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Use Long.parseLong(CharSequence,...) to avoid intermediate String creation

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Where possible, switch to the Long.parseLong variant that accepts a
start and end index for the supplied CharSequence, thus avoiding making
unnecessary copies of the String input.

Closes gh-30710
This commit is contained in:
Patrick Strawderman
2023-06-20 20:55:35 -07:00
committed by Sam Brannen
parent af1c06917d
commit 01e90bbd0e
4 changed files with 12 additions and 11 deletions
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6
spring-web/src/main/java/org/springframework/http/HttpRange.java
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@@ -147,9 +147,9 @@ public abstract class HttpRange {
Assert.hasLength(range, "Range String must not be empty");
int dashIdx = range.indexOf('-');
if (dashIdx > 0) {
long firstPos = Long.parseLong(range.substring(0, dashIdx));
long firstPos = Long.parseLong(range, 0, dashIdx, 10);
if (dashIdx < range.length() - 1) {
Long lastPos = Long.parseLong(range.substring(dashIdx + 1));
Long lastPos = Long.parseLong(range, dashIdx + 1, range.length(), 10);
return new ByteRange(firstPos, lastPos);
}
else {
@@ -157,7 +157,7 @@ public abstract class HttpRange {
}
}
else if (dashIdx == 0) {
long suffixLength = Long.parseLong(range.substring(1));
long suffixLength = Long.parseLong(range, 1, range.length(), 10);
return new SuffixByteRange(suffixLength);
}
else {