Question

If you take Java’s primitive numeric types, plus boolean, and compare it to C++ equivalent types, is there any difference what concerns the operators, like precedence rules or what the bit-manipulation operators do? Or the effect of parenthesis?

Asked another way, if I took a Java expression and tried to compile and run it in C++, would it always compile and always give the same result?

Answer 1

• For an expression like:

  a = foo() + bar();
  

  In Java, the evaluation order is well-defined (left to right). C++ does not specify whether foo() or bar() is evaluated first.

• Stuff like:

  i = i++;
  

  is undefined in C++, but again well-defined in Java.

• In C++, performing right-shifts on negative numbers is implementation-defined/undefined; whereas in Java it is well-defined.

• Also, in C++, the operators &, | and ^ are purely bitwise operators. In Java, they can be bitwise or logical operators, depending on the context.