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

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.

Code:

---

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, :D

Quote:

---

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.

Code:

---

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, :D

---

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

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.

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!

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

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

Code:

---

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

---

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