why can't I express 10 power 10

Question

I tried to express 10 power 10 but it didn't work, so I thought the problem is the range of the data type.

Finally I changed the type from int to __int64 but it isn't yet working.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int a, r, n;

    __int64  d;

    cin >> a; //  a=10
    cin >> r; //  r=10
    cin >> n; //  n=10

    d = a * pow (r, n - 1);

    cout << int(d);
}

My expected answer is 10000000000 but result of the operation is 1410065408

What's the problem?

The problem is that you converted the answer to an `int` with `cout << int(d)`, and the answer doesn't fit in an `int`. — user3386109, Jun 23 '20 at 00:58
In `` there is `std::int64_t`, which is a portable way of getting a 64 bit signed integer if one exists, instead of `__int64`. — François Andrieux, Jun 23 '20 at 01:00
Also, don't mix float and int. Use a for loop and multiply by ten ten times — Jeffrey, Jun 23 '20 at 01:11
@FrançoisAndrieux no, all overloads with integer exponent [has been removed since C++11](https://en.cppreference.com/w/cpp/numeric/math/pow). The only overload the accepts integral values also casts the value internally to floating-point types: *7) A set of overloads or a function template for all combinations of arguments of arithmetic type not covered by 1-3). **If any argument has integral type, it is cast to `double`. If any argument is `long double`, then the return type Promoted is also `long double`, otherwise the return type is always `double`.*** — phuclv, Mar 13 '21 at 03:39

Rohan Bari · Accepted Answer · 2020-06-23T13:14:04.163

1

You're trying to convert __int64 into int which can't hold values like 10000000000 in the cout statement.

Don't confuse between __int64 (holds 8-bytes data) and int (holds 4-bytes data). Simply remove the conversion int().

edited Jun 23 '20 at 13:14

answered Jun 23 '20 at 03:54

Rohan Bari

7,482
3
14
34

1

Thank you sir I really appreciate – Hamp Jun 23 '20 at 18:30

score -1 · Answer 2 · answered Jun 23 '20 at 04:10

-1

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int a,n;

    long long int  d; //change the type to long long int


    cin >> a;
    cin >> n; //  n=10

    d =  pow (a, n );

    cout << d;

}

The problem was the range of your data type. By changing to "long long int" you'll get the expected answer 10000000000.

answered Jun 23 '20 at 04:10

Sharath Chander P

151
8

`long long int` is not likely to be larger than `__int64` and a 64 bit integer is definitely large enough already anyway. You've removed the cast from the `cout` which is why you get the expected result. – François Andrieux Jun 23 '20 at 12:25
Thank you sir I really appreciate – Hamp Jun 23 '20 at 18:31
[Why should I not #include ?](https://stackoverflow.com/q/31816095/995714), [Why is “using namespace std;” considered bad practice?](https://stackoverflow.com/q/1452721/995714) – phuclv Mar 08 '21 at 07:07

score -3 · Answer 3 · answered Jun 23 '20 at 01:18

-3

Use powl(a,b) instead of pow(a,b) for large integers as in your case.

answered Jun 23 '20 at 01:18

Abhi Gupta

17
6

Thank you sir I really appreciate – Hamp Jun 23 '20 at 18:31
@Hamp But some intellectuals down voted my first answer – Abhi Gupta Jun 23 '20 at 19:22
because your "answer" is wrong. It has nothing to do with pow or powl. The issue lies in `int(d)` where an int can't store such a huge number – phuclv Mar 08 '21 at 07:01

why can't I express 10 power 10

3 Answers3