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Subject: Re: Bizar Question for programmers: very strange behaviour of my engine

Author: Robert Hyatt

Date: 14:59:27 01/18/06

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On January 17, 2006 at 18:08:59, Dann Corbit wrote:

>On January 17, 2006 at 18:01:16, Dann Corbit wrote:
>
>>On January 17, 2006 at 17:50:33, Dan Honeycutt wrote:
>>
>>>On January 17, 2006 at 14:59:15, Robert Hyatt wrote:
>>>
>>>...
>>>>
>>>>if (a && b) where if a is 0, b would not even be fetched usually...
>>>>
>>>
>>>I thought, in this case, b _must_ not be fetched.
>>>
>>>Best
>>>Dan H.
>>
>>3.5:    But what about the && and || operators?
>>        I see code like "while((c = getchar()) != EOF && c != '\n')" ...
>>
>>A:      There is a special "short-circuiting" exception for these
>>        operators: the right-hand side is not evaluated if the left-hand
>>        side determines the outcome (i.e. is true for || or false for
>>        &&).  Therefore, left-to-right evaluation is guaranteed, as it
>>        also is for the comma operator.  Furthermore, all of these
>>        operators (along with ?:) introduce an extra internal sequence
>>        point (see question 3.8).
>>
>>        References: K&R1 Sec. 2.6 p. 38, Secs. A7.11-12 pp. 190-1; K&R2
>>        Sec. 2.6 p. 41, Secs. A7.14-15 pp. 207-8; ISO Sec. 6.3.13,
>>        Sec. 6.3.14, Sec. 6.3.15; H&S Sec. 7.7 pp. 217-8, Sec. 7.8 pp.
>>        218-20, Sec. 7.12.1 p. 229; CT&P Sec. 3.7 pp. 46-7.
>
>A place where the distinction could be important for the case at hand:
>       !(bfZwart[kol+1]&bfBoven[rij])
>could be if rij is an invalid address (negative or bigger than the dimention).
>If the value of kol was always zero when rij was an invalid offset, then adding
>1 would cause the exception to stop happening.
>
>Now, the bitwise operators do NOT guarantee left to right evaluation.  Hence the
>"usually" I think.  But such an ordering is possible, just not required for
>bitwise operators.  Note the start of Semantics item 4 below...
>

I was thinking of the difference between "fetching" and "evaluating".  If you do
something like this:

if ((a < b) && (i++ < 0))

I is supposed to be unchanged unless a is less than b.  But that's a lousy way
to program, _and_ I'm not aware of a thing that says the compiler can't
pre-fetch i if it so chooses, and even increment it as well, just so it doesn't
store it back until it is sure that a is less than b...

I would write code with that in mind even if the ANSI C standard did dictate
when things can be fetched, because there are lots of other programming
languages that would blow up there as well..



>
>©ISO/IEC ISO/IEC 9899:1999 (E)
>6.5.13 Logical AND operator
>Syntax
>1 logical-AND-expression:
>inclusive-OR-expression
>logical-AND-expression && inclusive-OR-expression
>Constraints
>2 Each of the operands shall have scalar type.
>Semantics
>3 The && operator shall yield 1 if both of its operands compare unequal to 0;
>otherwise, it yields 0. The result has type int.
>4 Unlike the bitwise binary & operator, the && operator guarantees left-to-right
>evaluation; there is a sequence point after the evaluation of the first operand.
>If the first operand compares equal to 0, the second operand is not evaluated.
>6.5.14 Logical OR operator
>Syntax
>1 logical-OR-expression:
>logical-AND-expression
>logical-OR-expression || logical-AND-expression
>Constraints
>2 Each of the operands shall have scalar type.
>Semantics
>3 The || operator shall yield 1 if either of its operands compare unequal to 0;
>otherwise, it yields 0. The result has type int.
>4 Unlike the bitwise | operator, the || operator guarantees left-to-right
>evaluation; there is a sequence point after the evaluation of the first operand.
>If the first operand compares unequal to 0, the second operand is not evaluated.
>§6.5.14 Language 89



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