Question
#include<stdio.h>
int main()
{
int x, y, z;
x=y=z=1;
z = ++x || ++y && ++z;
printf("x=%d, y=%d, z=%d\n", x, y, z);
return 0;
}
What will be the output of the program?
#include<stdio.h>
int main()
{
int x, y, z;
x=y=z=1;
z = ++x || ++y && ++z;
printf("x=%d, y=%d, z=%d\n", x, y, z);
return 0;
}
Answer: Option A
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Step 1: x=y=z=1; here the variables x ,y, z are initialized to value '1'.
Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x
becomes 2. So there is no need to check the other side because ||(Logical OR) condition is
satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns
'1'. So the value of variable z is '1'
Step 3: printf("x=%d, y=%d, z=%d`setminus`n", x, y, z); It prints "x=2, y=1, z=1". here x is increemented
in previous step. y and z are not increemented.
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