Unexpected results when creating 32-bit int from byte array

Question

I'm having problems creating a 32-bit integer from a 4-byte array. In the following complete, minimal, and verifiable example, converting a byte array containing 0x00, 0x04, 0x04F, 0x45 results in 4F45 (expected result would be 44F45). I must be overlooking something fundamental here.... can anyone see where I went wrong?

void setup()
{
    Serial.begin(9600);
    while(!Serial);
}

void loop()
{
    uint8_t bytes[4] = {0x00, 0x04, 0x4F, 0x45}; //00044F45
    int32_t theInt = byteArrayToInt32(bytes);
    Serial.println(theInt, HEX); //Print 4F45
    delay(250);
}

uint32_t byteArrayToInt32(uint8_t* bytes)
{
    uint32_t i = 0;
    i |= bytes[0] << 24;
    i |= bytes[1] << 16;
    i |= bytes[2] << 8;
    i |= bytes[3];
    return i;
}

@jsotola I went ahead re-tested with it as uint32_t, no difference :( — You'reAGitForNotUsingGit
– You'reAGitForNotUsingGit, Commented Jun 10, 2019 at 2:53

jsotola · Accepted Answer · 2019-06-10 03:31:56Z

1

The problem is that you are trying to shift uint8_t more than 8 bits and then OR it with int32_t.

You have to cast the uint8_t to int32_t before doing the shift.

void setup()
{
    Serial.begin(9600);
    while(!Serial);
}

void loop()
{
    uint8_t bytes[4] = { 0x00, 0x04, 0x4F, 0x45}; //00044F45
    int32_t theInt = byteArrayToInt32(bytes);
    Serial.println(theInt, HEX); //Print 4F45
    delay(1000);
}

int32_t byteArrayToInt32(uint8_t* bytes)
{
    int32_t i = 0;
    i |= (int32_t) bytes[0] << 24;
    i |= (int32_t) bytes[1] << 16;
    i |= (int32_t) bytes[2] <<  8;
    i |= (int32_t) bytes[3];
    return i;
}

answered Jun 10, 2019 at 3:31

jsotola

1,5532 gold badges13 silver badges22 bronze badges

Note that if the most significant bit of bytes[0] is set, then (int32_t) bytes[0] << 24 will shift it into the sign bit, which is undefined behavior. It is safer to cast to a uint32_t, which can in turn be safely assigned to an int32_t.

Edgar Bonet
– Edgar Bonet

2019-06-10 09:14:34 +00:00
Commented Jun 10, 2019 at 9:14
Hmm, so doing as you suggested fixes the problem, however I'm very perplexed as to why the left shift by 8 bits worked originally, then?

You'reAGitForNotUsingGit
– You'reAGitForNotUsingGit

2019-06-10 14:26:11 +00:00
Commented Jun 10, 2019 at 14:26
it appears that the left shift function uses a 16 bit buffer .... possibly because it uses the same code for 16 bit shifts

jsotola
– jsotola

2019-06-10 18:53:52 +00:00
Commented Jun 10, 2019 at 18:53

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