For different computers, because their main frequency is different, the parameters of delay 1s are also different. The method of calculating delay is as follows:
Computer frequency: x (Hz)
The number of cycles of execution of a LOOP statement: y
The time required for delay: z (s)
Number of statements to be executed: a
z=y*(1/x)*a
Calculate the required number of execution statements to write the program.
Example: (The main frequency of the computer is 3GHz)
delay proc near push bx push cx mov bx,400h for1:mov cx,0ffffh for2:loop for2 dec bx jnz for1 pop cx pop bx ret delay endp
Extended knowledge:
Single-chip assembly jump instruction delays by one second
DELAY: MOV R7,#10 ;Delay 1S subroutine
DL1: MOV R6,#200-----1T
DL2: MOV R5,#248------1T
DJNZ R5,$
DJNZ R6,DL2
DJNZ R7,DL1
RETLP
RET
How does this delay? The crystal oscillator is 12MHZ and T=1us, DJNZ is a double-period instruction. It mainly delays the nested loop for one second. Let's look at the calculation result:
DJNZ R5,$-----------------------248*2=496500us
DJNZ R6,DL2----------------------（496+1+2）*200=99800us
DJNZ R7,DL1------------------------(99800+2+1)*10=998030~~1s;
Summarize
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