表达式是正确的还是错误的

Question

uint8_t ui8 = 255;
ui8 == (int16_t)-1

As far as I understand the standard :

Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.

ui8 would be converted/promoted to int16_t and then it's arithmetic value would be -1 . Unfortunately the compiler I am using says I am not right.

Answer 1

Whenever a uint8_t or any other small integer type is used as part of an expression, it gets integer promoted to type int . The same goes for int16_t , in case int is larger than 16 bits.

So your code is guaranteed to be equal to:

(int)255 == (int)-1

which is always false.

Because of integer promotions, both operands are already of the same type, so no balancing between operands takes place.

Answer 2

variable ui8 , with value 255 will be converted to type int for the comparison, where it still has value 255 .

Value (int16_t)-1 will also be converted to type int , where it still has value -1 .

255 compared against -1 is false .

Answer 3

ui8 would be converted/promoted to int16_t and then it's arithmetic value would be -1.

Yes , type of ui8 will be converted/promoted .(As mentioned by Lundin Sir !! in comment ) Its because of integer promotion thus ui8 will be promoted to type int16_t in case int is 16 bit but if on your machine int is larger than 16 bit then both will be promoted to int .

And as per your quote from standards , it says about type not the values. So ui8 will retain its value ie 255 .

And thus -

  ui8==(int16_t)-1 // may be (int)u8==(int)-1 depends on size of int on you machine

results in false .