What will be the output of the program?

#include<stdio.h>
int main()
{
    int a=100, b=200, c;
    c = (a == 100 || b > 200);
    printf("c=%d\n", c);
    return 0;
}

Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'.
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".

Step 1: int x=4, y, z; here variable x, y, z are declared as an integer type and variable x is initialized to 4.
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".

The integer value 2 is represented as 00000000 00000010 in binary system.

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.

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".

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".

The order of evaluation of arguments passed to a function call is unspecified.

Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.

In TurboC, the output will be 4, 3.

In GCC, the output will be 4, 4.

Since C is a machine dependent language sizeof(int) may return different values.

The output for the above program will be cd in Windows (Turbo C) and gh in Linux (GCC).

To understand it better, compile and execute the above program in Windows (with Turbo C compiler) and in Linux (GCC compiler).

Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable i, j are initialized to 3, 4 respectively.

The function addmult(i, j); accept 2 integer parameters.

Step 2: k = addmult(i, j); becomes k = addmult(3, 4)

In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;

kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.

ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.

return (kk, ll); It returns the value of variable ll only.

The value 12 is stored in variable 'k'.

Step 3: l = addmult(i, j); becomes l = addmult(3, 4)

kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.

ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.

return (kk, ll); It returns the value of variable ll only.

The value 12 is stored in variable 'l'.

Step 4: printf("%d, %d\n", k, l); It prints the value of k and l

Hence the output is "12, 12".

Step 1: int i; The variable i is declared as an global and integer type.

Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.

Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.

Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.

Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.

Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.

Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.

Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.

Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.

Hence the output is "4 3 4 3".