casting question! Is this roudning to the nearest integer?.:Resolved:.

[voidflux]

Hello everyone!
I'm reading some exericses and it wants me to round the number to the nearest integer. Okay this is quite easy I thought, but does this mean if i have:
34.6 to print 34
or does it mean
34.6 to print 35?
my code does the first thing, it prints 34 not 35.

double x = 34.6;
cout <<"x: " << static_cast<int>(x) << endl;

this just truncates the decimal number and prints 34! is this right?
Thanks for listening,

[Comreak]

Originally posted by voidflux
Hello everyone!
I'm reading some exericses and it wants me to round the number to the nearest integer. Okay this is quite easy I thought, but does this mean if i have:
34.6 to print 34
or does it mean
34.6 to print 35?
my code does the first thing, it prints 34 not 35.

double x = 34.6;
cout <<"x: " << static_cast<int>(x) << endl;

this just truncates the decimal number and prints 34! is this right?
Thanks for listening,

Nope, 34.6 rounded to the nearest integer is 35 (normally).

[transcendental]

It usually truncate off the decimal point part.

There is no way round about this. You would have to use floor() and ceil() in <cmath> header.

[voidflux]

Thanks guys! this book must have a typo then because this is only chapter 2 and they only introduced how to static_cast somthing, never how to use the cmath functions for rounding but thanks for the info!

[voidflux]

I just found out that you can round numbers by just using some basic ostream objects, such as setprecision and fixed, yay!

[sunburnt]

you can also round to the nearest integer in amazing mathemagical ways!

int RoundedValue = static_cast<int>(dblValue + .5);

Since casting to an int will just truncate the decimal part.