The ITInt8 class
Base class: none
Encapsulates 8-byte integer value. This class can be used in any HCL Informix® Object Interface for C++ client application, although only HCL Informix supports the int8 data type.
This class provides the following methods.
Method | Description |
---|---|
ITInt8() | Creates uninitialized instance of ITInt8. |
ITInt8 &operator=(<<type>) | Sets ITInt8 to the value of <type>,
where <type> is one of the following:
|
IsNull() | Returns TRUE if an object does not represent a valid 8-byte integer. |
Conversion operators | ITInt8 provides conversions to the value of one
of the following types:
|
Other operators | ITInt8 provides assignment comparison, and arithmetic operators. The results of arithmetic operations on ITInt8 objects might not fit into 8 bytes, in which case, the result would not be a valid ITInt8 object. |
In Version 2.70, you can use new constructors to create objects by using each of the built-in numeric types as initialization arguments. This eliminates the need to explicitly assign a numeric type that is not an int8 (for example, int) to an ITInt8 object before comparing it with an ITInt8 object.
The new constructors are:
ITInt8( const int );
ITInt8( const long );
ITInt8( const float );
ITInt8( const double );
ITInt8( const mi_int8 );
ITInt8( const ITString & );
#ifdef IT_COMPILER_HAS_LONG_LONG
ITInt8( const IT_LONG_LONG );
#endif
Before version 2.70, to initialize an ITInt8 object, the application
must assign a value to an ITInt8 object by using the assignment operator
(=), as follows:
int i = 100;
ITInt8 i8;
i8 = i;
if ( i8 == (ITInt8)i )
With Version 2.70 and later, the assignment can be replaced by
an ITInt8 constructor call:
int i = 100;
ITInt8 i8(i); // or ITInt8 i8(100);
if ( i8 == (ITInt8)i )