Mixed-type Expressions

Every expression produces a value, and each value produced has a particular type. Thus, when you add or subtract two integers, the result is an integer. But what if....

a = 5 * 3.5;

The CPU uses different circuitry for integer and floating-point calculations. To evaluate this expression, both operands must be type int, or, they both must be type double. If we convert both to int, we lose information; converting them to double does not.

When your compiler encounters an expression that uses different types, it determines the operand with the greatest information potential. It then creates temporary values of that type, initializing them with the other values. This is called promotion.

Assignment and Mixed Expressions

What is stored in a in the example shown above? That depends on the type of a. If the variable is other than double, the value is again, implicitly converted into the same type as the variable. Thus, while the value calculated is 17.5, if a has type int then only the 17 will be stored.

Widening conversions occur when the assignment causes a promotion, such as from int to double. These will always succeed (just as they do in Java or C#).
Narrowing conversions occur when the assignment has the potential for losing information, such as assigning from double to int.

Narrowing implicit assignment conversions are prohibited in Java and C#, but they are the default behavior in C++. To turn off such implicit narrowing conversions, C++11 added brace or list assignment; this makes C++ work more like Java and C#.

int a, b;
a = 5 * 3.5;          // 17; implicit narrowing conversion
b = {5 * 3.5};        // c++11+; compiler error