By Roger Hutty
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Extra info for Z80 Assembly Language Programming for Students
How then do we convert these codes to form the 8-bi t value 123 in a register or memory byte? First of all each digit's character code must be converted to the digit's value. Then the first digit's value is multiplied by 100, the second digit's value is multiplied by 10 and these two products are added to the third digit's value to give the total value of the number. So that for the number 123 the computation 1 x 100 + 2 x 10 + 3 would be performed. An equivalent, but more efficient, algorithm for computing the value of a 3-digit number is ((first digit x 10) + second digit) x 10 + last digit which can easily be extended to an algorithm which inputs an ndigit number.
The address of the stack bottom can be set by program instructions and there may be more than one stack. However, a system will usually have only one stack and the address of that stack bottom will be set by the operating system. Items may be added to the stack, or pushed on the stack, as we say, when the SP register will be decremented by two to point to the new top of the stack. Also, items may be removed from the stack, or popped off, as we say, when the SP register will be incremented by two to point to the new top of the stack.
1 Assuming that the contents of the accumulator and carry flag are A7H and 0, respectively, what will be their contents after the execution of an SRL A instruction? In all shift instructions, the bit which is moved out of the register or memory byte, be it to the left or right, is placed in the carry flag. This can be useful because the value of a bi t 49 which has been moved out can be checked by any of the carry conditional jumps, such as JP C,label and JR NC,label. 2 THE SRA INSTRUCTION The Shift Right Arithmetic instruction is the same as the SRL instruction except that, instead of bit 7 being reset to 0 it is set to what it was before the shift.
Z80 Assembly Language Programming for Students by Roger Hutty