Tag: limit

C Program To Find Odd Numbers Between Range using For Loop

Lets write a C program to generate odd numbers between 2 integer values input by the user using For loop.

Related Read:
Even or Odd Number: C Program
C Program to Generate Odd Numbers Between Two Integers

Note 1: An odd number is an integer that is not exactly divisible by 2.

Note 2: Odd numbers are of the form (2 * number + 1);

Note 3: Modular division( % ) returns remainder of division. For example, 10 / 2 = 5. But 10 % 2 = 0.

Video Tutorial: C Program To Find Odd Numbers Between Range using For Loop


[youtube https://www.youtube.com/watch?v=P85sziIyIdA]

YouTube Link: https://www.youtube.com/watch?v=P85sziIyIdA [Watch the Video In Full Screen.]


In above c program, we ask the user to input 2 integer value and store it in variables start and end. If value of start is greater than the value of end, then we swap the values.

For loop counter is initialized to start, and for loop executes until value of count is less than or equal to end. For each iteration of the for loop, count value increments by 1.

Inside for loop, for every value of count, we check if its not perfectly divisible by 2. If true, it’s a Odd number and we output that number to the console window.

Source Code: C Program To Find Odd Numbers Between Range using For Loop

 
#include<stdio.h>

int main()
{
    int start, end, temp, count;

    printf("Enter start and end value, to find odd numbers\n");
    scanf("%d%d", &start, &end);

    if(start > end)
    {
        temp  = start;
        start = end;
        end   = temp;
    }

    printf("Odd numbers between %d and %d are\n", start, end);

    for(count = start; count <= end; count++)
    {
        if(count % 2 != 0)
            printf("%d\n", count);
    }

    return 0;
}

Output 1
Enter start and end value, to find odd numbers
40
60
Odd numbers between 40 and 60 are
41
43
45
47
49
51
53
55
57
59

Output 2
Enter start and end value, to find odd numbers
60
40
Odd numbers between 40 and 60 are
41
43
45
47
49
51
53
55
57
59

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program To Find Range of Set of Numbers

Write a C program to find the range of a set of numbers entered through the keyboard. Range is the difference between the smallest and biggest number in the list.

Example: If biggest number in the list is 5 and smallest number in the list is 1. The difference between them is the range. i.e., 5 – 1 = 4. So range = 4.

Related Read:
while loop in C programming
if else statement in C
Relational Operators In C
C Program To Find Absolute Value of a Number

Expected Output for the Input

User Input:
Enter the limit
5
Enter 5 numbers
1
2
3
4
5

Output:
Small Number = 1
Big Number = 5
Range is 4

Logic To Find Range of Set of Numbers

First we ask the user to enter the length or the size of the list, and store it inside the variable limit. If the user enters the list size as 5, then we ask the user to enter 5 numbers.

Next we ask the user to enter the first number. We assign the first number entered by the user to variables small and big. Since user already entered 1 number into the list, we decrement the value of variable limit by 1.

Next we take remaining inputs inside the while loop. For each iteration of the while loop we decrement the value of variable limit by 1, until the value of limit is 0. Once value of limit is zero, control exits while loop.

For each input(inside the while loop), we check if the value entered is bigger than the value present in variable big. If its true, we assign the bigger number to variable big. We also check if the value entered is smaller than the value present in variable small. If its true, we assign the smaller number to variable small.

Once the control exits the while loop, variable big will have the biggest number in the list and variable small will have the smallest number in the list.

Finally, we use below formula to calculate range of the list:
range = big – small;

Video Tutorial: C Program To Find Range of Set of Numbers


[youtube https://www.youtube.com/watch?v=-yN6KsA8d-k]

YouTube Link: https://www.youtube.com/watch?v=-yN6KsA8d-k [Watch the Video In Full Screen.]

Source Code: C Program To Find Range of Set of Numbers

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int small, big, range, num, limit;

    printf("Enter the limit\n");
    scanf("%d", &limit);

    printf("Enter %d numbers\n", limit);
    scanf("%d", &num);

    small = big = num;

    limit = limit - 1;

    while(limit)
    {
        scanf("%d", &num);

        if(num > big)
        {
            big = num;
        }

        if(num < small)
        {
            small = num;
        }

        limit--;
    }

    range = big - small;

    printf("Small Number = %d\nBig Number = %d\n", small, big);
    printf("Range is %d\n", abs(range));

    return 0;
}

Output 1:
Enter the limit
5
Enter 5 numbers
0
-1
2
3
4
Small Number = -1
Big Number = 4
Range is 5

Output 2:
Enter the limit
6
Enter 6 numbers
10
9
8
7
6
5
Small Number = 5
Big Number = 10
Range is 5

Note: abs() returns the absolute value of a number. Absolute value is always positive.

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program to Generate Odd Numbers Between Two Integers

C program to generate odd numbers between 2 integer values input by the user.

Related Read:
Even or Odd Number: C Program
Even or Odd Number without using Modular Division: C Program

Note 1: An odd number is an integer that is not exactly divisible by 2.

Note 2: Odd numbers are of the form (2 * n + 1);

Note 3: Modular division( % ) returns remainder of division. For example, 10 / 2 = 5. But 10 % 2 = 0.

C Program to Generate Odd Numbers Between Two Integers

 
#include<stdio.h>

int main()
{
    int count, limit;

    printf("Enter start and end value to generate Odd numbers\n");
    scanf("%d%d", &count, &limit);

    printf("\nOdd numbers between %d and %d are:\n", count, limit);

    while(count <= limit)
    {
        if(count % 2 != 0)
        {
            printf("%d\n", count);
        }
        count++;
    }

    return 0;
}

Output 1
Enter start and end value to generate Odd numbers
10
30

Odd numbers between 10 and 30 are:
11
13
15
17
19
21
23
25
27
29

Output 2
Enter start and end value to generate Odd numbers
1
10

Odd numbers between 1 and 10 are:
1
3
5
7
9

C Program to Generate Odd Numbers Between Two Integers


[youtube https://www.youtube.com/watch?v=vPzV7pE0Djw]

YouTube Link: https://www.youtube.com/watch?v=vPzV7pE0Djw [Watch the Video In Full Screen.]


In above c program, we ask the user to input 2 integer value and store it in variables count and limit. While loop keeps executing until the start value i.e., count is less than or equal to the end value i.e., limit.

Inside while loop, for every value of count, we check if its not perfectly divisible by 2. If true, it’s a Odd number and we output that number to the console window. On every iteration of the loop we increment the value of count by 1.

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program to Generate Even Numbers Between Two Integers

C program to generate even numbers between 2 integer values input by the user.

Related Read:
Even or Odd Number: C Program
Even or Odd Number without using Modular Division: C Program

Note 1: An even number is an integer that is exactly divisible by 2.

Note 2: Even numbers are of the form 2 * n;

Note 3: Modular division( % ) returns remainder of division. For example, 10 / 2 = 5. But 10 % 2 = 0.

C Program to Generate Even Numbers Between Two Integers

 
#include<stdio.h>

int main()
{
    int count, limit;

    printf("Enter start value and end value to generate Even no's\n");
    scanf("%d%d", &count, &limit);

    printf("\nEven numbers between %d and %d are:\n", count, limit);

    while(count <= limit)
    {
        if(count % 2 == 0)
        {
            printf("%d\n", count);
        }
        count++;
    }

    return 0;
}

Output 1
Enter start value and end value to generate Even no’s
10
30

Even numbers between 10 and 30 are:
10
12
14
16
18
20
22
24
26
28
30

Output 2
Enter start value and end value to generate Even no’s
1
10

Even numbers between 1 and 10 are:
2
4
6
8
10

C Program to Generate Even Numbers Between Two Integers


[youtube https://www.youtube.com/watch?v=cE25_Fm7eMU]

YouTube Link: https://www.youtube.com/watch?v=cE25_Fm7eMU [Watch the Video In Full Screen.]


In above c program, we ask the user to input 2 integer value and store it in variables count and limit. While loop keeps executing until the start value i.e., count is less than or equal to the end value i.e., limit.

Inside while loop, for every value of count, we check if its perfectly divisible by 2. If true, it’s a Even number and we output that number to the console window. On every iteration of the loop we increment the value of count by 1.

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

Update:
In the above video I’ve said: “count should be exactly divisible by 0”.

I’m sorry about that. It should be – “Value inside variable count should be exactly divisible by 2.”

Cursor Object: MongoDB

Lets have a deeper look into the MongoDB cursor object.

cursor-object-mongodb

Documents in our collection
test database, names collection.

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> use test
switched to db test
> show collections
names
system.indexes
> db.names.find();
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }
{ "_id" : ObjectId("53be5d6104cc1cb0a7bfc3c3"), "name" : "Dev D" }
{ "_id" : ObjectId("53be5d6804cc1cb0a7bfc3c4"), "name" : "Emli" }
{ "_id" : ObjectId("53be5d8604cc1cb0a7bfc3c5"), "name" : "Farhan" }
{ "_id" : ObjectId("53be5d9204cc1cb0a7bfc3c6"), "name" : "Gangs" }
{ "_id" : ObjectId("53be5d9904cc1cb0a7bfc3c7"), "name" : "Hum" }
{ "_id" : ObjectId("53be5e3704cc1cb0a7bfc3c8"), "name" : 25 }

We have 8 documents with name in alphabetical order, and 1 document with name as 25. So in total we have 9 documents in our names collection.

Establishing Cursor

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> var cur = db.names.find();
 
> cur
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }
{ "_id" : ObjectId("53be5d6104cc1cb0a7bfc3c3"), "name" : "Dev D" }
{ "_id" : ObjectId("53be5d6804cc1cb0a7bfc3c4"), "name" : "Emli" }
{ "_id" : ObjectId("53be5d8604cc1cb0a7bfc3c5"), "name" : "Farhan" }
{ "_id" : ObjectId("53be5d9204cc1cb0a7bfc3c6"), "name" : "Gangs" }
{ "_id" : ObjectId("53be5d9904cc1cb0a7bfc3c7"), "name" : "Hum" }
{ "_id" : ObjectId("53be5e3704cc1cb0a7bfc3c8"), "name" : 25 }

Look at the contents of our cursor object cur.

hasNext() and next() methods on Cursor

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> var cur = db.names.find();
> cur.hasNext();
true
> cur.next()
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
> cur.next()
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
> cur.next()
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }

If there are any documents to iterate inside cursor object, then hasNext() will return true orelse it’ll return false. If hasNext() returns true, then we can iterate through the documents using next() method on the cursor object.

sort() method on Cursor Object

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> var cur = db.names.find();
 
> cur.sort({"name": -1});
{ "_id" : ObjectId("53be5d9904cc1cb0a7bfc3c7"), "name" : "Hum" }
{ "_id" : ObjectId("53be5d9204cc1cb0a7bfc3c6"), "name" : "Gangs" }
{ "_id" : ObjectId("53be5d8604cc1cb0a7bfc3c5"), "name" : "Farhan" }
{ "_id" : ObjectId("53be5d6804cc1cb0a7bfc3c4"), "name" : "Emli" }
{ "_id" : ObjectId("53be5d6104cc1cb0a7bfc3c3"), "name" : "Dev D" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
{ "_id" : ObjectId("53be5e3704cc1cb0a7bfc3c8"), "name" : 25 }

We could modify the cursor object using methods like sort(), limit() and skip(). In above example, we are modifying cursor object using sort() method, and we are sorting it in reverse lexicographical order on the name field.

limit() method on Cursor Object

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> var cur = db.names.find();
> cur.limit(3);
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }

We could limit the output/result using limit() method.

Chaining method on Cursor Object

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> var cur = db.names.find();
 
> cur.sort({"name": -1}).limit(5).skip(2);
{ "_id" : ObjectId("53be5d8604cc1cb0a7bfc3c5"), "name" : "Farhan" }
{ "_id" : ObjectId("53be5d6804cc1cb0a7bfc3c4"), "name" : "Emli" }
{ "_id" : ObjectId("53be5d6104cc1cb0a7bfc3c3"), "name" : "Dev D" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }

Here we chain the methods sort(), limit() and skip(). We are sorting in reverse lexicographical order on the name field, then skipping the first 2 documents and then limiting the result/output to 5 documents.

The order in which these 3 methods are applied are: First sort, then skip and then limit.
Also note that, these methods modify cursor object at the server side and not on client site.

Cursor Object: MongoDB


[youtube https://www.youtube.com/watch?v=_LexQW0vSlQ]

YouTube Link: https://www.youtube.com/watch?v=_LexQW0vSlQ [Watch the Video In Full Screen.]



explain() method on Cursor Object

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> db.names.find().explain();
{
        "cursor" : "BasicCursor",
        "isMultiKey" : false,
        "n" : 9,
        "nscannedObjects" : 9,
        "nscanned" : 9,
        "nscannedObjectsAllPlans" : 9,
        "nscannedAllPlans" : 9,
        "scanAndOrder" : false,
        "indexOnly" : false,
        "nYields" : 0,
        "nChunkSkips" : 0,
        "millis" : 0,
        "server" : "PC:27017",
        "filterSet" : false
}

explain() method shows that find() returns a basic cursor. More on explain() method in coming videos.

Programmatic way of printing Cursor Object content

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> var cur = db.names.find();
 
> while(cur.hasNext()) printjson(cur.next());
{ "_id" : ObjectId("53be5d4604cc1cb0a7bfc3c0"), "name" : "Alia" }
{ "_id" : ObjectId("53be5d5204cc1cb0a7bfc3c1"), "name" : "Bebo" }
{ "_id" : ObjectId("53be5d5904cc1cb0a7bfc3c2"), "name" : "Chameli" }
{ "_id" : ObjectId("53be5d6104cc1cb0a7bfc3c3"), "name" : "Dev D" }
{ "_id" : ObjectId("53be5d6804cc1cb0a7bfc3c4"), "name" : "Emli" }
{ "_id" : ObjectId("53be5d8604cc1cb0a7bfc3c5"), "name" : "Farhan" }
{ "_id" : ObjectId("53be5d9204cc1cb0a7bfc3c6"), "name" : "Gangs" }
{ "_id" : ObjectId("53be5d9904cc1cb0a7bfc3c7"), "name" : "Hum" }
{ "_id" : ObjectId("53be5e3704cc1cb0a7bfc3c8"), "name" : 25 }

While loop executes until cur.hasNext() returns true. Until cur.hasNext() is true, cur.next() keeps printing next document in the cursor cur.