Explain this code in this output?

Question

#include <stdio.h>

int main() {
int a=1;
int b;
b=a;
printf ("%d",--a+b--);
return 0;
}

Answer 1

Nithish

a = 1

b = a it means b = 1

So

--a will be 0 and b-- will be 1

--a+b-- = 0 + 1 = 1

Note -

--a is pre-decrement - here first decrement the value by 1 then assign

b-- is post-decrement - here first assign the value then decrement the value by 1

Answer 2

Thanks AJ

Answer 3

The answer is 1.
Explain about this

Answer 4

In this code a is initialised to 1 and b is equal to 1 as a is assigned to b

Now coming to print statement
--a + b--

In this above expression
--a // pre-decreament therefore first a gets decremented and used in the expression i.e a=0 is used

But in the case of
b-- //post-decreament therefore first value of b is used in the expression and then decrements i.e b=1 is used

Hence result will be
0 + 1 = 1

//Ans= 1