osx.m: avoid division by zero in startConfigureSheet.

When we set up a configuration sheet, we track the minimum overall
width that the controls will fit into (in a variable 'totalw'), and
separately, the minimum width needed by each of the left and right
columns containing control labels and actual controls ('leftw' and
'rightw'). If totalw > leftw+rightw at the end of the process, then we
must expand the two columns so that they have the right sum.

However, sometimes leftw+rightw can be zero, while totalw > 0. This
occurs if _no_ control in the box was of a type that used the left and
right columns for different things, so that the entire loop over the
controls only incremented totalw, and not leftw or rightw. For
example, in a puzzle such as Cube that defines no preferences of its
own, the only control in the preferences pane is midend.c's standard
"Keyboard shortcuts without Ctrl" preference, which is C_BOOLEAN and
only uses totalw.

In that situation, the code for proportionate distribution of the
excess divides by zero. So it needs a special case.

osx.m

@@ -1321,7 +1321,17 @@ struct frontend {
 	totalw = leftw + SPACING + rightw;
     if (totalw > leftw + SPACING + rightw) {
 	int excess = totalw - (leftw + SPACING + rightw);
-	int leftexcess = leftw * excess / (leftw + rightw);
+        /*
+         * Distribute the excess in proportion across the left and
+         * right columns of the sheet, by allocating a proportion
+         * leftw/(leftw+rightw) to the left one. An exception is if
+         * leftw+rightw == 0, which can happen if every control in the
+         * sheet was a C_BOOLEAN which only increments totalw; in that
+         * case it doesn't much matter what we do, so I just allocate
+         * the space half and half.
+         */
+	int leftexcess = (leftw + rightw == 0 ? excess / 2 :
+                          leftw * excess / (leftw + rightw));
 	int rightexcess = excess - leftexcess;
 	leftw += leftexcess;
 	rightw += rightexcess;