Share via

InkRecognitionModes Enumeration

  • Article

InkRecognitionModes Enumeration

Defines values that specify how the recognizer interprets the ink and determines the result string.

Declaration

[C++]

typedef enum InkRecognitionModes {
    IRM_None = 0,
    IRM_WordMode = 1,
    IRM_Coerce = 2,
    IRM_TopInkBreaksOnly = 4,
    IRM_Max = 8
} InkRecognitionModes;

[Microsoft® Visual Basic® 6.0]

Enum InkRecognitionModes
    IRM_None = 0
    IRM_WordMode = 1
    IRM_Coerce = 2
    IRM_TopInkBreaksOnly = 4
    <hidden> IRM_Max = 8
End Enum

Members

The following tables list the enumeration values.

Name Description
None Specifies that the recognizer applies no recognition modes.
WordMode Specifies that the recognizer treats the ink as a single word.

For example, if the context contains to get her, the recognizer returns together.

Note: Some compound words in the dictionary are treated as single words by recognizers of Latin script (for example, Los Angeles). In addition, certain factoids such as Date in U.S., UK, German, and French treat some multiword dates as single words (such as January 21, 2000).
Coerce Specifies that the recognizer coerces the result based on the factoid that you specified for the context.

For example, if you specified the Telephone factoid and the user enters the word hello, the recognizer may return a random phone number or an empty string. If you do not specify this flag, the recognizer returns hello as the result.
TopInkBreaksOnly Disables multiple segmentation.

This turns off the recognizer's ability to return recognition results based on more than one segment of the ink, where each segment corresponds to a word (in recognizers of Latin script) or a character (in recognizers of East Asian characters).

In other words, the word "together" always returns alternates based on "together" being a single word, and the recognizer does not consider that the string might also be "to get her" or some other variation with differing segmentation.

Turning on this flag enhances recognition speed.

Remarks

In C++, explicit casting is required when trying to set more than one flag at a time using the bitwise OR operator. A compilation error occurs if explicit casting is not used.