[basicTypes] Remove confusing statement about operator precedence.

The question was about boolean logic, and not precedences.

Author: hageboeck

exercises/basicTypes/basicTypes.cpp

@@ -45,7 +45,7 @@ int main() {
   bool condition1 = false;
   bool condition2 = true;
   print( alwaysTrue && condition1 && condition2 );
-  print( alwaysTrue || condition1 && condition2 );  // Q: Why does operator precedence render this expression useless?
+  print( alwaysTrue || condition1 && condition2 );  // Q: Is this expression useful?
   print( alwaysTrue && condition1 || condition2 );
   print(condition1 != condition1);                  // Q: What is the difference between this and the following expression?
   print(condition2 = !condition2);

exercises/basicTypes/solution/basicTypes.sol.cpp

@@ -45,8 +45,8 @@ int main() {
   bool condition1 = false;
   bool condition2 = true;
   print( alwaysTrue && condition1 && condition2 );
-  print( alwaysTrue || condition1 && condition2 );  // Q: Why does operator precedence render this expression useless?
-  print( alwaysTrue && condition1 || condition2 );  // A: "true || " is evaluated last. The expression therefore is always true.
+  print( alwaysTrue || condition1 && condition2 );  // Q: Is this expressing useful?
+  print( alwaysTrue && condition1 || condition2 );  // A: Not really. Since we use "true ||", it is always true.
   print(condition1 != condition1);                  // Q: What is the difference between this and the following expression?
   print(condition2 = !condition2);                  // A: The first is a comparison, the second a negation with subsequent assignment
   print( alwaysTrue && condition1 && condition2 );