Simplification
Both tests have the same meaning. They should be merged. Also, if "deleteLast()" deletes more than one char (which never happens in the current implementation, but it's not guaranteed), the new code is more correct. Change-Id: I216df7cd45f4a7d76d1d5d6aa7f8d7f9a6e35ea3main
parent
d2c69c7d12
commit
2939063947
|
@ -1423,10 +1423,8 @@ public class LatinIME extends InputMethodServiceCompatWrapper implements Keyboar
|
||||||
ic.setComposingText(textWithUnderline, 1);
|
ic.setComposingText(textWithUnderline, 1);
|
||||||
if (mWordComposer.size() == 0) {
|
if (mWordComposer.size() == 0) {
|
||||||
mHasUncommittedTypedChars = false;
|
mHasUncommittedTypedChars = false;
|
||||||
}
|
// Remaining size equals zero means we just erased the last character of the
|
||||||
if (1 == length) {
|
// word, so we can show bigrams.
|
||||||
// 1 == length means we are about to erase the last character of the word,
|
|
||||||
// so we can show bigrams.
|
|
||||||
mHandler.postUpdateBigramPredictions();
|
mHandler.postUpdateBigramPredictions();
|
||||||
} else {
|
} else {
|
||||||
// length > 1, so we still have letters to deduce a suggestion from.
|
// length > 1, so we still have letters to deduce a suggestion from.
|
||||||
|
|
Loading…
Reference in New Issue