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    <p>Arithmetic operations involving <code>float</code> results in <code>float</code>.</p> <pre><code>int + float = float int * float = float float * int = float int / float = float float / int = float int / int = int </code></pre> <p>For more detail answer. Look at what the section §5/9 from the C++ Standard says</p> <blockquote> <p>Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, <strong>which is also the type of the result</strong>.</p> <p>This pattern is called the usual arithmetic conversions, which are defined as follows:</p> <p>— If either operand is of type long double, the other shall be converted to long double. </p> <p>— Otherwise, if either operand is double, the other shall be converted to double.</p> <p>— Otherwise, if either operand is float, the other shall be converted to float.</p> <p>— Otherwise, the integral promotions (4.5) shall be performed on both operands.54) </p> <p>— Then, if either operand is unsigned long the other shall be converted to unsigned long. </p> <p>— Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int. </p> <p>— Otherwise, if either operand is long, the other shall be converted to long.</p> <p>— Otherwise, if either operand is unsigned, the other shall be converted to unsigned. </p> <p>[Note: otherwise, the only remaining case is that both operands are int ]</p> </blockquote>
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    1. POImplicit type conversion rules in C++ operators
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    1. CO...so long as the other type is neither `double` nor `long double`.
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    2. CO@Charles: Correct. I quoted the relevant section from the Standard to clarify further.
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