Add additional copy to explain the padding that we may encounter when…

… extracting the r and s components, and how the result must be padded to correctly fill the result.

src/main/java/io/fusionauth/jwt/ec/ECDSASignature.java

@@ -70,6 +70,7 @@ public byte[] derDecode(Algorithm algorithm) throws IOException {
     byte[] r = sequence[0].getPositiveBigInteger().toByteArray();
     byte[] s = sequence[1].getPositiveBigInteger().toByteArray();
 
+    // The length of the result is fixed and discrete per algorithm.
     byte[] result;
     switch (algorithm) {
       case ES256:
@@ -85,8 +86,26 @@ public byte[] derDecode(Algorithm algorithm) throws IOException {
         throw new IllegalArgumentException("Unable to decode the signature for algorithm [" + algorithm.name() + "]");
     }
 
+    // Because the response is not encoded, the r and s component must take up an equal amount of the resulting array.
+    // This allows the consumer of this value to always safely split the value in half based upon an index value since
+    // the result is not encoded and does not contain any meta-data about the contents.
     int len = result.length / 2;
 
+    // The extracted byte array of the DER encoded value can be left padded. For this reason, the component lengths
+    // may be greater than 'len' which is half of the result. So for example, if r is left padded, the length may be
+    // equal to 67 in ES512 even though len is only 66. This is why we must calculate the source position for reading
+    // when we copy the r byte array into the result. The same is potentially true for either component. We cannot
+    // make an assumption that the source position in r or s will be 0.
+    //
+    // Similarly, when the r and s components are not padded, but they are shorter than len, we need to pad the value
+    // to be right aligned in the result. This is why the destination position may not be 0 or len respectively for
+    // r and s.
+    //
+    // If s is 65 bytes, then the destination position in the 0 initialized resulting array needs to be len + 1 so that
+    // we write the final byte of s at the end of the result.
+    //
+    // For clarity, calculate each input to the arraycopy method first.
+
     int rSrcPos = r.length > len ? (r.length - len) : 0;
     int rDstPos = Math.max(0, len - r.length);
     int rLength = Math.min(r.length, len);