What is meant by . (dot) usage after a number in Fortran code?
For example:
x=a+b+45.-c-d
This means that it's a floating point constant, not integer. 45. is a shorter way to write 45.0.
Related
I want to print "845100400152152934331135470251" or "1071292029505993517027974728227441735014801995855195223534251"
but in C++ the max value of "Unsigned long long " is "18446744073709551615"
this is much less than which I want to print
please help me...
First of all, your problem is not about printing big numbers but storing them in variables (and maybe calculating on them).
On some compilers (GCC for example), you have variable types like int128 that can handle numbers up to 10^38 (more less).
If this doesn't solve the problem, you'll have to write your own arithmetic. For example, store numbers in strings and write functions that will calculate on them (addition and subtraction is rather easy, multiplying medium (as long as numbers aren't really huge), dividing by big integers hard). Alternatively you can look for already made big integer libraries (on the Internet, c++ doesn't have built-in one).
I am trying to convert a boost::multiprecision::cpp_dec_float_x to a boost::multiprecision::uintx_t. So basically a boost bigreal to a boost bigint, with regards to memory needed for this conversion not to be lossy.
Consider the following:
boost::multiprecision::cpp_dec_float_100 myreal(100); /* bigreal */
boost::multiprecision::uint256_t myint; /* bigint */
Designing memory allocation
I want to make a conversion from the first to the last. Consider that I kept in count the number of bits needed for this. Starting from a 256-bit integer I need a float able to store from 0 to 2^256-1. How many digits do I need for this? Exactly 256*log_10(2) ~= 77. So a 100 digit float is more than enough. So if I keep my real number lower than 2^256, I can convert it to a 256-bit integer.
How can I make the conversion considering that convert_to<> can only be used with built in types and static_cast<> raise errors (which is expected considering that the boost documentation does not mention such a context)? Thankyou
Do not care about data-loss
I do not care about data loss. For my purpose I will store (in the bigreal variable) an integer number (no decimal part). So I am just ok!
I don't know if this is you are looking for, try...
cpp_dec_float_100 myreal(100);
cpp_dec_float_100 int_part = myreal.backend().extract_integer_part();
The type is still cpp_dec_float_100, but only contains the integer part.
I hope this helps.
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Inputting large numbers in c++?
Arbitrary-precision arithmetic Explanation
I need to multiply two huge huge integers, like:
a=1212121212121212121212121212121212121212121212121212;
b=1212121212121212121212121212121212121212121212121212;
I think there are no data types in C and C++ to hold this huge an integer, so I thought to keep it as a string format like:-
char *number1="1212121212121212121212121212121212121212121212121212";
char *number2="1212121212121212121212121212121212121212121212121212";
during the time of multiplication I convert it into string with help of atoi() function like:
atoi(number1)*atoi(number2);
As usual the output of this multiplication will be obviously huge, so I need to change the output in string format.
I know there is an itoa() function which converts an integer to a string but it is not compatible with all compilers. Can any body tell me what I should do in this scenario?
I am using Ubuntu-10.04 and the g++ compiler.
Since C and C++ do not offer a native type that supports big numbers, it makes no sense to call atoi() to parse such numbers. atoi() returns a native int which is capped at 2,147,483,647 on 32-bit platforms.
You can use one of the numerous bignum libraries, like GMP for instance.
I think, the best variant besides using some math libraries is to split those numbers into int arrays with some fixed limit. Then just perform multiplication using basic math multiplication methods. And do not forget about overflows.
Multiplying the large numbers is very
difficult, however we can do it by
applying the logarithm of
multiplication of two numbers formula
and now we are going know how we
derived the product of two numbers’
logarithm.
Let us consider a, m and n are positive real numbers but a does not equal to 1 which means ‘a’ belongs to R+ – {1}. Logarithm of m and n to base a are x and y respectively by satisfying ax is equal to m and ay is equal to n condition.
loga (m.n) = x + y
As we already know x = loga m and y = loga n.
loga (m.n) = loga m + loga n
logarithm of multiplication of two values is equal to summation of the same values’ logarithms. The same logarithmic fundamental can now help us in multiplying the two large numbers by adding the logarithm of those values. If you don’t have a calculator, just take the logarithmic table help to perform this.
Using atoi() is also not helpful since the number itself won't fit in integer data type.
You have to simulate the method you did in elementary school.
121
*23
----
363
242*
----
2783
The implementation is left as an exercise. You would also need to know how to add big numbers.
So, for a very silly project in C++, we are making our own long integer class, called VLI (Very Long Int). The way it works (they backboned it, blame them for stupidity) is this:
User inputs up to 50 digits, which are input as string.
String is stored in pre-made Sequence class, which stores the string in an array, in reverse order.
That means, when "1234" is input, it gets stored as [4|3|2|1].
So, my question is this: How can I go about doing division using only these arrays of chars?
If the input answer is over 32 digits, I can't use ints to check for stuff, and they basically saying using long ints here is cheating.
Any input is welcome, and I can give more clarification if need be, thanks everyone.
Implement the long division algorithm you learned in grade school.
Start by implementing subtraction. Create a function which can string-subtract any number from the input. Then you should be able to detect whether the result is negative. Modify this function to allow the number to be string-shifted before you subtract…
Get your school math book out, you did manual division some years ago in school I suppose. It is exactly the same principle :)
Potatoswatter is correct. I wrote a Pascal program in the past that worked on arbitrary length numbers as strings, and it could calculate the square root as well.
Here is a reminder of technique for long division: Long Division to Decimal Places
It's a coding practice. I read these numbers as double from a file:
112233 445566
8717829120000 2.4
16000000 1307674.368
10000 2092278988.8
1234567 890123
After some computation, I should output some of them. I want to make them appear just the same as in the file, no filling zeros, no scientific notation, how could I achieve it? Do I have to read in as string then convert them?
Edit: Erm...Do you guys mean that there is actually no way for the program to know how the numbers look like originally?
If you want the output to be identical to the input, then yes, you need to read them in as strings and store the strings to be output later.
Why? When dealing with floating point numbers, the computer can't represent most decimal fractional parts exactly in binary. So in a number like 2.4, the internal representation won't be exactly 2.4, it will be slightly different. Most of the time, the C/C++ I/O libraries will take such a binary number and print 2.4, but for some numbers, it might print something like 2.40000000001 or 2.399999999.
So, that's why you want to keep the original strings around.
There's no built-in way to do this. Yes, you'll need to keep the original representation, and some link between that representation and the parsed number (a map<string, double> for example).
If you need to output the original numbers as they were, you might be better off storing the original strings somewhere. During the decimal -> binary -> decimal conversion, some precision may get lost, due to the precision limits of double. You may not end up printing the exact same decimal digits.
You read them as strings, record all the many finnicky aspects (do they have decimal parts, how long, any padding zeroes, etc, etc) and remember all those aspects for each field, convert them into numbers if you absolutely must, then use all the crazy little aspect you stored when formatting the numbers for output again. Crazy, but the only way to achieve literally what you're asking for.