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 vignesh1988i 2Oct2009 19:06

i want a complete explanation for this... please

pl.. explain a complete stack explanation for the below expression and say me the final value for j...

int i=16,j;
j=i-- + i-- + --i + i--;
printf("%d",j);
:(

Re: i want a complete explanation for this... please

This one has undefined behaviour, since it modifies the lvalue i without intervening sequence points.

See : http://en.wikipedia.org/wiki/Sequence_point.

 vignesh1988i 2Oct2009 20:12

Re: i want a complete explanation for this... please

but this is giving the value of 'i' is 60... but for me when am working out 'i' is only 57 for me...

however this operation should have surely made use of STACK data structure......:smug:

so only am asking the correct explanation with stack implementation...... plzzzz:)

 vignesh1988i 2Oct2009 20:33

Re: i want a complete explanation for this... please

sorry , 'j' is giving me 60.... pl. explain with stack

 xpi0t0s 2Oct2009 22:37

Re: i want a complete explanation for this... please

Any explanation would be completely compiler specific. This expression gives undefined behaviour - that is the only possible answer. The explanation for that is what Saswat has already said: it modifies an lvalue more than once without an intervening sequence point. So to turn this into sensible code that has predictable results you must add sequence points, for example:
Code:

```int i=16,j; j=i--; j+=i--; j+= --i; j+=i--; printf("%d",j);```
which gives the output 57, and i=12 at the end.

Don't know why you want a stack-based explanation; the expression won't necessarily use a stack. In Visual Studio 2008 the result of:
Code:

```int i=16,j; j=i-- + i-- + --i + i--; printf("%d",j);```
is 60, because it separates all side effects out from the equation, doing pre-decrements first and post-decrements after, and is therefore equivalent to:
Code:

```        int i=16,j;         --i;         j=i+i+i+i;         i--;         i--;```
(note this behaviour is Visual Studio 2008 specific and may not be reflected in other compilers)
and the assembly code generated by the expression is:
Code:

```        j=i-- + i-- + --i + i--; 004117C5  mov        eax,dword ptr [i] 004117C8  sub        eax,1 004117CB  mov        dword ptr [i],eax 004117CE  mov        ecx,dword ptr [i] 004117D1  add        ecx,dword ptr [i] 004117D4  add        ecx,dword ptr [i] 004117D7  add        ecx,dword ptr [i] 004117DA  mov        dword ptr [j],ecx 004117DD  mov        edx,dword ptr [i] 004117E0  sub        edx,1 004117E3  mov        dword ptr [i],edx 004117E6  mov        eax,dword ptr [i] 004117E9  sub        eax,1 004117EC  mov        dword ptr [i],eax 004117EF  mov        ecx,dword ptr [i] 004117F2  sub        ecx,1 004117F5  mov        dword ptr [i],ecx```
which as you can see involves no stack operations.

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