Use character literals in String.indexOf(). #1555

Fixes `LengthOneStringInIndexOf` inspection violations in test code.

Description:
>Reports String literals of length one being used as a parameter in String.indexOf() or String.lastIndexOf() calls. These String literals may be replaced by equivalent character literals, gaining some performance enhancement.
This commit is contained in:
Michal Kordas 2015-08-18 22:01:03 +02:00 committed by Roman Ivanov
parent efa16e17bf
commit 4f6d7e415d
1 changed files with 3 additions and 3 deletions

View File

@ -143,8 +143,8 @@ public abstract class BaseCheckTestSupport
String actual = lnr.readLine();
assertEquals("error message " + i, expected, actual);
String parseInt = removeDeviceFromPathOnWindows(actual);
parseInt = parseInt.substring(parseInt.indexOf(":") + 1);
parseInt = parseInt.substring(0, parseInt.indexOf(":"));
parseInt = parseInt.substring(parseInt.indexOf(':') + 1);
parseInt = parseInt.substring(0, parseInt.indexOf(':'));
int lineNumber = Integer.parseInt(parseInt);
Integer integer = Arrays.asList(aWarnsExpected).contains(lineNumber) ? lineNumber : 0;
assertEquals("error message " + i, (long) integer, lineNumber);
@ -203,7 +203,7 @@ public abstract class BaseCheckTestSupport
private static String removeDeviceFromPathOnWindows(String path) {
String os = System.getProperty("os.name", "Unix");
if (os.startsWith("Windows")) {
return path.substring(path.indexOf(":") + 1);
return path.substring(path.indexOf(':') + 1);
}
return path;
}