C Programming - Expressions - Discussion

Can any one explain how we take binary form for -2.

int i=-1, j=2, k=0, m;
m = ++i && ++j && ++k;

While evaluating the above statement ++i becomes 0 so right hand side operators to logical and operand j and k doesn't gets executed and m value equals to 0.

Hi,
I am trying to explain this question.

Here value assigned to i=-3,j=2,k=0;
In next step m = ++i && ++j && ++k;

Here,we are using pre-increment operator .so first value incremented by one
And now value of i,j and k are -2,3,1 respectively..

Now expressions is -2 && 3 && 1

And we know that minus value stored in memory in its 2's compliment so stored value for -2 is 2's compliment of its binary
11111111 11111110(in 16 bit compiler) and 3 would stored as 00000000 00000011 and 1 would stored as 00000000 00000001 .

Now expression evaluated in this way
m = 11111111 11111110 && 00000000 00000011 && 00000000 00000001

In this expression 11111111 11111110 && 00000000 00000011 will evaluated non zero value. so it becomes TRUE and return 1 .

Now expression is m = 1 && 00000000 00000001
1 && 0000000 00000001 it also evaluated non zero value.so it becomes TRUE and return 1 so this i initialized to m

Hence, output is -2,3,1,1

#include<stdio.h>
int main()
{
int i=-1, j=2, k=0, m;
m = ++i && ++j && ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
If we changed the value of i= -1 then output we got
0, 2, 0, 0 (GCC compiler).

Anybody can suggest us how this compiler give the output and
How precedence operator work.

m=-2 && 3 && 1;

I am not getting this statement. Please explain me.

Please make me understand.

Not able to understand that. Kindly explain again.

@above:

In case of step 2:

m= -2 && 3 && 1;

All those are not equal to zero.so those values(-2,3,1) are true. In AND operation if all trues then result would be true.

In 'C', TRUE value can be represented with value '1' and false as '0'.

so it returns '1' value and assigned to variable 'm'.