Nav Nav - 25 days ago 6
C++ Question

Purpose of a ".f" appended to a number?

I saw "1/3.f" in a program, and wondered what the ".f" was for. So tried my own program:

#include<iostream>
using namespace std;
int main()
{
cout<<(float)1/3<<endl;
cout<<1/3.f<<endl;
cout<<1/3<<endl;
}


Is the .f used like a cast? Any place where I can read more about this interesting syntax?

Answer Source

Without the .f the number gets interpreted as an integer, hence 1/3 is (int)1/(int)3 => (int)0 instead of the desired (float)0.333333. The .f tells the compiler to interpret the literal as a floating point number of type float. There are other such constructs such as for example 0UL which means a (unsigned long)0, whereas a plain 0 would be an (int)0.

The .f is actually two components, the . which indicates that the literal is a floating point number rather than an integer, and the f suffix which tells the compiler the literal should be of type float rather than the default double type used for floating point literals.

Disclaimer; the "cast construct" used in the above explanation is not an actual cast, but just a way to indicate the type of the literal.

If you want to know all about literals and the suffixes you can use in them, you can read the C++ standard, (1997 draft, C++11 draft, C++14 draft) or alternatively, have a look at a decent textbook, such as Stroustrup's The C++ Programming Language.

As an aside, in your example (float)1/3 the literals 1 and 3 are actually integers, but the 1 is first cast to a float by your cast, then subsequently the 3 gets implicitly cast to a float because it is a righthand operand of a floating point operator. (The operator is floating point because its lefthand operand is floating point.)