I've been using c++ for over 3 months and I can't figure out the exact maximum value of numbers in long long int.
Does anyone know the value?
It depends on the system. The C++ standard only guarantees that the minimum size for long long int will be 64-bits. This is also by far the most common size.
With a 64-bit size, the maximum number that can be represented will be 2^63 - 1, which equals 9223372036854775807. The reason for this exact size, is that we need half of the bit combinations for the negative numbers, then one for 0 and the rest for the positive numbers.
The max value on a specific system can also be checked programmatically with:
#include <iostream>
#include <limits>
int main() {
std::cout << std::numeric_limits<long long int>::max();
}
Output:
9223372036854775807
long long int doesn't have a fixed maximum value specified by C++ language - it depends on the platform.
Use std::numeric_limits<long long>::max() (from <limits> header) to get its maximum value.
The guarantee by the C++ Standard for the long long modifier is that it will have a width of at least 64 bits. (see here).
The exact width of the type is, however, dependant on the particular platform, so you might get more thatn 64 bits for your types.
To check the maximum and minimum number a long long inttype can hold on your machine and implementation, use the <limits> library with std::numeric_limits<long long>. You can read more about that here.
Related
Is it possible to define 999e999 value without using the char type?
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Thanks in advance.
Is it possible to define 999e999 value without using the char type?
No, that's not possible using intrinsic c++ data types. That's a way to big number that could be held in either a unsigned long long type in c++.
A long double type would enable you to use 10 based exponents as large as you want, for modern FPU architectures.
What can be achieved with your current CPU architecture can be explored using the std::numeric_limits facilities like this:
#include <iostream>
#include <limits>
int main() {
std::cout<< "max_exponent10: " << std::numeric_limits<long double>::max_exponent10 << std::endl;
}
Output:
max_exponent10: 4932
See the online demo
You have to use a 3rd party library (like GMP) or write your own algorithms to deal with big numbers like that.
In most (If not all) implementations, that constant is just too big to be represented as a unsigned long long or long double (Though some may just have it be floating point infinity).
You may instead be interested in std::numeric_limits<T>::infinity() (for float, double or long double) or std::numeric_limits<T>::max() instead.
I've tried defining it even with unsigned long long, but the compiler keeps giving me constant too big error.
Of course it does. A long long is typically 64 bits long, which gives you log(2^64) ≅ 19 decimal digits of precision. 999e999 ≅ (10^3)^1000, so is on the order of 3000 decimal digits long, or nearly 10,000 bits long. So 999e999 isn't just too big for a long long, it's too big by an enormous margin.
Is it possible to define 999e999 value without using the char type?
Sure. You could define an integer-like type based on an array of some sort of integers, like long long. You'd still need to write a set of operators to work with your new giant type, though. Also, most of the time when you're working with numbers that large, you don't need an exact representation, which is why floating point types like float and double are useful.
thanks for checking this question.
so i wonder how can i get output 2^64, if i input is 2^64.
in unsigned long long int, it just only reach 2^64-1 == 18446744073709551615
the point is, when input number == 18446744073709551616
the output will be "2^64"
but code that i have is :
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
unsigned long long int a;
cin >> a;
if (a == pow(2,64))
{
cout << "2^64";
}
}
so the problem is, if i input : 18446744073709551616
it will no output. how can i make the output "2^64"?
unsigned long long is 64-bits or larger. That means in some machines, it is going to be just 64-bits. In this case, you will have overflow problem in your code.
Check ULLONG_MAX (#include <climits>)
Maximum value for an object of type unsigned long long int 18446744073709551615 (2^64-1) or greater*
the actual value depends on the particular system and library implementation, but shall reflect the limits of these types in the target platform.
(From http://www.cplusplus.com/reference/climits/ ).
This means your target platform supports 64 bit unsigned long long values. Thus your limit is 2^64-1.
You can try using a big integer library, like this to work around the limitation.
The largest data type in c++ differs from compiler to compiler...but generally unsigned long long int is of course considered large!!
So to solve ur problem..,better change the if condition to pow(2,64)-1...
Other than that if you really wanna implement that condition in your project, u add a "Do u mean condition..like do u mean 'thenumber+1' ?? "and proceed..;)
So I think I'm a bit confused. I'm searching information about the limits of the differents types of integers. I've seen that the limit for unsigned long int is 4294967295 but when I do:
cout << numeric_limits<unsigned long int>::max() << endl;
I'm getting:
18446744073709551615
And if I'm not wrong this number is the limit of unsigned long long, isn'it? So what is happening?
Thank you
I've seen that the limit for unsigned long int is 4294967295
Whoever told you that was wrong.
The limit for unsigned long int will usually be that on systems for which the type is 32-bit.
But yours is evidently 64-bit, so you have a different limit.
this number is the limit of unsigned long long, isn'it?
Again, you're making assumptions about type width.
The width of types varies across compilers/platforms.
If you want to use types with a fixed size, then those do exist.
The standard only defines lower bounds for the limits of integers. For example, the lower bound for the maximum that an unsigned long can represent is 4294967295.
std::numeric_limits<unsigned long>::max() gives the implementation-defined maximum value an unsigned long can represent (i.e. what the current implementation aka compiler/linker/etc actually supports).
This means it is required that
std::numeric_limits<unsigned long>::max() gives a value that is 4294967295 or more. There is nothing preventing it giving a larger result. However, an implementation that gives a smaller result is non-compliant with the standard.
Note that, when moving between compilers, the only guarantee is "4294967295 or more". If one implementation gives a larger value, there is no guarantee that another implementation will.
For the most part, the standard actually says nothing whatsoever about the number of actual bits used to represent the basic integral types, like unsigned long.
The value 18446744073709551615 is consistent with a 64-bit unsigned long, in practice.
Similar stories, albeit with different values, for other integral types (int, char, short, long, etc).
I'm working on a relatively simple problem based around adding all the primes under a certain value together. I've written a program that should accomplish this task. I am using long type variables. As I get up into higher numbers (~200/300k), the variable I am using to track the sum becomes negative despite the fact that no negative values are being added to it (based on my knowledge and some testing I've done). Is there some issue with the data type or I am missing something.
My code is below (in C++) [Vector is basically a dynamic array in case people are wondering]:
bool checkPrime(int number, vector<long> & primes, int numberOfPrimes) {
for (int i=0; i<numberOfPrimes-1; i++) {
if(number%primes[i]==0) return false;
}
return true;
}
long solveProblem10(int maxNumber) {
long sumOfPrimes=0;
vector<long> primes;
primes.resize(1);
int numberOfPrimes=0;
for (int i=2; i<maxNumber; i++) {
if(checkPrime(i, primes, numberOfPrimes)) {
sumOfPrimes=sumOfPrimes+i;
primes[numberOfPrimes]=long(i);
numberOfPrimes++;
primes.resize(numberOfPrimes+1);
}
}
return sumOfPrimes;
}
Integers represent values use two's complement which means that the highest order bit represents the sign. When you add the number up high enough, the highest bit is set (an integer overflow) and the number becomes negative.
You can resolve this by using an unsigned long (32-bit, and may still overflow with the values you're summing) or by using an unsigned long long (which is 64 bit).
the variable I am using to track the sum becomes negative despite the fact that no negative values are being added to it (based on my knowledge and some testing I've done)
longs are signed integers. In C++ and other lower-level languages, integer types have a fixed size. When you add past their maximum they will overflow and wrap-around to negative numbers. This is due to the behavior of how twos complement works.
check valid integer values: Variables. Data Types.
you're using signed long, which is usually 32 bit, which means -2kkk - 2kkk, you can either use unsigned long, which is 0-4kkk, or use 64 bit (un)signed long long
if you need values bigger 2^64 (unsigned long long), you will need to use bignum math
long is probably only 32 bits on your system - use uint64_t for the sum - this gives you a guaranteed 64 bit unsigned integer.
#include <cstdint>
uint64_t sumOfPrimes=0;
You can include header <cstdint> and use type std::uintmax_t instead of long.
I'm new to Windows development and I'm pretty confused.
When I compile this code with Visual C++ 2010, I get an error "constant too large." Why do I get this error, and how do I fix it?
Thanks!
int _tmain(int argc, _TCHAR* argv[])
{
unsigned long long foo = 142385141589604466688ULL;
return 0;
}
The digit sequence you're expressing would take about 67 bits -- maybe your "unsigned long long" type takes only (!) 64 bits, your digit sequence won't fit in its, etc, etc.
If you regularly need to deal with integers that won't fit in 64 bits you might want to look at languages that smoothly support them, such as Python (maybe with gmpy;-). Or, give up on language support and go for suitable libraries, such as GMP and MPIR!-)
A long long is 64 bits and thus holds a maximum value of 2^64, which is 9223372036854775807 as a signed value and 18446744073709551615 as an unsigned value. Your value is bigger, hence it's a constant value that's too large.
Pick a different data type to hold your value.
You get the error because your constant is too large.
From Wikipedia:
An unsigned long long's max value is at least 18,446,744,073,709,551,615
Here is the max value and your value:
18,446,744,073,709,551,615 // Max value
142,385,141,589,604,466,688 // Your value
See why your value is too long?
According to http://msdn.microsoft.com/en-us/library/s3f49ktz%28VS.100%29.aspx, the range of unsigned long long is 0 to 18,446,744,073,709,551,615.
142385141589604466688 > 18446744073709551615
You have reached the limit of your hardware to represent integers directly.
It seems that beyond 64 bits (on your hardware) requires the integer to be simulated by software constructs. There are several projects out there that help.
See BigInt
http://sourceforge.net/projects/cpp-bigint/
Note: Others have misconstrued that long long has a limit of 64 bits.
This is not accurate. The only limitation placed by the language are:
(Also Note: Currently C++ does not support long long (But C does) It is an extension by your compiler (coming in the next version of the standard))
sizeof(long) <= sizeof(long long)
sizeof(long long) * CHAR_BITS >= 64 // Not defined explicitly but deducible from
// The values defined in limits.h
For more details See:
What is the difference between an int and a long in C++?