What will be the output of the program?

#include<stdio.h>
int fun(int, int);
typedef int (*pf) (int, int);
int proc(pf, int, int);

int main()
{
    printf("%d\n", proc(fun, 6, 6));
    return 0;
}
int fun(int a, int b)
{
   return (a==b);
}
int proc(pf p, int a, int b)
{
   return ((*p)(a, b));
}

When an automatic structure is partially initialized remaining elements are initialized to 0(zero).

True, we can find the size of short integer and long integer using the sizeof() operator.
Example:

#include<stdio.h>
int main()
{
    short int i = 10;
    long int j = 10;
    printf("short int is %d bytes.,\nlong int is %d bytes.",
            sizeof(i),sizeof(j));
    return 0;
}

Output:
short int is 2 bytes.
long int is 4 bytes.

The macro function CUBE(x) (x*x*x) calculates the cubic value of given number(Eg: 10³.)

Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.

Step 2: a = CUBE(b++); becomes

=> a = b++ * b++ * b++;

=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.

=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)

Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.

Hence the output of the program is 27, 6.

Step 1: int k=35; The variable k is declared as an integer type and initialized to 35.

Step 2: k = func1(k=func1(k=func1(k))); The func1(k) increement the value of k by 1 and return it. Here the func1(k) is called 3 times. Hence it increements value of k = 35 to 38. The result is stored in the variable k = 38.

Step 3: printf("k=%d\n", k); It prints the value of variable k "38".

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions.

Step 2: printf("%u, %u\n", a+1, &a+1);

The base address(also the address of the first element) of array is 65472.

For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480

Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496

Hence the output of the program is 65480, 65496

In TurboC, the output will be 2, 1 because the size of the pointer is 2 bytes in 16-bit platform.

But in Linux, the output will be 4, 1 because the size of the pointer is 4 bytes.

This difference is due to the platform dependency of C compiler.

Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively.

Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. )

Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters.

Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i.

Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j.

Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function.

Step 7: printf("%d, %d", i, j); It prints the value of variable i and j.

Hence the output is 25, 4.

Step 1: int no=5; The variable no is declared as integer type and initialized to 5.

Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter.

The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);.

The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely.

Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.

printf("%e,", d); Here '%e' specifies the "Scientific Notation" format. So, it prints the 2.25 as 2.250000e⁺⁰⁰⁰.

printf("%f,", d); Here '%f' specifies the "Decimal Floating Point" format. So, it prints the 2.25 as 2.250000.

printf("%g,", d); Here '%g' "Use the shorter of %e or %f". So, it prints the 2.25 as 2.25.

printf("%lf,", d); Here '%lf' specifies the "Long Double" format. So, it prints the 2.25 as 2.250000.

sizeof(x) returns the size of x in bytes.
float *p is a pointer to a float.

In 16 bit compiler, the pointer size is always 2 bytes.
In 32 bit compiler, the pointer size is always 4 bytes.