block is copied one byte at a time, and the data movement and looping logic utilizes 16-bit operations.

<!--This code is not intended to be optimized, but to illustrate the variety of instructions available on the CPU. Still, it must be correct.-->
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<pre>
 
 
 
 
 
 
 
 
 
 
 
1000
1000
1000  78
1001  B1
1002  C8
1003  1A
1004  77
1005  13
1006  23
1007  0B
1008  78
1009  B1
100A  C2 03 10
100D  C9
</pre>
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<syntaxhighlight lang="nasm">
; memcpy --
; Copy a block of memory from one location to another.
;
; Entry registers
;       BC - Number of bytes to copy
;       DE - Address of source data block
;       HL - Address of target data block
;
; Return registers
;       BC - Zero

            org     1000h       ;Origin at 1000h
memcpy      public
            mov     a,b         ;Copy register B to register A
            ora     c           ;Bitwise OR of A and C into register A
            rz                  ;Return if the zero-flag is set high.
loop:       ldax    d           ;Load A from the address pointed by DE
            mov     m,a         ;Store A into the address pointed by HL
            inx     d           ;Increment DE
            inx     h           ;Increment HL
            dcx     b           ;Decrement BC   (does not affect Flags)
            mov     a,b         ;Copy B to A    (so as to compare BC with zero)
            ora     c           ;A = A | C      (are both B and C zero?)
            jnz     loop        ;Jump to 'loop:' if the zero-flag is not set.   
            ret                 ;Return
</syntaxhighlight>
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