Exercise "Example is better than precept."
- (Proverb)
6.
What will be the output of the program?
#include<stdio.h>
int main()
{
const char *s = "";
char str[] = "Hello";
s = str;
while(*s)
printf("%c", *s++);
return 0;
}
Answer: C
Explanation:
Step 1 : const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.
Step 2 : char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".
Step 3 : s = str; The value of the variable str is assigned to the variable s . Therefore str contains the text "Hello".
Step 4 : while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s .
Hence the output of the program is "Hello".
7.
What will be the output of the program?
#include<stdio.h>
int get();
int main()
{
const int x = get();
printf("%d", x);
return 0;
}
int get()
{
return 20;
}
Answer: D
Explanation:
Step 1 : int get(); This is the function prototype for the funtion get() , it tells the compiler returns an integer value and accept no parameters.
Step 2 : const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".
The function get() returns the value "20".
Step 3 : printf("%d", x); It prints the value of the variable x .
Hence the output of the program is "20".
8.
What will be the output of the program (in Turbo C)?
#include<stdio.h>
int fun(int *f)
{
*f = 10;
return 0;
}
int main()
{
const int arr[5] = {1, 2, 3, 4, 5};
printf("Before modification arr[3] = %d", arr[3]);
fun(&arr[3]);
printf("\nAfter modification arr[3] = %d", arr[3]);
return 0;
}
A. Before modification arr[3] = 4 B. Error: cannot convert parameter 1 from const int * to int * C. Error: Invalid parameter D. Before modification arr[3] = 4
Answer: D
Explanation:
Step 1 : const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized to
arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5
Step 2 : printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4).
Step 3 : fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10.
A const variable can be indirectly modified by a pointer.
Step 4 : printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10).
Hence the output of the program is
Before modification arr[3] = 4
After modification arr[3] = 10
9.
What will be the output of the program?
#include<stdio.h>
int main()
{
const int i=0;
printf("%d\n", i++);
return 0;
}
A. 10 B. 11 C. No output D. Error: ++needs a value
Answer: E
Explanation:
This program will show an error "Cannot modify a const object".
Step 1 : const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).
Step 2 : printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".
Because, we cannot modify a const variable.
10.
What will be the output of the program?
#include<stdio.h>
int main()
{
const c = -11;
const int d = 34;
printf("%d, %d\n", c, d);
return 0;
}
A. Error B. -11, 34 C. 11, 34 D. None of these
Answer: D
Explanation:
Step 1 : const c = -11; The constant variable 'c' is declared and initialized to value "-11".
Step 2 : const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.
Step 3 : printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.
Hence the output of the program is -11, 34
