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# Execution of x++

 P: n/a I was surprised to learn in class today that given a pointer p, the statement: *p++; dereferences first and then increments the pointer. So say p points to an array location array[4]. During execution p would be dereferenced and THEN increment p to point to array[5]. Our professor says that when operators have equal precedence, you read from right to left. So I'm left wondering, what is the rule for determining when exactly a post-increment executes? What if it were the statement: *(p++); Same thing? -Mike- Jul 22 '05 #1
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 P: n/a Debaser wrote: I was surprised to learn in class today that given a pointer p, the statement: *p++; dereferences first and then increments the pointer. So say p points to an array location array[4]. During execution p would be dereferenced and THEN increment p to point to array[5]. Our professor says that when operators have equal precedence, you read from right to left. So I'm left wondering, what is the rule for determining when exactly a Think about the operator as if it were a function: * increment (p) The value returned is one thing, the secondary effect on p is another. -- Salu2 Jul 22 '05 #2

 P: n/a Debaser wrote: I was surprised to learn in class today that given a pointer p, the statement: *p++; dereferences first and then increments the pointer. That's incorrect. There's no such thing as "first" and "then" in this case. Relative precedence of operators '*' and post-'++' means only one thing: '++' applies to 'p' and '*' applies to the result of 'p++', which is the original value of 'p'. The only requirement that must be satisfied is that these operators receive these correct values as parameters. Which one will be applied first in this case is absolutely unpredictable. So say p points to an array location array[4]. During execution p would be dereferenced and THEN increment p to point to array[5]. No. Not necessarily. It can be done the way you describe. Or it can be done in completely opposite order. For example, 'p' could be incremented first, and only then the old value of 'p' (previously stored somewhere) could be dereferenced. Our professor says that when operators have equal precedence, you read from right to left. That's correct. But the important thing your professor didn't seem to tell you is that precedence only defines the grouping of operators and their operands. It does not define the temporal ordering of operations. So I'm left wondering, what is the rule for determining when exactly a post-increment executes? As long as you are dealing with built-in operators, there's no such rule. In C++ language two actions are ordered in time if and only if they are separated by at least one sequence point. Actions that are not separated by sequence points can be executed in arbitrary order (as long as their semantics remains unchanged). Expression '*p++' has no sequence points inside, which means that nothing can be said about what happens "first" and what happens "second". What if it were the statement: *(p++); Same thing? Yes, exactly the same thing. Parentheses affect grouping of operators and their operands. They have no effect on the order of operations. Read about sequence points and remember a simple rule: no sequence point - no ordering. -- Best regards, Andrey Tarasevich Jul 22 '05 #3

 P: n/a Andrey Tarasevich wrote: ... Our professor says that when operators have equal precedence, you read from right to left. That's correct. But the important thing your professor didn't seem to tell you is that precedence only defines the grouping of operators and their operands. It does not define the temporal ordering of operations. ... Oops. That's, of course, incorrect. When you are dealing with _binary_ operators of equal precedence, the order in which you "read" is defined by _associativity_ of those operators. With arithmetical operators, for example, you read from left to right. When you are dealing with _unary_ operators, the simple rule is that postfix operators always have higher precedence than prefix operators. That's how it works in case of '*p++'. -- Best regards, Andrey Tarasevich Jul 22 '05 #4

 P: n/a "Andrey Tarasevich" wrote in message news:10*************@news.supernews.com... Debaser wrote: <> Our professor says that when operators have equal precedence, you read from right to left. That's correct. But the important thing your professor didn't seem to tell you is that precedence only defines the grouping of operators and their operands. It does not define the temporal ordering of operations. Huh? Not all operators of equal precendence are "read" from right to left. That "read" in there makes me wonder what is meant. Certainly not how they are evaluated. Operators are evaluated any way the compiler wishes (with no sequence point). What do you suppose this professor means? -- Gary Jul 22 '05 #5

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