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How many will be the value of the expression (++x * ++x * ++x) for x = 5?

i run the following programme under vc, gcc and tc:

#include <stdio.h>
#define CUBIC(x) ((x) * (x) * (x))

int main()
{
  int x = 5;
  int y = CUBIC(++x);
  printf("y=%d\n",y);
}

and get the results separately as follow:
y=392                    under vc and gcc
y=512                    under tc

why?
if y = CUBIC(x++), i get 125 under all the above compilers. why? i had thought the result will be depend on compiler and will be different.
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bourbaki
Asked:
bourbaki
1 Solution
 
tinchosCommented:
I would have thought that the result would be 512, no idea of why is 392

I'll explain why I thought of 512

As you have a define, then the preprocessor would replace it before compiling it, so your code would be like this. (After preprocessing)

int main()
{
 int x = 5;
 int y = ( (++x) * (++x) * (++x) );             // Important Line
 printf("y=%d\n",y);
}

When it gets to "Important Line", it does the following

the first ++x                      so x = 6
then the second ++x         so x = 7
then the third ++x             so x = 8

then it evaluates y = 8 * 8 * 8;

then y = 512;

The reason for Cubic( x++ ) = 125 is that in that case

first y is assigned and afterwards x is incremented, so the y value is not influenced by the autoincrement operators

Hope this helps

Tincho
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dennis_georgeCommented:
hi bourbaki ,

   First i must tell you there is some difference in the expression evaluation in various compilers....

TC handles the expression differently and VC handles it differently.....

In TC environment if  one variable is occured many times in a expression then the final value of that vairable is substituted in all the places of its occurence......

The #define will replace  CUBIC(++x) to

y = (++x) * (++x) * (++x) ;

So in TC first the value(final valule) of x is  calcuated and then it is substituted in all its occurances....
so the final value of x after this expression will be 8

so y = 8 * 8 * 8 ;  = 512....


Now in VC the value  of x(or any variable) is done at the time of postfix evaluation....

NOTE:: In Computer expressions are normally evaluated by postfix evalutaion technique...


The postfix expression for (++x) * (++x) * (++x)

is   (++x)(++x)*(++x)*

so first (++x)(++x)* is calculated so the value of x is calculated here... here x is incremented twice so x = 7

(7)(7)* = 49....

Now the expression becomes (49)(++x) * and the value of x = 7;

Now again calculate (49)(++x)*   after increment x becomes 8
so (49)(8)* = 392...


So i hope u got some idea...

Dennis
 
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efnCommented:
The question was why the answers were different in one case and the same in the other case.  The answer is simply that that's how the compilers were implemented.

With code like this, the language definition does not specify the order of the operations, so different compilers can do different things or the same thing without failing to conform to the standard.

So as a practical matter, it is better to avoid code like this than try to understand what it will do, because what it will do can legitimately vary from one compiler to another.

See also:

http://www.experts-exchange.com/Programming/Programming_Languages/C/Q_20638918.html

--efn
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bourbakiAuthor Commented:
hi Dennis,
   Thank you.
    Can the result be 6*7 when (++x)(++x)* is calculated? If x is increased and used its value,6,and then increased again and used its value, 7, and at last ...6*7.
    Can you tell me whether the analysis above is ture?

bourbaki
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dennis_georgeCommented:
Hi,

  No i don't think so...... when the compiler evaluate the value of a postfix expression, like (++x)(++x)*, it first calculates its value. Actually the postfix evaluation takes place in stack. both (++x) will be in stack. so before calculating the result it pops out the variables and calculates its value...so after two increments the value of x will be 7. so the compiler will calculate 7 * 7 not 6 * 7..... because during calculation of result it does not substitute the value after getting each variable but only after getting both the operands....


Dennis
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