在C ++中比较uint8_t和uint16_t

Question

i would like to test if a value of an input variable passed one byte size or know so i write this code

{
    uint8_t x;
    cout << " enter x" << endl;
    cin  >> hex >> x;
    cout << hex << x;
    uint8_t y ;  
    y = x >> 4 ;
    cout << hex << y;
    if ( y == 0 )
    {
        cout << " succes, coorect value for x";
    }
    if (y >  0)
    {
        /*here i supoosed that x = 0xfff and when shifting, y would be 0xff but y is uint8 so it's just for compare
        std::cout << "fail, ";
    }
    return 0;
}

I would like to know how to test if the user type two or more byte in uint8. And tell him to retype the value of just one byte.that's why i tried to compare uint8_t to uint16_t.

Answer 1

The problem with the above code is that you are testing for a negative value in an unsigned integer . The if(y < 0) part will always be false.

Now, from my understanding you want to filter values that are greater than 1Byte . You know, that for a given byte you have 2^8 - 1 range of values, that is, the unsigned number lies in the range [0-255] . Thus, your problem is reduced to a simple check, that is, if the input value is greater than 255, throw away that input . (Ofc you have to check for < 0 values)

There are other solutions to this but without any context I can't provide a precise answer. Hope the above simplification helps.

Answer 2

In order to check if a value is in the range of a small type, you have to assign it to a bigger type first. You simply cannot do the check with the small type itself because it can never contain that too huge value you want to check for.

uint16_t tempX;
uint8_t x;
cin >> hex >> tempX;
if (tempX > 0xff)
    cout << "error";
else
    x = (uint8_t)tempX;