如何将char转换为unsigned int？

Question

I have a char array that is really used as a byte array and not for storing text. In the array, there are two specific bytes that represent a numeric value that I need to store into an unsigned int value. The code below explains the setup.

char* bytes = bytes[2];
bytes[0] = 0x0C; // For the sake of this example, I'm 
bytes[1] = 0x88; // assigning random values to the char array.

unsigned int val = ???; // This needs to be the actual numeric 
                        // value of the two bytes in the char array.  
                        // In other words, the value should equal 0x0C88;

I can not figure out how to do this. I would assume it would involve some casting and recasting of the pointers, but I can not get this to work. How can I accomplish my end goal?

UPDATE

Thank you Martin B for the quick response, however this doesn't work. Specifically, in my case the two bytes are 0x00 and 0xbc . Obviously what I want is 0x000000bc . But what I'm getting in my unsigned int is 0xffffffbc .

The code that was posted by Martin was my actual, original code and works fine so long as all of the bytes are less than 128 (.ie positive signed char values.)

Answer 1

unsigned int val = (unsigned char)bytes[0] << CHAR_BIT | (unsigned char)bytes[1];

This if sizeof(unsigned int) >= 2 * sizeof(unsigned char) (not something guaranteed by the C standard)

Now... The interesting things here is surely the order of operators (in many years still I can remember only +, -, * and / ... Shame on me :-), so I always put as many brackets I can). [] is king. Second is the (cast) . Third is the << and fourth is the | (if you use the + instead of the | , remember that + is more importan than << so you'll need brakets)

We don't need to upcast to (unsigned integer) the two (unsigned char) because there is the integral promotion that will do it for us for one, and for the other it should be an automatic Arithmetic Conversion .

I'll add that if you want less headaches:

unsigned int val = (unsigned char)bytes[0] << CHAR_BIT;
val |= (unsigned char)bytes[1];

Answer 2

unsigned int val = (unsigned char) bytes[0]<<8 | (unsigned char) bytes[1];

Answer 3

The byte ordering depends on the endianness of your processor. You can do this, which will work on big or little endian machines. (without ntohs it will work on big-endian):

unsigned int val = ntohs(*(uint16_t*)bytes)

Answer 4

unsigned int val = bytes[0] << 8 + bytes[1];

Answer 5

I think this is a better way to go about it than relying on pointer aliasing:

union {unsigned asInt; char asChars[2];} conversion;
conversion.asInt = 0;
conversion.asChars[0] = 0x0C;
conversion.asChars[1] = 0x88;
unsigned val = conversion.asInt;