Merge #11133: Document assumptions that are being made to avoid divis…

…ion by zero

55509f1 Document assumptions that are being made to avoid division by zero (practicalswift)

Pull request description:

  Document assumptions (via `assert(…)`:s) that are being made to avoid division by zero.

  Rationale:
  * Make it clear to human reviewers and non-human static analyzers that what might look like potential division by zero cases are written the way they are intentionally (these cases are currently flagged by various static analyzers).

Tree-SHA512: bbb67b1370afd8f39bda35f9e3a20f4325f017d94cc1bfac3b0d36c9f34c2d95a9efe11efe44db29fb4aadd25d8276d8f0e03c8806ac64f0d21d821912e13b8e

src/policy/fees.cpp

@@ -503,6 +503,7 @@ void TxConfirmStats::removeTx(unsigned int entryHeight, unsigned int nBestSeenHe
         }
     }
     if (!inBlock && (unsigned int)blocksAgo >= scale) { // Only counts as a failure if not confirmed for entire period
+        assert(scale != 0);
         unsigned int periodsAgo = blocksAgo / scale;
         for (size_t i = 0; i < periodsAgo && i < failAvg.size(); i++) {
             failAvg[i][bucketindex]++;

src/qt/coincontroldialog.cpp

@@ -582,6 +582,7 @@ void CoinControlDialog::updateLabels(WalletModel *model, QDialog* dialog)
     QString toolTipDust = tr("This label turns red if any recipient receives an amount smaller than the current dust threshold.");
 
     // how many satoshis the estimated fee can vary per byte we guess wrong
+    assert(nBytes != 0);
     double dFeeVary = (double)nPayFee / nBytes;
 
     QString toolTip4 = tr("Can vary +/- %1 satoshi(s) per input.").arg(dFeeVary);

src/wallet/wallet.cpp

@@ -2704,6 +2704,7 @@ bool CWallet::CreateTransaction(const std::vector<CRecipient>& vecSend, CWalletT
 
                     if (recipient.fSubtractFeeFromAmount)
                     {
+                        assert(nSubtractFeeFromAmount != 0);
                         txout.nValue -= nFeeRet / nSubtractFeeFromAmount; // Subtract fee equally from each selected recipient
 
                         if (fFirst) // first receiver pays the remainder not divisible by output count