Captain Obvious

StM0n

ever asked him, what that should do... and what i++ means... sry, i'm curious about it :)

(yes|no|maybe)*

leppie

st0le wrote:

The fact that it makes sense

Makes sense in the fact that it compiles and does not break anything? ;P

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Robert Surtees

Perhaps I'm being Captain Obvious but, in plain C at least, the result is undefined. The compiler I'm using at the moment does not leave the loop as i is never incremented.

Doc Lobster

I wrote a test application (the code works in VC++ 8, and i stays 0 in VC#), but it seems that I had a mental blockade to enter i = i++; It took me two i = i+1; until I could force my fingers to do that. Is the result really undefined by ANSI or whoevers specification?

PIEBALDconsult

Doc Lobster wrote:

Is the result really undefined by ANSI or whoevers specification?

I think i = i++ might be OK, but apparently i = ++1 + 1 is not. I'm looking in the ISO/IEC 9899:TC2 document, section 6.5 on page 67. See footnote 71 " 6.5 Expressions 1 An expression is a sequence of operators and operands that specifies computation of a value, or that designates an object or a function, or that generates side effects, or that performs a combination thereof. 2 Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.71) 3 The grouping of operators and operands is indicated by the syntax.72) Except as specified later (for the function-call (), &&, ||, ?:, and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified. 4 Some operators (the unary operator ~, and the binary operators <<, >>, &, ^, and |, collectively described as bitwise operators) are required to have operands that have integer type. These operators yield values that depend on the internal representations of integers, and have implementation-defined and undefined aspects for signed types. 5 If an exceptional condition occurs during the evaluation of an expression (that is, if the result is not mathematically defined or not in the range of representable values for its type), the behavior is undefined. 6 The effective type of an object for an access to its stored value is the declared type of the object, if any.73) If a value is stored into an object having no declared type through an lvalue having a type that is not a character type, then the type of the lvalue becomes the effective type of the object for that access and for subsequent accesses that do not modify 71) This paragraph renders undefined statement expressions such as i = ++i + 1; a[i++] = i; while allowing i = i + 1; a[i] = i; 72) The syntax specifies the precedence of operators in the evaluation of an expression, which is the same as the order of the major subclauses of this subclause, highest precedence first. Thus, for example, the expressions allowed as the operands of the binary + operator (6.5.6) are those expressions defined in 6.5.1 through 6.5.6. The exceptions are cast expressions

PIEBALDconsult

It works with Borland C/C++ 5.5

CPallini

With gcc (version 3.4.4) it doesn't work, i remaining 0 forever. :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke

Lost User

i wonder what would happen in other languages?? *writes java application* we've got a Infinite loop here...i=0, all the way :omg: :confused:

-st0le [st0le'n'stuff softwarez!] http://st0lenc0des.googlepages.com/

modified on Tuesday, February 19, 2008 12:24 AM

Brady Kelly

C# is the same, so, it seems this behaviour is predictable and I will have to read PIEBALD's lengthy excerpt.

CPallini · modified on Tuesday, February 19, 2008 12:24 AM

Java behaves the same way, i.e. loops indefinitely (or at least, until the hammer comes down). :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke

liquidplasmaflow

What's happening makes sense. The operator in question is the postfix incrementor; it increments the variable, but it returns the pre-incrementation value. So, it increments itself, but it's being assigned its original value.

Jitse

Isn't the question just whether the runtime environment first assigns the old value. I really thought this would just eventually increment the variable. Because it should normally first assign the old value to i, and then increment i. Why should it ever store the old value somewhere first, then increment i, then put that old value back in i? I know I'm wrong, cuz as said before C# gives an infinite loop, but still it's weird.

PIEBALDconsult

The code I wrote to test it is:

# include <stdio.h>

int
main
(
int argc
,
char* argv[]
)
{
int result = 0 ;

result = result++ ;

printf ( "%d" , result ) ;

return ( result ) ;

}

(This is the same code I used to test Borland C/C++ 5.5) compiling this using DEC C V6.0-001 on OpenVMS Alpha V7.3-2 yields the warning " In this statement, the expression "result=result++" modifies the variable "result" more than once without an intervening sequence point. This behavior is undefined. " but it compiles and returns 1 when executed.

liquidplasmaflow

How could it do that? The operator's function has to end (return) before the assignment occurs; that means the incrementation has to happen first.

PIEBALDconsult

But not so with gcc version 3.2 (mingw special 20020817-1)

Luc Pattyn

leppie wrote:

it compiles and does not break anything

Two big qualities indeed. :)

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Luc Pattyn

liquidplasmaflow wrote:

the incrementation has to happen first

says who? the autoincrement is not necessary in the expression evaluation, and hence it can be scheduled before or after the assignment operator, that is why the net result is undefined. :)

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Lost User

I agree. If the code would look like this: a = i++; Then a would be assigned the value of i and i would be incremented after that, so I would expect in given case that in first iteration i would be assigned value of 0 then i would be incremented. Weird indeed.

liquidplasmaflow

C++'s operator ++ is a function. Last I checked, a function can't do anything after it returns :) int& operator++(int& argument, int) { int return_value = argument; argument += 1; return return_value; } Am I right?

Luc Pattyn

Hi, the autoincrement operator may be implemented as a function, I don't think it has to; in C it typically is not. And even when it is a function, it could be inlined automatically, and the instructions then can be rescheduled by the compiler, so there is no way you can predict which one (the final store, or the autoincrement) will occur last and hence prevail; all that in accordance with the legalese language specs PIEBALD showed us. :)

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