My computer cant do math, can your's do ?

[omar shaaban]

Code: Select all

int main()
{
    long double t=5.0f/30.0f;
long double us=1.0f/30.0f;
 
 
    cout <<int(((t)/us))<< endl;
    return 0;
}

in the code above the answer should be 5 but instead its 4

[Radikalizm]

You're probably having some floating point errors which reduces the result to 4 after the integer conversion
It also depends a lot on the compiler you're using, if I remember correctly the MSVC compiler treats a long double like a regular double for x86 architectures, so you get 64-bit precision instead of 80-bit

[hendu]

Remove the int cast and be happy.

[ACE247]

and remove the f's and the (((t)/us)) -> (t/us) they're just not necessary with this.

Because calculation of a variable put in as a equation already occurs, when the t is declared and stored as 5/30th instead of stored as an 'instruction' of 5.0 divided by 30.
The equation does not look like (((5.0f/30.0f)/(1.0f/30.0f)) but like (0.166.../0.033...)
If I'm correct that is...

[docWild]

Write it out as binary operations, then it makes perfect sense. Check out the IEEE floating point standards for proper explanations. Integer casting and division has its place, but it can also be the source of the worst kind of bug.. that which generates no errors.

[serengeor]

this works:

Code: Select all

#include <iostream>
int main()
{
long double t=5.0/30.0;
long double us=1.0/30.0;
 
 
    std::cout <<(int) ( (double) (t/us) )<< std::endl;
    return 0;
}