char []到uint32_t无法正常工作

Question

I am using the following to convert a char[4] to a uint32_t .

frameSize = (uint32_t)(frameSizeBytes[0] << 24) | (frameSizeBytes[1] << 16) | (frameSizeBytes[2] << 8) | frameSizeBytes[3];

frameSize is a uint32_t variable, and frameSizeBytes is a char[4] array. When the array contains, for example, the following values (in hex)

00 00 02 7b

frameSize is set to 635, which is the correct value. This method also works for other combinations of bytes, with the exception of the following

00 00 9e ba

for this case, frameSize is set to 4294967226, which, according to this website , is incorrect, as it should be 40634 instead. Why is this behavior happening?

Answer 1

Your char type is signed in your specific implementation and undergoes integer promotion with most operators. Use a cast to unsigned char where the signed array elements are used.

EDIT: actually as pointed out by Olaf in the comment, you should actually prefer casts to unsigned int (assuming common 32-bit unsigned int ) or uint32_t to avoid potential undefined behavior with the << 24 shift operation.

Answer 2

To keep things tidy I'd suggest an inline function along the lines of:

static inline uint32_t be_to_uint32(void const *ptr)
{
    unsigned char const *p = ptr;
    return p[0] * 0x1000000ul + p[1] * 0x10000 + p[2] * 0x100 + p[3];
}

Note: by using an unsigned long constant this code avoids the problem of unsigned char being promoted to signed int and then having the multiplication/shift cause integer overflow (an annoying historical feature of C). Of course you could also use ((uint32_t)p[0]) << 24 as suggested by Olaf.