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 known powers of 2: it’s much slower than and for remainder, … Read more
Your factorial() function returns 0 for any input because of how you defined your range. The range builtin starts at 0 unless otherwise defined so: is doing: on the first iteration so for all subsequent iterations: will always be 0 Start your range at 1 like so you can see this more explicityly: or instead … Read more
You already got the answer, no need to deal with the decimals if you assign it to an integer. In your comment you say that you are working with decimals, then Math.Floor is a possibility. ie:
Use cdq / idiv for signed 32-bit / 32-bit => 32 bit division,xor edx,edx / div for unsigned. With the dividend in EAX to start with, and the divisor specified as an operand to DIV or IDIV. If you zero EDX/RDX instead of sign-extending into EDX:EAX before idiv, you can get a large positive result for -5 / 2, for example. Using the “full … Read more
I keep seeing this form often: div 3, 15 this is not any valid INTEL mneumonic! To divide 15 by 3: For the second error, you cannot move a 16 bit register into a 32 bit register like that. You need to use one of the following: Or:
I sincerely hope this will help future searchers when googling for this common question.
For some number y and some divisor x compute the quotient (quotient) and remainder (remainder) as:
For some number y and some divisor x compute the quotient (quotient) and remainder (remainder) as:
The two operands (1 and 3) are integers, therefore integer arithmetic (division here) is used. Declaring the result variable as double just causes an implicit conversion to occur after division. Integer division of course returns the true result of division rounded towards zero. The result of 0.333… is thus rounded down to 0 here. (Note that the processor … Read more
Will result always be the floor of the division? What is the defined behavior? Not quite. It rounds toward 0, rather than flooring. 6.5.5 Multiplicative operators 6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.88) If the quotient a/b is representable, the expression (a/b)*b … Read more