and so on all the way counting back down. they actually go backwards, if you noted it above. When you force -4 to be unsigned bit pattern, that is why you get a large value. eg 8 bits is 2^8 -1 max value, 256-1 -> 255 same for 16 and 32 and 64 bit ints) It must be this, or the overflow to zero and negation algorithm will not work in 32 bits.īut what is that value above in unsigned? well its 2^32 -1 - 3. all the way up, but it overflows:īut the 1 is lost in the overflow, its. The algorithm for 2's complement, which is how negative numbers are stored, is binary not +1.ġ111 1111 //255 unsigned is -1 in signed!ġ+1 is 0 carry the 1. When you cast it to int it is no longer 8 bits, it is 32 (which I can see by the value of -4 which is about 2^32). Or go old school and avoid the insane mess cout makes of simple tasks. There should be a way clear the high order bits with an & (add &255) for example, but oddly, I was unable to get the casting happy due to cout being cout. You can force cast the bits from one to the other, to see both values, but at that moment in time, its signed or not, and has one of the 2 values at that moment. They are the same number, but if you look at the ranges, the large value DOES NOT EXIST in the range for signed values for that type - so it can only be one of the two, the large unsigned value, or the small negative value - not both. if you promote a negative value to a larger, negative type, some of the high order bits get set in the process of making it negative, and that makes it a gigantic value in your case. Here you are seeing a combined effect of using different sized types and using signed and unsigned. But when you tell the compiler that it is signed, it knows that 128 is really a negative number, or that 253 is a negative number, etc.

The same bits exist for both (8 bits of 1s and 0s).

eg for a simple byte, you have -127 to 127 or 0-255.

You will see something (now that you are looking for it!): the unsigned version's biggest value is about twice that of the signed value. If you look up a table of ranges (what numbers the variables of different types in c++ can be) This is critical, if you do not understand it, you need to stop until you do. So how is it possible when these values are different, that the output is the same when I convert them to hex?

C convert char to hex store to file