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    <p>Without the <code>.f</code> the number gets interpreted as an integer, hence <code>1/3</code> is <code>(int)1/(int)3</code> => <code>(int)0</code> instead of the desired <code>(float)0.333333</code>. The <code>.f</code> tells the compiler to interpret the literal as a floating point number of type float. There are other such constructs such as for example <code>0UL</code> which means a <code>(unsigned long)0</code>, whereas a plain <code>0</code> would be an <code>(int)0</code>.</p> <p>The <code>.f</code> is actually two components, the <code>.</code> which indicates that the literal is a floating point number rather than an integer, and the <code>f</code> suffix which tells the compiler the literal should be of type float rather than the default double type used for floating point literals.</p> <p>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.</p> <p>If you want to know all about literals and the suffixes you can use in them, you can read the C++ standard, (<a href="http://open-std.org/jtc1/sc22/open/n2356/lex.html#lex.literal" rel="nofollow noreferrer">1997 draft</a>, <a href="http://www.open-std.org/JTC1/SC22/WG21/docs/papers/2012/n3337.pdf#page=37" rel="nofollow noreferrer">C++11 draft</a>, <a href="http://www.open-std.org/JTC1/SC22/WG21/docs/papers/2013/n3797.pdf#page=37" rel="nofollow noreferrer">C++14 draft</a>, <a href="http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2017/n4659.pdf#page=43" rel="nofollow noreferrer">C++17 draft</a>) or alternatively, have a look at a decent textbook, such as Stroustrup's <a href="http://www.stroustrup.com/4th.html" rel="nofollow noreferrer">The C++ Programming Language</a>.</p> <p>As an aside, in your example <code>(float)1/3</code> the literals <code>1</code> and <code>3</code> 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.)</p>
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    1. POPurpose of a ".f" appended to a number?
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    1. COThanks...I'm surprised that this kind of syntax exists in-spite of casting syntax being available. Any place where I can read more about it? Tried googling for "floating point literal", but couldn't find any useful info.
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    2. COTo be pedantic `3.` would be interpreted as a double, not as an int.
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    3. CO@peoro indeed, I should say without the `.f`, fixed now, thanks
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