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