hey all ! i am pragensh and joined today for follwoing reason
can you help me how this output came ?
x=3;
x-=--x-x--;
printf("x=%d",x);
here answer is coming as 1 how come ?
can you please help me to find the reason ? or
step wise explaination
Putting two unary operations like that on the same line has undefined behavior, it's not ever a good idea. However, in your case, it looks like your compiler evaluates left to right. However, parenthesis come first. I substitute for x when I put it's value in parenthesis, i.e. --(3), (2)--, or (1).
x = 3;
x = x - (--x - x--)
x = x - (--(3) - x--)
x = x - (2 - x--)
x = x - (2 - (2)--)
x = x - (2 - 2)
x = (1) - 0
x = 1 - 0
x = 1
Putting two unary operations like that on the same line has undefined behavior, it's not ever a good idea. However, in your case, it looks like your compiler evaluates left to right. However, parenthesis come first. I substitute for x when I put it's value in parenthesis, i.e. --(3), (2)--, or (1).
x = 3;
x = x - (--x - x--)
x = x - (--(3) - x--)
x = x - (2 - x--)
x = x - (2 - (2)--)
x = x - (2 - 2)
x = (1) - 0
x = 1 - 0
x = 1
thank you sir but i am get back to you with some doubt regarding the same topic
sir i think this is how the steps will go,but can u explain me your logic in steps (1) & (2)
x = 3;
x = x - (--x - x--)
x = x - (--(3) - x--)
x = x - (2 - x--)
x = x - (2 - (2)--)
x = x - (2 - 1) /* (sir according to postfix operation) */ .. (1)
x = (2) - 1 (2)
x = 2-1
x = 1
