The role of the single chip position addressing area, how to judge the direct addressing and bit add

A: The bit-addressing area is located between 20F and 2Fh in the data memory. The registers in this interval can be bit-addressed, and they can be bit-manipulated and bit-arithmetic.

The main thing to pay attention to is the relationship between the address of the register "itself" and the bit address corresponding to the "bit" of the register, which is the most error-prone point. The bit address corresponding to 20h is from 20h to 27h. Namely:

The zero bit address of 20h is 20h, the first bit address is 21h, and so on. The bit address for 21h (I mean the register, not the first bit address for 20h) is from 28h to 2fh; 22h (register) corresponding to the bit address 30h to 37h; And so on.

Got it? It can be seen that the bit address of a certain bit of one register may coincide with the direct address of another register, but the instructions for bitwise address operation are different from those for direct address operation, so there is no confusion.

In use, you can use the bit address to store a certain switch quantity (only two values, 0 and 1), to do logical operations on the value of the bit address, or to pass it to a special function register, etc.

There are a number of bitwise operations in the special function registers. Set, zero, etc. This allows you to operate only on that bit without affecting the other bits.

The address used to store the content of one bit, for example, some switches 0 and 1 only need one bit address to be placed. On the other hand, a byte address is an address that stores 8 bits, which is one byte of content.

Yes, C is a special register--- a bit of the Carry Flag in the Program Status Register (PSW) - --- Cy;

Since c is just one bit, then the mov c,20h statement is bit addressing;

In the MCU instruction set, the only transmission instruction that can be bit-addressed is the instruction containing C, which is easy to remember and naturally easy to judge;

Answer: A is an accumulator, 8 bits, so it is directly addressed, and c is an overflow bit, which only occupies 1 bit, so it is bit addressing.

First of all, C is not a register, just a bit in the register.

Second, when your assembly statement writes the destination address as a bit address, the assembler knows to translate it according to the bit-addressing instruction when it converts.

The 16 byte units of the single-chip microcomputer from 20h to 2fh, a total of 128 bits can be addressed by bit, corresponding to the bit addressing space from 00h to 7fh, so the unit address with a bit address of 40h is 40h 8+20h=28h.

In addition, there are 16 special function registers that can be addressed, just check ** on the line, 88h corresponding to the byte address is also 88h. In the same way, 2ah-20h=0ah, 0ah*8=50h.

The 20H 2FH of the RAM on the chip of the single-chip microcomputer, which is 16 bytes, can be addressed according to the "bits" of the precautionary light.

There are a total of 128 "bit addresses", which are 00h and 7fh.

For example, the bit address 4EH is the 6th bit of the "byte unit with a byte address of 29h".

Addressing in a microcontroller is the form of operands provided in the microcontroller instructions. That is, the way of finding operands or addresses where operands are located.

In the 51 series microcontrollers, there are four types of memory space for storing data: internal RAM, special function register SFR, external RAM, and program memory ROM. Among them, the internal RAM and SFR are uniformly addressed, and the external RAM and program memory are addressed separately.

In order to distinguish the address space where the operands in the instruction are located, different addressing methods are used for data operations on different memory settings.

The answer is stupid.

The correct answer is as follows:

51What is the address range of the internal memory of the single-chip microcomputer that can be addressed by bits?

It's 20h 2fh.

What is the bit address space range?

It's 00h 7fh.

In addition, the question is "Internal memory of microcontrollers."

So, here's the sentence:

Special reed-carrying function register address: 80h 7fh (21 discrete distribution)] do not need to be written. Defend the rental belt.

Besides, the address is incorrect, it should be 80h ffh.

There is a processor integrated in the single-chip microcomputer, and different CPUs have their own different instructions, so there are some special bit processing instructions in the single-chip microcomputer, and the addressing mode adopted by some bit processing instructions is the bit addressing mode

The so-called bit addressing is essentially the same as the general memory unit addressing, except that each unit can only store one binary bit, and each bit unit has a bit address

For example, in the MCS51 microcontroller, it takes the 16 bytes of 20h 2FH in the internal RAM area as the bit addressing area, 8 bits per byte, a total of 128 bit units, starting from the lowest bit of the 20h byte unit, and the general starting address starts from 0, so the bit address is 00 7fh specific instructions such as mov c, the source operand in 04h.

In the bit-addressing mode, the D4 bit of the byte unit 20h is accessed, which is 20h 4

There are seven addressing methods for single-chip microcomputers, which are:1Register addressing e.g

mova,r0;2.Direct addressing for example:

mova,4fh;3.Register indirect addressing.

mova,@r04.Immediate number addressing.

mova,#0ffh5.Changeable address addressing (typically a lookup command) MOVA, @a+dptr6Relative addressing.

sjmp$7.Bit addressing.

mov60h,40h;Yes is directly addressed; (Use the address of the memory unit directly to find the corresponding content).

mova,#58hmov

ro,#0b0hmov

ro,amov

0b0h, 28h and the result after executing the above instructions is: (0b0h)=p3=28h; Because in the 8051 microcontroller, the P3 port is a special function register (also accounting for a memory), and its address is exactly 0b0h; At the end of the program, 0b0h is assigned an immediate number of 28h;

mov60h,40h

It is a direct addressing mode mode MOV

a,#58hmov

ro,#0b0hmov

ro,amov

0b0h, 28h result: (0b0h)=28h; p3=58h (this is not so sure) seven methods of internal RAM can be used, and the external data memory can only transmit data to the A accumulator.

Well, this is not exactly what the book says, and the direct addressing he says is movpsw, 25 says that the first operand is addressed directly, and mova, 68h says the latter 68h

The way this number is addressed, there are often this kind of in books, and if it is not exact, it is not true, and it is a question of how you understand it, but you can think of it when you know all of this.