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  • Question
  • Range of double is -1.7e-38 to 1.7e+38 (in 16 bit platform - Turbo C under DOS)


  • Options
  • A. True
  • B. False

  • Correct Answer
  • False 

    Explanation
    False, The range of double is -1.7e+308 to 1.7e+308.

    Declarations and Initializations problems


    Search Results


    • 1. A long double can be used if range of a double is not enough to accommodate a real number.

    • Options
    • A. True
    • B. False
    • Discuss
    • 2. Range of float id -2.25e+308 to 2.25e+308

    • Options
    • A. True
    • B. False
    • Discuss
    • 3. Size of short integer and long integer would vary from one platform to another.

    • Options
    • A. True
    • B. False
    • Discuss
    • 4. If the definition of the external variable occurs in the source file before its use in a particular function, then there is no need for an extern declaration in the function.

    • Options
    • A. True
    • B. False
    • Discuss
    • 5. Which of the following correctly represents a long double constant?

    • Options
    • A. 6.68
    • B. 6.68L
    • C. 6.68f
    • D. 6.68LF
    • Discuss
    • 6. A float is 4 bytes wide, whereas a double is 8 bytes wide.

    • Options
    • A. True
    • B. False
    • Discuss
    • 7. Size of short integer and long integer can be verified using the sizeof() operator.

    • Options
    • A. True
    • B. False
    • Discuss
    • 8. What will be the output of the program?
      #include<stdio.h>
      int main()
      {
          int i=4, j=-1, k=0, w, x, y, z;
          w = i || j || k;
          x = i && j && k;
          y = i || j &&k;
          z = i && j || k;
          printf("%d, %d, %d, %d\n", w, x, y, z);
          return 0;
      }
      

    • Options
    • A. 1, 1, 1, 1
    • B. 1, 1, 0, 1
    • C. 1, 0, 0, 1
    • D. 1, 0, 1, 1
    • Discuss
    • 9. What will be the output of the program?
      #include<stdio.h>
      int main()
      {
          char ch;
          ch = 'A';
          printf("The letter is");
          printf("%c", ch >= 'A' && ch <= 'Z'? ch + 'a' - 'A':ch);
          printf("Now the letter is");
          printf("%c\n", ch >= 'A' && ch <= 'Z'? ch : ch + 'a' - 'A');
          return 0;
      }
      

    • Options
    • A. The letter is a
      Now the letter is A
    • B. The letter is A
      Now the letter is a
    • C. Error
    • D. None of above
    • Discuss
    • 10. What will be the output of the program?
      #include<stdio.h>
      int main()
      {
          int x=12, y=7, z;
          z = x!=4 || y == 2;
          printf("z=%d\n", z);
          return 0;
      }
      

    • Options
    • A. z=0
    • B. z=1
    • C. z=4
    • D. z=2
    • Discuss


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