What do bitrate operations in the Intel 8085 assembly do?

Question

I am trying to explain to myself the following assembly code 8085

I have this code that requests two numbers (from the virtual keyboard inside the IDE) and displays them on LEDs 7 and 8:

.ORG 0000 CALL DATA MOV C,A CALL PRNT CALL DATA MOV E,A CALL PRNT MVI D,00H MOV L,E MVI H,00H MOV A,C DCR A JUMP: DAD D DCR A JNZ JUMP MOV A,L DAA JMP IMPR RET DATA: MVI A,00000000B OUT 00H IN 00H RLC RLC RLC RLC ANI F0H MOV B,A MVI A,00000000B OUT 00H IN 00H ANI 0FH ORA B RET IMPR: MOV B,A ANI F0H RLC RLC RLC RLC CALL NUMZ OUT 06H MOV A,B ANI 0FH CALL NUMZ OUT 07H RET NUMZ: CPI 00H JNZ ONE MVI A,01110111B JMP EXIT ONE: CPI 01H JNZ TWO MVI A,01000100B JMP EXIT TWO: CPI 02H JNZ THREE MVI A,00111110B JMP EXIT THREE: CPI 03H JNZ FOUR MVI A,01101110B JMP EXIT FOUR: CPI 04H JNZ FIVE MVI A,01001101B JMP EXIT FIVE: CPI 05H JNZ SIX MVI A,01101011B JMP EXIT SIX: CPI 06H JNZ SEVEN MVI A,01111011B JMP EXIT SEVEN: CPI 07H JNZ EIGHT MVI A,01000110B JMP EXIT EIGHT: CPI 08H JNZ NINE MVI A,01111111B JMP EXIT NINE: CPI 09H JNZ SAL MVI A,01001111B JMP EXIT EXIT: RET 

I do not include PRNT because it is not important for my question.

I understand .ORG 0000 , which is the beginning of the program - it's like BEGIN in Pascal.

CALL DATA is a routine that fills the Battery with binary zeros and displays them (?) In port 0 (00H in hexadecimal format), then it receives (from the virtual keyboard) a number, and then rotates it to the left of the shift.

My question is why? What is the point of this? What is the use? I read about it on Wikipedia, but I still don't get it. What does he do in this code and why is it needed?

Answer 1

The DATA routine loads two decimal ASCII characters and forms them into a two-digit BCD value . It shifts the first character remaining 4 bits, saving only the LS 4 bit, and then puts the LS 4 bits of the second character in the LS 4 bit of the result.

In C, this would be roughly equivalent:

 char c = getchar(); // get first ASCII decimal character char result = (c << 4) & 0xf0; // shift into MS nybble of result and mask c = getchar(); // get second ASCII decimal characters result = result | (c & 0x0f); // mask and inset into LS nybble of result return result; 

Note that masking everything except LS nybble decimal ASCII gives you its decimal equivalent, for example. ASCII '4' = 0x34 => 0x04.

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To make this clear, I drew a diagram that shows, step by step, what happens when the user enters the number “69”, that is, ASCII '6', followed by ASCII '9', as the two characters are masked and combined to give Representation of BCD number 69: