What is the maximum value for an int32?

Answer 1

Well, it has 32 bits and hence can store 2^32 different values. Half of those are negative.

The solution is 2,147,483,647

And the lowest is −2,147,483,648.

(Notice that there is one more negative value.)

Answer 2

First write out 47 twice, (you like Agent 47, right?), keeping spaces as shown (each dash is a slot for a single digit. First 2 slots, then 4)

--47----47

Think you have 12 in hand (because 12 = a dozen). Multiply it by 4, first digit of Agent 47's number, i.e. 47, and place the result to the right of first pair you already have

12 * 4 = 48
--4748--47 <-- after placing 48 to the right of first 47

Then multiply 12 by 3 (in order to make second digit of Agent 47's number, which is 7, you need 7 - 4 = 3) and put the result to the right of the first 2 pairs, the last pair-slot

12 * 3 = 36
--47483647 <-- after placing 36 to the right of first two pairs

Finally drag digits one by one from your hand starting from right-most digit (2 in this case) and place them in the first empty slot you get

2-47483647 <-- after placing 2
2147483647 <-- after placing 1

There you have it! For negative limit, you can think of that as 1 more in absolute value than the positive limit.

Practise a few times, and you will get the hang of it!

Answer 3

The most correct answer I can think of is Int32.MaxValue.

Answer 4

That's how I remembered 2147483647:

• 214 - because 2.14 is approximately pi-1
• 48 = 6*8
• 64 = 8*8

Write these horizontally:

214_48_64_
and insert:
   ^  ^  ^
   7  3  7 - which is Boeing's airliner jet (thanks, sgorozco)

Now you've got 2147483647.

Hope this helps at least a bit.

Answer 5

Anyway, take this regex (it determines if the string contains a non-negative Integer in decimal form that is also not greater than Int32.MaxValue)

[0-9]{1,9}|[0-1][0-9]{1,8}|20[0-9]{1,8}|21[0-3][0-9]{1,7}|214[0-6][0-9]{1,7}|2147[0-3][0-9]{1,6}|21474[0-7][0-9]{1,5}|214748[0-2][0-9]{1,4}|2147483[0-5][0-9]{1,3}|21474836[0-3][0-9]{1,2}|214748364[0-7]

Maybe it would help you to remember.

Answer 6

2^(x+y) = 2^x * 2^y

2^10 ~ 1,000
2^20 ~ 1,000,000
2^30 ~ 1,000,000,000
2^40 ~ 1,000,000,000,000
(etc.)

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512

So, 2^31 (signed int max) is 2^30 (about 1 billion) times 2^1 (2), or about 2 billion. And 2^32 is 2^30 * 2^2 or about 4 billion. This method of approximation is accurate enough even out to around 2^64 (where the error grows to about 15%).

If you need an exact answer then you should pull up a calculator.

Handy word-aligned capacity approximations:

• 2^16 ~= 64 thousand // uint16
• 2^32 ~= 4 billion // uint32, IPv4, unixtime
• 2^64 ~= 16 quintillion (aka 16 billion billions or 16 million trillions) // uint64, "bigint"
• 2^128 ~= 256 quintillion quintillion (aka 256 trillion trillion trillions) // IPv6, GUID