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, never overload them - it breaks a hugely important assumption about the control flow of the program.
Relevant standard quotations
Builtin &&
and ||
have guaranteed short-circuit behavior
§5.14 ?1
Unlike &
, &&
guarantees left-to-right evaluation: the second operand is not evaluated if the first operand is false
.
§5.15 ?1
Unlike |
, ||
guarantees left-to-right evaluation; moreover, the second operand is not evaluated if the first operand evaluates to true
.
If overloaded, they behave as "regular" binary operators (no short-circuit or guaranteed ordering of evaluation)
§13.5 ?9
Operators not mentioned explicitly in subclauses 13.5.3 through 13.5.7 act as ordinary unary and binary operators obeying the rules of 13.5.1 or 13.5.2.
and &&
and ||
are not mentioned explicitly in these subclauses, so regular §13.5.2 holds:
§13.5.2 ?1
A binary operator shall be implemented either by a non-static member function (9.3) with one parameter or by a non-member function with two parameters. Thus, for any binary operator @
, x@y
can be interpreted
as either x.operator@(y)
or operator@(x,y)
.
with no special provision for evaluating only one side or in a particular order.
(all quotations from the C++11 standard)
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