Yes, it’s guaranteed, otherwise such operators would lose much of their usefulness. Important notice: this is valid only for the builtin && and ||; if some criminal overloads them, they are treated as “regular” overloaded binary operators, so in this case both operands are always evaluated, and in unspecified order as usual. For this reason, … Read more
Yes, the MSDN article is in error, at least with respect to standard C and C++1. Having said that, let me start with a note about terminology: in the C++ standard, they (mostly–there are a few slip-ups) use “evaluation” to refer to evaluating an operand, and “value computation” to refer to carrying out an operation. … Read more
It depends on the argument type, the called function’s calling convention, the archtecture and the compiler. On an x86, the Pascal calling convention evaluates arguments left to right whereas in the C calling convention (__cdecl) it is right to left. Most programs which run on multiple platforms do take into account the calling conventions to … Read more
It’s NOT, AND, OR, from highest to lowest according to the documentation on Operator precedence Here is the complete precedence table, lowest precedence to highest. A row has the same precedence and groups from left to right 0. := 1. lambda 2. if – else 3. or 4. and 5. not x 6. in, not … Read more
The comma operator has the lowest precedence of all C/C++ operators. Therefore it’s always the last one to bind to an expression, meaning this: a = b, c; is equivalent to: (a = b), c; Another interesting fact is that the comma operator introduces a sequence point. This means that the expression: a+b, c(), d … Read more
The operator precedence in the C/C++ language in not defined by a table or numbers, but by a grammar. Here is the grammar for conditional operator from C++0x draft chapter 5.16 Conditional operator [expr.cond]: conditional-expression: logical-or-expression logical-or-expression ? expression : assignment-expression The precedence table like this one is therefore correct when you use assignment on … Read more
In terms of evaluation sequence, it is easier to think of x*f(x) as if it was: operator*(x, f(x)); so that there are no mathematical preconceptions on how multiplication is supposed to work. As @dan04 helpfully pointed out, the standard says: Section 1.9.15: “Except where noted, evaluations of operands of individual operators and of subexpressions of … Read more
Use JSR-303 validation groups. If no groups are specified a constraint is part of the Default Bean Validation group (see: javax.validation.groups.Default). Create an interface to be your “Extended” (or whatever you want to call it) group: public interface Extended{} Now create an interface that will have the javax.validation.GroupSequence annotation. @GroupSequence({Default.class, Extended.class}) public interface MySequence {} … Read more
The syntax is MemberExpression : PrimaryExpression FunctionExpression MemberExpression [ Expression ] MemberExpression . IdentifierName new MemberExpression Arguments new foo().bar cannot be parsed as new (foo().bar) because foo().bar is not a MemberExpression. Moreover, new foo() cannot be parsed as new (foo()), for the same reason. Conversely, new foo.bar is parsed as new (foo.bar) because foo.bar is … Read more
x << y is syntactic sugar for a function call to operator<<(x, y). You will remember that the c++ standard places no restriction on the order of evaluation of the arguments of a function call. So the compiler is free to emit code that computes x first or y first. From the standard: §5 note … Read more