Renju Mathew

 Profile Answers by Renju Mathew 

• Jun 29th, 2007

 

Hi Pankaj

The precedence order of the operators will be x-- , --x , - , =- respectively
As the first two operations are executed the expression
will be as follows;
x -= 7 - 7....(the x-- operation will be evaluated on the next line only)

Now the subtraction is done and the statement will be , x -= 0
Now -= operation will execute. ie x = x - 0 ie x = 7.

On the next line this value will be decremented by one because of x-- statement

Thus the value 6 will be printed..(7-1)

sirilux

 Profile Answers by sirilux 

• Jan 11th, 2010

 

Some compilers may take values from left to right also..in that case
x=8
x-=--x-x--;
// --x is 7 and x-- is 7(latest values of x)
so 7-7 is 0
x-=0;
x=x-0(latest value of x is 6
so x is 6.

bjkrishna

 Profile Answers by bjkrishna 

• Jul 5th, 2010

 

x-=--x-x--;

This is a special case.

Here --x and x-- are evaluated first and then final value of x is substituted in the expression regardless of precedence and associativity.

After --x, x becomes 7.

After x--, x becomes 6.

Now the final value in x is 6. This is substituted in the expression.

So 

x-=6-6

x-=0

x=x-0

which gives x=6.

krischan

 Profile Answers by krischan 

• Nov 23rd, 2010

 

--x is pre decrement so x becomes 7.
x-- is post decrement no need to worry about this until the expression is evaluated.
first x will have 7. second x will have 7. dont forget about post dec.
7-7 is 0
now the expression is x-=0
we can write it as x=7-0=>7
now the expression is evaluated totally. so post decrement will be done.
x--;x=7-1;
6