What are they? Why do we need them? Casts What are they? Why do we need them? Copyright 2005-2007 Curt Hill

Mixed Mode Arithmetic Multiple types in one arithmetic expression 3.5 * 2 A real and integer constant Most programming languages allow some form of mixed mode C and C++ are quite liberal Java somewhat more restrictive Copyright 2005-2007 Curt Hill

Better Example Suppose: int a=2, b; double d; cin >> b >> d; d = d*b – a; a = b*d – a; What is the difference? One loses information and the other does not! Copyright 2005-2007 Curt Hill

Conversions C++ will perform certain conversions automatically Converting an integer value to a double value is automatic Doubles are stronger This conversion does not lose anything The conversion from double to int will also be done automatically Even though information is lost Copyright 2005-2007 Curt Hill

When is the conversion done? Consider the following: int a=2, b=3; double d = 3; d = a/b; What do we get? From an algebra view: 2/3 should be .6666… Actually we get zero Copyright 2005-2007 Curt Hill

Why zero? C++ only looks at the immediate pair of operands and refuses to look ahead So it first does a/b, which is 2/3 Since both are integers, we do integer division This gives a zero Copyright 2005-2007 Curt Hill

Observations C++ is very short sighted Even though the resulting expression of: d = a/b*d; is double, it does no conversion until needed It avoids some thornier problems by doing so Copyright 2005-2007 Curt Hill

So what is the fix? How do we get 2/3 instead of zero? A cast A cast is a conversion of type from one type to another The variable type is not changed! We extract the value and change the type of that value Copyright 2005-2007 Curt Hill

Styles of Casts The compiler will do a cast automatically for mixed mode arithmetic For operators convert weaker to stronger For assignments convert to receiving type C style cast C++ style cast Copyright 2005-2007 Curt Hill

The C Cast The form of a cast is the type in parenthesis This is a unary operator, it takes the operand on its right and converts the value into the appropriate type: (double)a/b It has very high precedence so it occurs before the division Copyright 2005-2007 Curt Hill

The C++ Cast C++ introduced a cast that looks like a function call or constructor double(a)/b Use the type name like a function name Precedence is not an issue This can only be used for types with a single name Use (long double)a/b for multiple names Copyright 2005-2007 Curt Hill

The 2/3 problem solved We have two ways to accomplish what we wanted: d = double(a)/b; d = (double)a/b; d = a/double(b); These work, but these do not: d = (double)(a/b); d = double(a/b); The damage is already done Copyright 2005-2007 Curt Hill

Rounding How is a double converted into an int? Truncation! (int)0.99999 is zero! How do we round? Simple: add 0.5 prior to the cast (int)(0.9999 + 0.5) is (int)(1.4999) is 1 Copyright 2005-2007 Curt Hill