operators... precedence and associativity

[arunkumar]

hi!
i am trying to run a use multiple operators in a single line... but i am finding it difficult to catch the flow of execution.

int x=3;
x-=--x-x--;

output is 1.

plz help me with "the compiler execution steps" of the instruction mentioned above...

thanks in advance...
Regards,
Arunkumar

[mspy2plus]

hello.
this is what your code does, in order:
(x)(3) means the value of the expression x is 3
assigns 3 to x
(x)( 3 ) = ((x)(3) - ( (x-1)(2)-( x )(2) )(0))(3)
x = x - 1
x = x - 1
x == 1

so, to simplify,
x=3;
x-=0;
x--;
x--;
also, this is good as an example, but try not to use this in the future.

[helios]

x-=--x-x--

The result of this expression is undefined.

On one compiler, it could be the same as

x=x-(x-1-x);
x=x-2;

While on a different compiler, it could be

temp0=x;
x=x-1;
x=temp0-(x-x);
x=x-1;

The order of evaluation is undefined, so it's never a good idea to reference a variable that is being decremented or incremented, more that once in the same expression.