C Programming - Expressions - Discussion

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

-2 && 3 && 1
It logical operators it checks from left to right.
Here logical operator is AND so first check -2 it is non zero value so it proceed for 3 same non zero value, than proceed for 1 it is non zero again so all are true so answer is 1.

For eg if it is m=-2 && 3 && 0.
For last 0 it is treated as false so (3&&0) give false ie.0 and again (-2&&0) gives zero so m=0.

Eg:
i=-1, j=3, k=-1;
m=++i && ++j && k++;
Then it evaluate i=0 then it doesn't proceed for left expression for j or k.
Output is m=0 i=0 j=3 k=-1.

But,
if i=-2, j=3, k=-1;
m=++i && ++j && k++;
Then output is m=1 i=-1 j=4 k=0.
Here k is post increment so in k++ it have value -1 but at next line it gets value 0.

And if i=-2, j=3, k=-1;
m=++i && ++j && ++k;
Then output is m=0 i=-1 j=4 k=0.
Here k is pre increment so m is 0 because k is false.

@Lokesh Jain.

if i=-2, j=3, k=-1;
m=++i && ++j && k++;
Then output is m=1 i=-1 j=4 k=0.
Here k is post increment so in k++ it have value -1 but at next line, it gets value 0.

And if i=-2, j=3, k=-1;
m=++i && ++j && ++k;
Then output is m=0 i=-1 j=4 k=0.
Here k is pre-increment so, m is 0 because k is false.
The compiler is giving m=1 how is it?

Hi;

AND IS LOGICAL OPERATOR.

It's operation having true value when both the values are true. i.e both value are non zero,
In our example both values are non zero.

m=-2&&3&&1.

Here -2 =non zero number and 3=non zero mean not equal to zero and 1 is also non zero therefore first and operation gives true and finally, we get true i.e m=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.

@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'.

And operator is like compulsion means if all the conditions are true it will return 1 (true) and if any one condition is false it will return 0 (false). And basically declaration is int (integer) so it ranges from - to +.

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.

int main()
{
int i=-2, j=3, k=-1,m;
m=++i && ++j && k++;
printf("%d, %d, %d, %d\n", i, j, k, m);

What should be its output? Please explain.

int main()
{
int i=-3, j=2, k=0, m;
m = ++i ||++j && ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
What will be the output?