long-integer

I have a pretty basic question, but I am not sure if I understand the concept or not. Suppose we have: int a = 1000000; int b = 1000000; long long c = a * b; When I run this, c shows negative value, so I changed also a and b to long long and then everything was fine. So why do I have to change a and b , when their values are in range of int and their product is assigned to c (which is long long )? I am using C/C++ The int s are not promoted to long long before multiplication, they remain int s and the product as well. Then the product is cast to long long , but too late, overflow has struck.

I'm looking for an efficient method of generating large numbers (that includes floating point types!) in Swift, with arbitrary ranges (which may even be UInt.max or Int.max ) All the existing questions I've seen either crash for large values ( UInt.max ) or don't support ranges. I know that you can read from /dev/urandom for random bytes, but that doesn't help restrict these values to a given interval (and I'm pretty sure looping until it does isn't efficient). Martin R Here is a possible solution for UInt , Int and Double which works with the full range of those types. It is written as

When I parse this little piece of JSON { "value" : 9223372036854775807 } that's what I get { hello: 9223372036854776000 } Is there any way to parse it properly? Not with built-in JSON.parse. You'll need to parse it manually and treat values as string (if you want to do arithmetics with them there is bignumber.js ) You can use Douglas Crockford JSON.js library as a base for your parser. EDIT: I created a package for you :) var JSONbig = require('json-bigint'); var json = '{ "value" : 9223372036854775807, "v2": 123 }'; console.log('Input:', json); console.log(''); console.log('node.js bult-in

In a recent homework assignment I've been told to use long variable to store a result, since it may be a big number. I decided to check will it really matter for me, on my system (intel core i5/64-bit windows 7/gnu gcc compiler) and found out that the following code: printf("sizeof(char) => %d\n", sizeof(char)); printf("sizeof(short) => %d\n", sizeof(short)); printf("sizeof(short int) => %d\n", sizeof(short int)); printf("sizeof(int) => %d\n", sizeof(int)); printf("sizeof(long) => %d\n", sizeof(long)); printf("sizeof(long int) => %d\n", sizeof(long int)); printf("sizeof(long long) => %d\n",

Here's an excerpt from Sun's Java tutorials : A switch works with the byte , short , char , and int primitive data types. It also works with enumerated types (discussed in Classes and Inheritance) and a few special classes that "wrap" certain primitive types: Character , Byte , Short , and Integer (discussed in Simple Data Objects). There must be a good reason why the long primitive data type is not allowed. Anyone know what it is? I think to some extent it was probably an arbitrary decision based on typical use of switch. A switch can essentially be implemented in two ways (or in principle, a

This question already has an answer here: Integer division: How do you produce a double? 10 answers if I have something like: long x = 1/2; shouldn't this be rounded up to 1? When I print it on the screen it say 0. It's doing integer division, which truncates everything to the right of the decimal point. Integer division has its roots in number theory. When you do 1/2 you are asking how many times does 2 equal 1? The answer is never, so the equation becomes 0*2 + 1 = 1, where 0 is the quotient (what you get from 1/2) and 1 is the remainder (what you get from 1%2). It is right to point out that