operator evaluation question

[linq936]

Hi,
I see the following description in <<Expert C Programming>>,

the order is defined for && and || and a couple of other operators.
These 2 evaluate their operands in a strict left-to-right oder,
stopping when the result is known.

Is this really true? I think the order is compiler dependent.

Thanks.

 

[Richard Heathfield]

(e-mail address removed) said:

Hi,
I see the following description in <<Expert C Programming>>,

the order is defined for && and || and a couple of other operators.
These 2 evaluate their operands in a strict left-to-right oder,
stopping when the result is known.

Is this really true? I think the order is compiler dependent.

The Standard mandates the order and the short-cut evaluation.
Implementations have no leeway in this regard.

 

[Chris Dollin]

Hi,
I see the following description in <<Expert C Programming>>,

the order is defined for && and || and a couple of other operators.
These 2 evaluate their operands in a strict left-to-right oder,
stopping when the result is known.

Is this really true? I think the order is compiler dependent.

Not for && and || and ?: and the comma-operator it isn't.

 

[Joe Wright]

Hi,
I see the following description in <<Expert C Programming>>,

the order is defined for && and || and a couple of other operators.
These 2 evaluate their operands in a strict left-to-right oder,
stopping when the result is known.

Is this really true? I think the order is compiler dependent.

Thanks.

Perhaps confusing order and precedence. The language specifies the
precedence of operators; c = a + b for example. + has higher precedence
than = and so (a+b) will be evaluated before the assignment to c.

Order of evaluation is a different issue. Given x = f() + g() makes no
guarantee that f() is evaluated before g(). Compiler decides.