Modbus设备4字节转32位无符号整数技术求助

Let's break this down clearly, since the original documentation lacks concrete byte-level examples. First, let's align with your sensor's setup: Register 2 is the low 16 bits, Register 3 is the high 16 bits of the 32-bit unsigned (u32) pressure value.

Core Conversion Formula #

The final 32-bit unsigned pressure value (in Pa) is calculated as:

u32_pressure = (uint16_high) * 65536 + uint16_low

Where:

• uint16_low: The unsigned 16-bit equivalent of Register 2's value
• uint16_high: The unsigned 16-bit equivalent of Register 3's value

How to get uint16_low/uint16_high from signed 16-bit reads: #

If your Modbus client reads registers as signed 16-bit integers (int16):

• If the register value is positive: uint16_val = reg_value
• If the register value is negative: uint16_val = reg_value + 65536 (since 2¹⁶ = 65536, this converts the two's complement negative value to its unsigned counterpart)

If your client reads registers directly as unsigned 16-bit integers (uint16), you can skip the sign conversion step and use the raw values directly.

Step-by-Step Verification with Your Documentation Example #

Your docs use these values:

• Register 2 (low 16 bits) signed value: -30072
• Register 3 (high 16 bits) signed value: -65535

Let's apply the formula:

1. Convert Register 2 to uint16: -30072 + 65536 = 35464
2. Convert Register 3 to uint16: -65535 + 65536 = 1
3. Calculate high 16 bit contribution: 1 * 65536 = 65536
4. Merge values: 35464 + 65536 = 101000 Pa

This matches the result in your docs—perfect.

Concrete 4-Byte Example #

Let's map this to actual 4-byte data (the raw bytes you'd receive over Modbus):
Each 16-bit register uses big-endian formatting per Modbus standards. So:

• Register 3 (high 16 bits) is 0x0001, which translates to bytes 0x00 followed by 0x01
• Register 2 (low 16 bits) is 0x8A88, which translates to bytes 0x8A followed by 0x88

The full 4-byte sequence you'd receive is 0x00 0x01 0x8A 0x88. When read as signed 16-bit registers:

• Register 3 (0x0001) → 1 (positive, no conversion needed)
• Register 2 (0x8A88) → -30072 (two's complement for -30072), so convert to uint16: 35464

Applying the formula gives the same 101000 Pa result.

Pseudocode for Implementation #

Here's a simple function to handle the conversion, regardless of whether your register reads are signed or unsigned:

def modbus_regs_to_u32_pressure(reg2_value, reg3_value):
    # Convert signed int16 to uint16 if needed
    def to_uint16(val):
        return val if val >= 0 else val + 65536
    
    uint16_low = to_uint16(reg2_value)
    uint16_high = to_uint16(reg3_value)
    
    return uint16_high * 65536 + uint16_low

Testing this with your example:

print(modbus_regs_to_u32_pressure(-30072, -65535))  # Output: 101000

内容的提问来源于stack exchange，提问作者Baptiste