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
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 ;
3 and 6 are both Int, and dividing one Int by another gives an Int: that’s why you get back 0. To get a non-integer value you need to get the result of the division to be a non-integer value. One way to do this is convert the Int to something else before dividing it, … Read more
The performance difference comes from the efficiency of 128-bit divisions/modulus with GCC/Clang in this specific case. Indeed, on my system as well as on godbolt, sizeof(long long) = 8 and sizeof(__int128_t) = 16. Thus operation on the former are performed by native instruction while not the latter (since we focus on 64 bit platforms). Additions, … Read more
The reason is that unsigned division by 2^n can be implemented very simply, whereas signed division is somewhat more complex. unsigned int u; int v; u / 4096 is equivalent to u >> 12 for all possible values of u. v / 4096 is NOT equivalent to v >> 12 – it breaks down when … Read more
Do floating point division then convert to an int. No extra modules needed. Python 3: >>> int(-1 / 2) 0 >>> int(-3 / 2) -1 >>> int(1 / 2) 0 >>> int(3 / 2) 1 Python 2: >>> int(float(-1) / 2) 0 >>> int(float(-3) / 2) -1 >>> int(float(1) / 2) 0 >>> int(float(3) / … Read more
Ideally, we would like to have two operations div and mod, satisfying, for each b>0: (a div b) * b + (a mod b) = a 0 <= (a mod b) < b (-a) div b = -(a div b) This is, however, a mathematical impossibility. If all the above were true, we would have … Read more
Consistency. You’ll need to follow some very basic and seemingly irrelevant explanations to understand it. In school you have learned division with a remainder. And you have done calculations like this: 8 ÷ 4 = 2 R 0 7 ÷ 4 = 1 R 3 6 ÷ 4 = 1 R 2 5 ÷ 4 … Read more
Editor’s note: this is not actually what compilers do, and gives the wrong answer for large positive integers ending with 9, starting with div10(1073741829) = 107374183 not 107374182 (Godbolt). It is exact for inputs smaller than 0x40000005, though, which may be sufficient for some uses. Compilers (including MSVC) do use fixed-point multiplicative inverses for constant … Read more
If your modulus / divisor is a known constant, and you care about performance, see this and this. A multiplicative inverse is even possible for loop-invariant values that aren’t known until runtime, e.g. see https://libdivide.com/ (But without JIT code-gen, that’s less efficient than hard-coding just the steps necessary for one constant.) Never use div for … Read more