Chenjp commented on code in PR #782:
URL: https://github.com/apache/tomcat/pull/782#discussion_r1857791659


##########
java/org/apache/catalina/servlets/DefaultServlet.java:
##########
@@ -1231,10 +1231,25 @@ private static boolean validate(ContentRange range) {
                 (range.getEnd() >= 0) && (range.getStart() <= range.getEnd()) 
&& (range.getLength() > 0);
     }
 
-    private static boolean validate(Ranges.Entry range, long length) {
-        long start = getStart(range, length);
-        long end = getEnd(range, length);
-        return (start >= 0) && (end >= 0) && (start <= end);
+    private static boolean validate(Ranges ranges, long length) {
+        List<long[]> rangeContext = new ArrayList<long[]>();
+        for (Ranges.Entry range : ranges.getEntries()) {
+            long start = getStart(range, length);
+            long end = getEnd(range, length);
+            if (start < 0 || start > end) {
+                // illegal entry
+                return false;
+            }
+            // see rfc9110 #status.416
+            for (long[] e : rangeContext) {
+                if (Long.min(end, e[1]) - Long.max(start, e[0]) >= 0) {

Review Comment:
   Replace with another ***more readable way***:
   ```java
       long start = getStart(range, length);
       long end = getEnd(range, length);
       if (start < 0 || start > end) {
           // illegal entry
           return false;
       }
       // see rfc9110 #status.416
       // invalid if range entries is overlap (equivalent to intersection is 
not empty).
       for (long[] r : rangeContext) {
           long s2 = r[0];
           long e2 = r[1];
           // Given [s1,e1] and [s2,e2]: if intersection is empty, then { s1>e2 
|| s2>e1 }
           // logically equivalent to: If not { s1>e2 || s2>e1 }, then 
intersection is not empty.
           if (start <= e2 && s2 <= end) {
               // isOverlap
               return false;
           }
       }
   ```
   
   ```math
   $$\begin{align}
     [s_1, e_1] \cap [s_2, e_2] = \emptyset \to {(s_1 \gt e_2 \lor s_2 \gt 
e_1)} \\\\
     \iff \lnot{(s_1 \gt e_2 \lor s_2 \gt e_1)} \to [s_1, e_1] \cap [s_2, e_2] 
\neq \emptyset \\\\
     \iff {s_1 \ge e_2 \land s_2 \ge e_1} \to [s_1, e_1] \cap [s_2, e_2] \neq 
\emptyset \\\\
     \iff s_1 \ge e_2 \land s_2 \ge e_1 \to {isOverlap}
   \end{align}$$
   ```



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