The // operator explicitly floors the result. Quoting the Binary arithmetic operations documentation: the result is that of mathematical division with the ‘floor’ function applied to the result. Flooring is not the same thing as rounding to 0; flooring always moves to the lower integer value. See the math.floor() function: Return the floor of x, … Read more
The current answers all focus on decimal digits, when applying the “add all digits and see if that divides by 3”. That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And “converting” to hex is a lot easier than converting to decimal. Pseudo-code: … Read more
The scanf call uses the wrong format string. You are reading an int so you need it to be: scanf(“%d”, &fahrenheit); The expression 5/9 is evaluated using integer division. In fact the compiler can work it out at compile time. That expression evaluates to 0. You need to perform floating point division. For instance: 5.0/9 … Read more
It’s because you’re doing integer division. Divide by a double or a float, and it will work: double scale = ( n / 1024.0 ) * 255 ; Or, if you want it as a float, float scale = ( n / 1024.0f ) * 255 ;
The integer division there is just taking the floor of the number obtained at the end. 10/3 -> floor(3.33) -> 3 -10/3 -> floor(-3.33) -> -4 (Why it floors) The modulo operation on the other hand is following the mathematical definition.
you are looking for the modulo operator: a % b for example: >>> 26 % 7 5 Of course, maybe they wanted you to implement it yourself, which wouldn’t be too difficult either.
Cast the operands to floats: float ans = (float)a / (float)b;
You need to cast one or the other to a float or double. int x = 1; int y = 3; // Before x / y; // (0!) // After ((double)x) / y; // (0.33333…) x / ((double)y); // (0.33333…) Of course, make sure that you are store the result of the division in a … Read more