Page 32 - DCAP404 _Object Oriented Programming

Unit 2: Beginning of OOP Language

If x had been assigned a value of –12 and y had been assigned 2, then the value of x/y would Notes
still be –6 but the value of a x%y would be 0. Similarly, if x and y had both been assigned
negative values (–12 and –2, respectively), then the value of x/y would be 3 and the value of x%y
would be –2.
x = ((x/y)*y) + (x%y)

will be satisfied in each of the above cases. Most versions of C++ will determine the sign of the
remainder in this manner, though this feature is unspecified in the formal definition of the
language.
Here is an illustration of the results that are obtained with floating-point operands having
different signs. Let y1 and y2 be floating-point variables whose assigned values are 0.70 and
3.50. Several arithmetic expressions involving these variables are shown below, together with
their resulting values.

Expression Value
y1+y2 2.72
y1–y2 –4.28
y/y2 –0.2728

Operands that differ in type may undergo type conversion before the expression takes on its
final value. In general, the final result will be expressed in the highest precision possible,
consistent with the data type of the operands. The following rules apply when neither operand
is unsigned.
1. If both operands are floating-point types whose precisions differ (e.g., a float and a double),
the lower-precision operand will be converted to the precision of the other operand, and
the result will be expressed in this higher precision. Thus, an operation between a float
and double will result in a double; a float and a long double will result in a long double;
and a double and a long double will result in a long double. (Note: In some versions of
C++, all operands of type float are automatically converted to double.)
2. If one operand is a floating-point type (e.g., float, double or long double) and the other is
a char or an int (including short int or long int), the char or int will be converted to the
floating-point type and the result will be expressed as such. Hence, an operation between
an int and a double will result in a double.
3. If neither operand is a floating-point type but one is long int, the other will be converted
to long int and the result will be long int. Thus, an operation between a long int and an int
will result in a long int.
4. If neither operand is a floating-point type or a long int, then both operands will be
converted to int (if necessary) and the result will be int. Thus, an operation between a short
into and an int will result in an int.

Task What does the following expression evaluate to?
6 + 5 * 4 % 3

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