Arduino Arithmetic error negative result

问题:

I am trying to times 52 by 1000 and i am getting a negative result

int getNewSum = 52 * 1000; 

but the following code is ouputting a negative result: -13536

回答1:

An explanation of how two's complement representation works is probably better given on Wikipedia and other places than here. What I'll do here is to take you through the workings of your exact example.

The int type on your Arduino is represented using sixteen bits, in a two's complement representation (bear in mind that other Arduinos use 32 bits for it, but yours is using 16.) This means that both positive and negative numbers can be stored in those 16 bits, and if the leftmost bit is set, the number is considered negative.

What's happening is that you're overflowing the number of bits you can use to store a positive number, and (accidentally, as far as you're concerned) setting the sign bit, thus indicating that the number is negative.

In 16 bits on your Arduino, decimal 52 would be represented in binary as:

0000 0000 0011 0100

(2^5 + 2^4 + 2^2 = 52)

However, the result of multiplying 52 by 1,000 -- 52,000 -- will end up overflowing the magnitude bits of an int, putting a '1' in the sign bit on the end:

*----This is the sign bit. It's now 1, so the number is considered negative.
1100 1011 0010 0000

(typically, computer integer arithmetic and associated programming languages don't protect you against doing things like this, for a variety of reasons, mostly related to efficiency, and mostly now historical.)

Because that sign bit on the left-hand end is set, to convert that number back into decimal from its assumed two's complement representation, we assume it's a negative number, and then first take the one's complement (flipping all the bits):

0011 0100 1101 1111

-- which represents 13,535 -- and add one to it, yielding 13,536, and call it negative: -13,536, the value you're seeing.

If you read up on two's complement/integer representations in general, you'll get the hang of it.

In the meantime, this probably means you should be looking for a bigger type to store your number. Arduino has unsigned integers, and a long type, which will use four bytes to store your numbers, giving you a range from -2,147,483,648 to 2,147,483,647. If that's enough for you, you should probably just switch to use long instead of int.

回答2:

Matt's answer is already a very good in depth explanation of what's happening, but for those looking for a more TL;dr practical answer:

Problem:

This happens quite often for Arduino programmers when they try to assign (= equal sign) the result of an arithmetic (usually multiplication) to a normal integer (int). As mentioned, when the result is bigger than the memory size compiler has assigned to the variables, overflowing happens.

Solution 1:

The easiest solution is to replace the int type with a bigger datatype considering your needs. As this tutorialspoint.com tutorial has explained there are different integer types we can use:

1. int:
   • 16 bit: from -32,768 to 32,767
   • 32 bit: from -2,147,483,648 to 2,147,483,647
2. unsigned int: from 0 to 65,535
3. long: from 2,147,483,648 to 2,147,483,647
4. unsigned long: from 0 to 4,294,967,295

Solution 2:

This works only if you have some divisions with big enough denominators, in your arithmetic. In Arduino compiler multiplication is calculated prior to the division. so if you have some divisions in your equation try to encapsulate them with parentheses. for example if you have a * b / c replace it with a * (b / c).