Varying behavior for possible loss of precision

That’s because b += 1.0; is equivalent to b = (int) ((b) + (1.0));. The narrowing primitive conversion (JLS 5.1.3) is hidden in the compound assignment operation. JLS 15.26.2 Compound Assignment Operators (JLS Third Edition): A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T … Read more

Lua operators, why isn’t +=, -= and so on defined?

This is just guesswork on my part, but: 1. It’s hard to implement this in a single-pass compiler Lua’s bytecode compiler is implemented as a single-pass recursive descent parser that immediately generates code. It does not parse to a separate AST structure and then in a second pass convert that to bytecode. This forces some … Read more

Do the &= and |= operators for bool short-circuit?

From C++11 5.17 Assignment and compound assignment operators: The behavior of an expression of the form E1 op = E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once. However, you’re mixing up logical AND which does short-circuit, and the bitwise AND which never does. The text snippet &&=, … Read more

Why doesn’t Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=)

Reason The operators &&= and ||= are not available on Java because for most of the developers these operators are: error-prone useless Example for &&= If Java allowed &&= operator, then that code: bool isOk = true; //becomes false when at least a function returns false isOK &&= f1(); isOK &&= f2(); //we may expect … Read more

Shortcut “or-assignment” (|=) operator in Java

The |= is a compound assignment operator (JLS 15.26.2) for the boolean logical operator | (JLS 15.22.2); not to be confused with the conditional-or || (JLS 15.24). There are also &= and ^= corresponding to the compound assignment version of the boolean logical & and ^ respectively. In other words, for boolean b1, b2, these … Read more

Operator precedence with JavaScript’s ternary operator

Use: h.className = h.className + (h.className ? ‘ error’ : ‘error’) You want the operator to work for h.className. Better be specific about it. Of course, no harm should come from h.className += ‘ error’, but that’s another matter. Also, note that + has precedence over the ternary operator: JavaScript Operator Precedence

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