problems with pow()

[rush5]

Hello,
for example int his code, pow(a,2) doesn't give the same value of a*a

{int a;
a=pow(60,2);
int b=60*60;
cout <<a<<" "<<b;
}

the output here is
3599 3600
can someone explain me why?
Thank you in advance

[Yawzheek]

I honestly have no idea. I put it into Visual Studio and correctly received 3600 3600. No warnings or anything. That's... strange.

[Chervil]

Possibly the implementation may use something like
exp(2*log(60))
which will give an approximate result using floating-point values.

    std::cout.precision(20);
    std::cout << exp(2*log(60)) << '\n';

Output:

3599.999999999998181

[rush5]

the interesting thing is that when I make a program:

{
cout <<pow(60,2);
}

the output is:

3600

cannot understand this c++...
I think that pow doesn't work well with integers

[rush5]

also look this program that calculates volume of a sphere:

int main()
{
    int a;
    cin >>a;
    cout <<"VOLUME = ";
    printf ("%.3f",a*a*a*4*3.14159/3);
    cout<<endl;
}

for input 1523
the output is

VOLUME = 1304460194.187

and in this program:

int main()
{
    int a;
    cin >>a;
    cout <<"VOLUME = ";
    printf ("%.3f",pow(a,3)*4*3.14159/3);
    cout<<endl;
}

the output is

VOLUME = 14797486501.627

which appears to be right...
again, why is there difference?

[Chervil]

Most of the unexpected results arise from truncation which occurs either when converting from floating-point to integer,

1 2 3 4 5

int d = 3599.99999999999; cout << d << '\n'; int e = round(3599.99999999999); cout << e << '\n';

3599 3600

or when exceeding the capacity of an integer.

1523*1523*1523*4 gives
14130570668
which requires more than 32 bits, and is truncated to
1245668780
The difference is 3*2^32

That leads to the incorrect result 1304460194.187

Recommendations. Use type double for floating-point work. Avoid mixing integer and floating-point types in the same calculation.

Constants such as 4 or 3 are integers, use 4.0 and 3.0 which are type double instead.


edit:
note also, cout << 3599.99999999999 << '\n'; result is

3600

Since the default precision for cout is 6 significant digits, the displayed result is rounded before being shown, the extra digits are still there, they are simply not displayed.

[rush5]

ok, thank You very much