How is the flag determined when temperature sensor data is stored in hexadecimal

The suffix H is hexadecimal.

Conventional discrete temperature sensors (including sensitive elements) are capable of exact relationships between power and power output data between addresses not shown in Table 2, with the output data in binary form, such as hexadecimal form.

dht11 reads data in 5 octet binaries.

My suggestion is to convert the data on the lower computer, and then send the converted data to the upper computer.

0000h ffffh, a total of 16 4 states, storage capacity of 16 4 = 65536 = 64k.

Last address = storage space size + first address -1 and because the first address is 0000h, the last address is equal to the storage space size -1.

So the last address is equal to 0x8000-1=0x7fff.

32k=32768=0x8000 The actual address is calculated from the 0x0000, so the last address of 32k should be: 0x8000-1=0x7fff In simple terms.

For example, if there is a 32kb of memory, if you allocate one byte per memory unit, which is a byte, 8 bits, then you need to have 32*1024 such memory cells.

32k=32768=0x8000

The actual address is calculated from the 0x0000, so the end address of 32k is 0x8000-1=0x7fff

32k=2 15, which means that there are 15 address lines, the starting number is 0000h, and the last address is ffffh

32kb=32*1024b=2^15b

Memory cells are measured in bytes (b), so 2 15 memory cells are required and expressed as binary bit cell addresses, which is converted from 0-111 1111 1111 1111 to hexadecimal is 0-7fffh

* The computer stores data in binary form, and the binary is a byte every 8 bits, as in your example

Decimal binary.

Hexadecimal A 2 F 3

The rightmost bit is the lowest bit, marked as bit 0, and the bit to the left is bit 1, and so on all the way to the 15th bit of this data.

The 0 7 bits 11110011 are the low-level bytes of this data (because all the tags it contains are smaller than the other bytes), and the 10100010 are the high-digit bytes (8 15 bits).

The smallest unit of computer storage is bytes, and the storage order (little-endian mode) is low-bit bytes first, high-bit bytes last

11110011, 10100010 i.e. hexadecimal tokens are f3, a2 two bytes. The computer will also automatically load f3 to the 0 7 bit of the variable when reading these two bytes to the integer variable, and a2 to the high byte of the variable.

This is a convention, because there are two modes, that is, big-endian mode and small-endian mode, the above kind of low-endian mode is called low-endian mode, and large-endian mode saves high-bit bytes first, if you encounter the CPU and operating system of big-endian mode, then your data will be saved as A2, F3.

Most CPUs store data in little-endian mode, and the way data is stored in memory and other storage media (flash, hard disk, CD-ROM, etc.) is the same.

Computer Storage:The smallest unit is the byte, and the storage order (little-endian mode) is that the low-bit byte comes first and the high bit comes last.

Hexadecimal:The rightmost bit is the lowest bit, marked as bit 0, and the bit to the left is bit 1, and so on all the way to the 15th bit of this data.

Deposit systemIt refers to a system in a computer composed of various storage devices, control components, and equipment (hardware) and algorithms (software) that manage information scheduling for storing programs and data. The main memory of the computer can not meet the requirements of fast access speed, large storage capacity and low cost at the same time, and there must be multi-level hierarchical memory with slow speed and large to small capacity in the computer, so as to form a storage system with acceptable performance with the optimal control and scheduling algorithm and reasonable cost.

Hexadecimal(English name: hexadecimal), is a representation of data in a computer. It's not the same as the representation in our daily lives.

It consists of 0-9, a-f, and the letters are not case sensitive. The correspondence with decimal is: 0-9 corresponds to 0-9; a-f corresponds to 10-15; Numbers in base can be represented by numbers of 0 (n-1) and letters a-f over 9.

"11110011, 10100010 is hexadecimal.

There is something wrong with this sentence, it should be said that it is "sixteen".

The big and small end format, this depends on the specific CPU, the high and low are the small end, the high and low are the big end!

Computer memory is arranged bytes, and the input and output streams are also arranged in byte order. int, float, double, how to arrange these multi-byte data, different computers deal with it differently. It is divided into big-endian and little-endian order.

PC is little-endian order.

With the method of the consortium, it is possible to understand the internal arrangement:

#include

union u u;

int main ()

Shared memory, = 0x0000a2f3

Output Byte Sequence:

f3 a2 00 00

This memory has a capacity of 64k, and the answer is (c).

Because the address code is from 0000-ffff, the largest address number is ffff, which is converted into a decimal number of 65536, which is the maximum addressing range of the memory. The capacity of memory refers to the maximum number of units of information that can be stored in a piece of memory. Here it is 65,536, so in a computer, the capacity of this memory can be expanded up to 64k.

In the MCS-51, the program memory is addressed by a 16-bit program counter (PC), which has 64KB addressing capability, that is, it can be arbitrarily addressed in the 64KB address space.

Among them, it has 4KB of on-chip program memory space, and the address is 000H-0FFH (Note: 8031 does not have on-chip program memory); The off-chip program memory space can be expanded to a maximum of 64KB, and the address is 0000h-ffffh, and the on-chip and off-chip are uniformly addressed.

From 0000 to ffffh, bytes are addressed.

The number of bytes is: ffffh 0000 + 1

10000h

The 64K capacity is: 64K bytes = 64KB.

That's how it is calculated, FFFF is converted to decimal is 65535, plus 0000 is 65536, 65536b=(65536 1024)kb=64kb, so the answer is c

If the address is 16 bits, then the range that the address can represent is 2 to the 16th power, which is the addressing space of 64k. The storage capacity depends on how big the storage content of an address is, and the storage space corresponding to an address in the question is a byte, so the total capacity is 64KB. Read one with caution, you can't answer it as 64k,

It can be directly converted, and in hexadecimal system, 16 4 b = (2 6) * 2 10 b = 64kb

Serial ports are sent in hexadecimal, there are two ways, after receiving the data, you can convert the hexadecimal to the decimal system, for example, the decimal 10 you send, when the transmission is hexadecimal 0x0a, after receiving the data into decimal 0ah = 10.

When sending, the data is converted into BCD code, the decimal 10 is sent, the hexadecimal code is 0x10, and the display is 10h when the data is received.

Just display the received data in hexadecimal format.

1k=2 to the 10th power,,, then 64k=2 to the 16th power, i.e., sixteen digits, then from 0000 to ffff