Print precise value of pow() function in c++

Question

my output is coming wrong. I guess i'm wrong with casting. please help me out.

    int n;
    cin>>n;
    unsigned long long int a,s;
    cin>>a;
    s=(2*pow(10,n)+a);

But when I am giving large n like 17 or 18 then my output which is s is not coming as expected. see image for output e.g: when n=17, a=67576676767676788 then s=267576676767676800 which ideally should be 2*10^17 + 67576676767676788

What is the largest value that can be stored in a `long long int`? — Scott Hunter, Jan 10 '20 at 20:40
@ScottHunter sir because pow() can be used in c as well as c++ — Tejas Jadhav, Jan 10 '20 at 20:41
there is an inflation of question incorrectly flagged as C. Why do you tag a language different from the one you are using? — 463035818_is_not_an_ai, Jan 10 '20 at 20:41
`cin` is C++ not C. Answers differ depending on what language your write your code in — 463035818_is_not_an_ai, Jan 10 '20 at 20:42
`pow` can be used in a lot of other languages, but you didn't tag them. — Scott Hunter, Jan 10 '20 at 20:42
@ScottHunter i guess long long int can store upto 18 digits but I'm not very sure — Tejas Jadhav, Jan 10 '20 at 20:42
Even if you can pass integer arguments to [`std::pow`](https://en.cppreference.com/w/cpp/numeric/math/pow) it's still a floating-point function, with all the drawbacks that have (see e.g. [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken)). — Some programmer dude, Jan 10 '20 at 20:42
What do you mean by "not coming as expected"? You should show us the value of `s` that you got, so that we can see if you are just seeing a floating point issue or something more catastrophic. — jjramsey, Jan 10 '20 at 20:51
If you change your last line to `auto s = (2 * powl(10, n) + a);` (and remove `s` declaration) it will work as you expect, just change cout precision to display it correctly. — Hawky, Jan 10 '20 at 20:51
You can solve the problem by writing a function to compute 10^n as an unsigned long long, e.g. `unsigned long long power10(int n) { unsigned long long result = 1; for (int i=0; i — user3386109, Jan 10 '20 at 21:03

score 8 · Answer 1 · edited Jun 27 '22 at 21:07

8

First you have to understand what is going on.

To be able to use std::pow, compiler silently converts integer types to double and returned value is a double too.
Note that double has 16 significant digits (in decimal representation).
When you do assignment, conversion of double to long long int is silently performed
unsigned long long int - if this type has 64 bits the maximum power of 10 is 19

Now if you want to exceed this limitation you should use an external library. gmp is quite nice.

If it is acceptable to have a limitation from range of unsigned long long int just implement your own power function.

edited Jun 27 '22 at 21:07

Dhruv

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answered Jan 10 '20 at 20:53

Marek R

32,568
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Or use `long double`, although I do not know if it is is required or just "may be" bigger than double – Hawky Jan 10 '20 at 21:01
1

@Hawky `sizeof(long double) >= sizeof(double)` per the standard so it may make no difference, and doesn't on winodws. – NathanOliver Jan 10 '20 at 21:11
@NathanOliver I checked just now, sizeof(long double): msvc 8, gcc 16, clang 16. – Hawky Jan 10 '20 at 21:17

Print precise value of pow() function in c++

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