C Programming - Expressions
- Expressions - General Questions
- Expressions - Find Output of Program
- Expressions - True / False Questions
- Expressions - Yes / No Questions
#include<stdio.h>
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;
}
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.
Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).
Hence the output is "-2, 3, 1, 1".
#include<stdio.h>
int main()
{
printf("%x\n", -2<<2);
return 0;
}
Negative numbers are represented in 2's complement method.
1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).
2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).
Therefore, in binary we represent -2 as: 11111111 11111110.
After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.
#include<stdio.h>
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;
}
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.
Step 2: m = ++i || ++j && ++k; here (++j && ++k;) this code will not get executed because ++i has non-zero value.
becomes m = -2 || ++j && ++k;
becomes m = TRUE || ++j && ++k; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of variable 'i' only increemented by '1'(one). The variable j,k are not increemented.
Hence the output is "-2, 2, 0, 1".
#include<stdio.h>
int main()
{
int x=12, y=7, z;
z = x!=4 || y == 2;
printf("z=%d\n", z);
return 0;
}
Step 1: int x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.
Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.
Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".
#include<stdio.h>
int main()
{
static int a[20];
int i = 0;
a[i] = i ;
printf("%d, %d, %d\n", a[0], a[1], i);
return 0;
}
Step 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will be automatically initialized to value '0'(zero).
Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Step 4: Hence the output is "0, 0, 0".