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Home C Programming Arrays See What Others Are Saying!
  • Question
  • What will be the output of the program?
    #include<stdio.h>
    
    int main()
    {
        void fun(int, int[]);
        int arr[] = {1, 2, 3, 4};
        int i;
        fun(4, arr);
        for(i=0; i<4; i++)
            printf("%d,", arr[i]);
        return 0;
    }
    void fun(int n, int arr[])
    {
        int *p=0;
        int i=0;
        while(i++ < n)
            p = &arr[i];
        *p=0;
    }
    


  • Options
  • A. 2, 3, 4, 5
  • B. 1, 2, 3, 4
  • C. 0, 1, 2, 3
  • D. 3, 2, 1 0

  • Correct Answer
  • 1, 2, 3, 4 

    Explanation
    Step 1: void fun(int, int[]); This prototype tells the compiler that the function fun() accepts one integer value and one array as an arguments and does not return anything.

    Step 2: int arr[] = {1, 2, 3, 4}; The variable a is declared as an integer array and it is initialized to

    a[0] = 1, a[1] = 2, a[2] = 3, a[3] = 4

    Step 3: int i; The variable i is declared as an integer type.

    Step 4: fun(4, arr); This function does not affect the output of the program. Let's skip this function.

    Step 5: for(i=0; i<4; i++) { printf("%d,", arr[i]); } The for loop runs untill the variable i is less than '4' and it prints the each value of array a.

    Hence the output of the program is 1,2,3,4


    More questions

    • 1. Which of the following statements correct about k used in the below statement?
      char ****k;

    • Options
    • A. k is a pointer to a pointer to a pointer to a char
    • B. k is a pointer to a pointer to a pointer to a pointer to a char
    • C. k is a pointer to a char pointer
    • D. k is a pointer to a pointer to a char
    • Discuss
    • 2. What will be the output of the program?
      #include<stdio.h>
      
      void fun(void *p);
      int i;
      
      int main()
      {
          void *vptr;
          vptr = &i;
          fun(vptr);
          return 0;
      }
      void fun(void *p)
      {
          int **q;
          q = (int**)&p;
          printf("%d\n", **q);
      }
      

    • Options
    • A. Error: cannot convert from void** to int**
    • B. Garbage value
    • C. 0
    • D. No output
    • Discuss
    • 3. What will be the output of the program?
      #include<stdio.h>
      #include<string.h>
      
      int main()
      {
          static char s[] = "Hello!";
          printf("%d\n", *(s+strlen(s)));
          return 0;
      }
      

    • Options
    • A. 8
    • B. 0
    • C. 16
    • D. Error
    • Discuss
    • 4. What will be the output of the program?
      #include<stdio.h>
      
      int main()
      {
          int i=4, j=8;
          printf("%d, %d, %d\n", i|j&j|i, i|j&&j|i, i^j);
          return 0;
      }
      

    • Options
    • A. 4, 8, 0
    • B. 1, 2, 1
    • C. 12, 1, 12
    • D. 0, 0, 0
    • Discuss
    • 5. What will you do to treat the constant 3.14 as a float?

    • Options
    • A. use float(3.14f)
    • B. use 3.14f
    • C. use f(3.14)
    • D. use (f)(3.14)
    • Discuss
    • 6. What will be the output of the program if value 25 given to scanf()?
      #include<stdio.h>
      
      int main()
      {
          int i;
          printf("%d\n", scanf("%d", &i));
          return 0;
      }
      

    • Options
    • A. 25
    • B. 2
    • C. 1
    • D. 5
    • Discuss
    • 7. In the statement expression1 >> expression2. if expression1 is a signed integer with its leftmost bit set to 1 then on right shifting it the result of the statement will vary from computer to computer

    • Options
    • A. True
    • B. False
    • Discuss
    • 8. What do the following declaration signify?
      int *ptr[30];

    • Options
    • A. ptr is a pointer to an array of 30 integer pointers.
    • B. ptr is a array of 30 pointers to integers.
    • C. ptr is a array of 30 integer pointers.
    • D. ptr is a array 30 pointers.
    • Discuss
    • 9. Range of float id -2.25e+308 to 2.25e+308

    • Options
    • A. True
    • B. False
    • Discuss
    • 10. What will be the output of the program (in Turbo C under DOS)?
      #include<stdio.h>
      
      int main()
      {
          char huge *near *far *ptr1;
          char near *far *huge *ptr2;
          char far *huge *near *ptr3;
          printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3));
          return 0;
      }
      

    • Options
    • A. 4, 4, 8
    • B. 2, 4, 4
    • C. 4, 4, 2
    • D. 2, 4, 8
    • Discuss


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