How is an integer stored in memory?

Question

This is most probably the dumbest question anyone would ask, but regardless I hope I will find a clear answer for this.

My question is - How is an integer stored in computer memory?

In c# an integer is of size 32 bit. MSDN says we can store numbers from -2,147,483,648 to 2,147,483,647 inside an integer variable.

As per my understanding a bit can store only 2 values i.e 0 & 1. If I can store only 0 or 1 in a bit, how will I be able to store numbers 2 to 9 inside a bit?

More precisely, say I have this code int x = 5; How will this be represented in memory or in other words how is 5 converted into 0's and 1's, and what is the convention behind it?

Answer 1

It's represented in binary (base 2). Read more about number bases. In base 2 you only need 2 different symbols to represent a number. We usually use the symbols 0 and 1. In our usual base we use 10 different symbols to represent all the numbers, 0, 1, 2, ... 8, and 9.

For comparison, think about a number that doesn't fit in our usual system. Like 14. We don't have a symbol for 14, so how to we represent it? Easy, we just combine two of our symbols 1 and 4. 14 in base 10 means 1*10^1 + 4*10^0.

1110 in base 2 (binary) means 1*2^3 + 1*2^2 + 1*2^1 + 0*2^0 = 8 + 4 + 2 + 0 = 14. So despite not having enough symbols in either base to represent 14 with a single symbol, we can still represent it in both bases.

In another commonly used base, base 16, which is also known as hexadecimal, we have enough symbols to represent 14 using only one of them. You'll usually see 14 written using the symbol e in hexadecimal.

For negative integers we use a convenient representation called twos-complement which is the complement (all 1s flipped to 0 and all 0s flipped to 1s) with one added to it.

There are two main reasons this is so convenient:

• We know immediately if a number is positive of negative by looking at a single bit, the most significant bit out of the 32 we use.

• It's mathematically correct in that x - y = x + -y using regular addition the same way you learnt in grade school. This means that processors don't need to do anything special to implement subtraction if they already have addition. They can simply find the twos-complement of y (recall, flip the bits and add one) and then add x and y using the addition circuit they already have, rather than having a special circuit for subtraction.

Answer 2

Binary works as follows (as your 32 bits).

   1  1  1  1 | 1  1  1  1 | 1  1  1  1 | 1  1  1  1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1 | 1 1 1 1

2^ 31 30 29 28  27 26 25 24  23 22 21 20  19 18 17 16......................................0
   x

x = sign bit (if 1 then negative number if 0 then positive)

So the highest number is 0111111111............1 (all ones except the negative bit), which is 2^30 + 2 ^29 + 2^28 +........+2^1 + 2^0 or 2,147,483,647.

The lowest is 1000000.........0, meaning -2^31 or -2147483648.

Answer 3

Is this what high level languages lead to!? Eeek!

As other people have said it's a base 2 counting system. Humans are naturally base 10 counters mostly, though time for some reason is base 60, and 6 x 9 = 42 in base 13. Alan Turing was apparently adept at base 17 mental arithmetic.

Computers operate in base 2 because it's easy for the electronics to be either on or off - representing 1 and 0 which is all you need for base 2. You could build the electronics in such a way that it was on, off or somewhere in between. That'd be 3 states, allowing you to do tertiary math (as opposed to binary math). However the reliability is reduced because it's harder to tell the difference between those three states, and the electronics is much more complicated. Even more levels leads to worse reliability.

Despite that it is done in multi level cell flash memory. In these each memory cell represents on, off and a number of intermediate values. This improves the capacity (each cell can store several bits), but it is bad news for reliability. This sort of chip is used in solid state drives, and these operate on the very edge of total unreliability in order to maximise capacity.

Answer 4

This is not a dumb question at all.

Let's start with uint because it's slightly easier. The convention is:

• You have 32 bits in a uint. Each bit is assigned a number ranging from 0 to 31. By convention the rightmost bit is 0 and the leftmost bit is 31.
• Take each bit number and raise 2 to that power, and then multiply it by the value of the bit. So if bit number three is one, that's 1 x 2³. If bit number twelve is zero, that's 0 x 2¹².
• Add up all those numbers. That's the value.

So five would be 00000000000000000000000000000101, because 5 = 1 x 2⁰ + 0 x 2¹ + 1 x 2² + ... the rest are all zero.

That's a uint. The convention for ints is:

• Compute the value as a uint.
• If the value is greater than or equal to 0 and strictly less than 2³¹ then you're done. The int and uint values are the same.
• Otherwise, subtract 2³² from the uint value and that's the int value.

This might seem like an odd convention. We use it because it turns out that it is easy to build chips that perform arithmetic in this format extremely quickly.