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Java Training Classes > Java Training in Chennai > Kamal

Kamal Java trainer in Chennai

Kamal

locationImg Chennai International Airport, Chennai
10 yrs of Exp
students 1 student
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Education

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Computer Expert

Online Classes
Student's home
I m having 10+ years of experience. I teach programming in a unique way that one can understand. I teach how to make logic and then how to program so that you can write any given program yourself. I have participated in many programming contests and I like making game programs and any kind of program/projects. You can also learn all these. 100% satisfaction assured.

Languages Spoken

Hindi

English

Education

Bangalore university 2005

Master of Computer Applications (M.C.A.)

Address

Chennai International Airport, Chennai, India - 600027

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Teaches

Java Training Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Java Training Classes

10

Teaches

Core Java

Certification training offered

No

BCA Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BCA Tuition

10

BCA Subject

C Language Programming, Java Programming, Programming in C++

Type of class

Crash Course, Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

C Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in C Language Classes

10

C++ Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in C++ Language Classes

10

Proficiency level taught

Basic C++

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

10

Board

International Baccalaureate, State, ISC/ICSE, CBSE, IGCSE

Subjects taught

Computer Science, Information Technology

Taught in School or College

Yes

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

10

Board

International Baccalaureate, State, ISC/ICSE, CBSE, IGCSE

Subjects taught

Computer Science, Information Technology

Taught in School or College

Yes

Reviews

No Reviews yet!

Answers by Kamal

Answered on 15/09/2016 Learn IT Courses/Programming Languages/C Language

Ask a Question

Post a Lesson

how to calculate address of n dimensional matrix

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a, a, .. . in 2-D array elements are logically in the form of matrix having row and column index and... ...more
An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a[0], a[1], .. . in 2-D array elements are logically in the form of matrix having row and column index and can be represented as a[0][0], a[0][1] etc. and they are stored row wise in the memory, because the memory is linear. Now a 3-D array is nothing but a logical structure which can be accessed by 3 index numbers. If the array is of size 3 in all the dimensions(int a[3][3][3] then it is stored in the memory in the following order a[0][0][0], a[0][0][1], a[0][0][2], a[0][1][0], a[0][1][1], a[0][1][2], a[0][2][0], a[0][2][1], a[0][2][2], a[1][0][0], a[1][0][1] and so on. Now to find the address of any element of the array based on the given index number and given dimension size, given element size(data type size) and given base address: Suppose the array is of n-dimension having the size of each dimension as S1,S2,S3. . .. Sn And the element size is ES, Base Address is BA And the index number of the element to find the address is given as i1, i2, i3, . . . .in Then the formula will be: Address of A[i1][ i2][ i3]. . . .[ in] = BA + ES*[ i1*( S2* S3* S4 *. . ..* Sn) + i2*( S3* S4* S5 *.. .. * Sn) + .. . . .. + in-2*( Sn-1*Sn) + in-1*Sn + in ] Example-1: An int array A[100][50] is stored at the address 1000. Find the address of A[40][25]. Solution: BA=1000, ES=2, S1=100, S2=50, i1=40, i2=25 Address of A[40][25]=1000+2*[40*50 + 25]=1450 Example-2: An int array A[50][40][30] is stored at the address 1000. Find the address of A[40][20][10]. Solution: BA=1000, ES=2, S1=50, S2=40, S3=30, i1=40, i2=20, i3=10 Address of A[40][20][10]=1000+2*[40*40*30 + 20*30 + 10]=98220
Answers 21 Comments
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Answered on 15/09/2016 Learn IT Courses/Programming Languages/C Language

Ask a Question

Post a Lesson

how to calculate address of n dimensional matrix

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). e.g in 1-D array you the elements are linearly arranged and can be addressed as a, a,... in 2-D array elements are logically in the form of matrix having row and column index and... ...more
An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). e.g in 1-D array you the elements are linearly arranged and can be addressed as a[0], a[1],... in 2-D array elements are logically in the form of matrix having row and column index and can be represented as a[0][0], a[0][1] etc. and they are stored row wise in the memory, because the memory is linear. Now a 3-D array is nothing but a logical structure which can be accessed by 3 index numbers. If the array is of size 3 in all the dimensions(int a[3][3][3] then it is stored in the memory in the following order a[0][0][0], a[0][0][1], a[0][0][2], a[0][1][0], a[0][1][1], a[0][1][2], a[0][2][0], a[0][2][1], a[0][2][2], a[1][0][0], a[1][0][1] and so on. Now to find the address of any element of the array based on the given index number and given dimension size, given element size(data type size) and given base address: Suppose the array is of n-dimension having the size of each dimension as S1,S2,S3....Sn And the element size is ES, Base Address is BA And the index number of the element to find the address is given as i1, i2, i3, ....in Then the formula will be: Address of A[i1][ i2][ i3]....[ in] = BA + ES*[ i1*( S2* S3* S4*....* Sn) + i2*( S3* S4* S5*....* Sn) + ...... + in-2*( Sn-1*Sn) + in-1*Sn + in ] Example-1: An int array A[100][50] is stored at the address 1000. Find the address of A[40][25]. Solution: BA=1000, ES=2, S1=100, S2=50, i1=40, i2=25 Address of A[40][25]=1000+2*[40*50 + 25]=1450 Example-2: An int array A[50][40][30] is stored at the address 1000. Find the address of A[40][20][10]. Solution: BA=1000, ES=2, S1=50, S2=40, S3=30, i1=40, i2=20, i3=10 Address of A[40][20][10]=1000+2*[40*40*30 + 20*30 + 10]=98220
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Teaches

Java Training Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Java Training Classes

10

Teaches

Core Java

Certification training offered

No

BCA Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BCA Tuition

10

BCA Subject

C Language Programming, Java Programming, Programming in C++

Type of class

Crash Course, Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

C Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in C Language Classes

10

C++ Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in C++ Language Classes

10

Proficiency level taught

Basic C++

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

10

Board

International Baccalaureate, State, ISC/ICSE, CBSE, IGCSE

Subjects taught

Computer Science, Information Technology

Taught in School or College

Yes

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

10

Board

International Baccalaureate, State, ISC/ICSE, CBSE, IGCSE

Subjects taught

Computer Science, Information Technology

Taught in School or College

Yes

No Reviews yet!

Answers by Kamal

Answered on 15/09/2016 Learn IT Courses/Programming Languages/C Language

Ask a Question

Post a Lesson

how to calculate address of n dimensional matrix

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a, a, .. . in 2-D array elements are logically in the form of matrix having row and column index and... ...more
An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a[0], a[1], .. . in 2-D array elements are logically in the form of matrix having row and column index and can be represented as a[0][0], a[0][1] etc. and they are stored row wise in the memory, because the memory is linear. Now a 3-D array is nothing but a logical structure which can be accessed by 3 index numbers. If the array is of size 3 in all the dimensions(int a[3][3][3] then it is stored in the memory in the following order a[0][0][0], a[0][0][1], a[0][0][2], a[0][1][0], a[0][1][1], a[0][1][2], a[0][2][0], a[0][2][1], a[0][2][2], a[1][0][0], a[1][0][1] and so on. Now to find the address of any element of the array based on the given index number and given dimension size, given element size(data type size) and given base address: Suppose the array is of n-dimension having the size of each dimension as S1,S2,S3. . .. Sn And the element size is ES, Base Address is BA And the index number of the element to find the address is given as i1, i2, i3, . . . .in Then the formula will be: Address of A[i1][ i2][ i3]. . . .[ in] = BA + ES*[ i1*( S2* S3* S4 *. . ..* Sn) + i2*( S3* S4* S5 *.. .. * Sn) + .. . . .. + in-2*( Sn-1*Sn) + in-1*Sn + in ] Example-1: An int array A[100][50] is stored at the address 1000. Find the address of A[40][25]. Solution: BA=1000, ES=2, S1=100, S2=50, i1=40, i2=25 Address of A[40][25]=1000+2*[40*50 + 25]=1450 Example-2: An int array A[50][40][30] is stored at the address 1000. Find the address of A[40][20][10]. Solution: BA=1000, ES=2, S1=50, S2=40, S3=30, i1=40, i2=20, i3=10 Address of A[40][20][10]=1000+2*[40*40*30 + 20*30 + 10]=98220
Answers 21 Comments
Dislike Bookmark

Answered on 15/09/2016 Learn IT Courses/Programming Languages/C Language

Ask a Question

Post a Lesson

how to calculate address of n dimensional matrix

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). e.g in 1-D array you the elements are linearly arranged and can be addressed as a, a,... in 2-D array elements are logically in the form of matrix having row and column index and... ...more
An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). e.g in 1-D array you the elements are linearly arranged and can be addressed as a[0], a[1],... in 2-D array elements are logically in the form of matrix having row and column index and can be represented as a[0][0], a[0][1] etc. and they are stored row wise in the memory, because the memory is linear. Now a 3-D array is nothing but a logical structure which can be accessed by 3 index numbers. If the array is of size 3 in all the dimensions(int a[3][3][3] then it is stored in the memory in the following order a[0][0][0], a[0][0][1], a[0][0][2], a[0][1][0], a[0][1][1], a[0][1][2], a[0][2][0], a[0][2][1], a[0][2][2], a[1][0][0], a[1][0][1] and so on. Now to find the address of any element of the array based on the given index number and given dimension size, given element size(data type size) and given base address: Suppose the array is of n-dimension having the size of each dimension as S1,S2,S3....Sn And the element size is ES, Base Address is BA And the index number of the element to find the address is given as i1, i2, i3, ....in Then the formula will be: Address of A[i1][ i2][ i3]....[ in] = BA + ES*[ i1*( S2* S3* S4*....* Sn) + i2*( S3* S4* S5*....* Sn) + ...... + in-2*( Sn-1*Sn) + in-1*Sn + in ] Example-1: An int array A[100][50] is stored at the address 1000. Find the address of A[40][25]. Solution: BA=1000, ES=2, S1=100, S2=50, i1=40, i2=25 Address of A[40][25]=1000+2*[40*50 + 25]=1450 Example-2: An int array A[50][40][30] is stored at the address 1000. Find the address of A[40][20][10]. Solution: BA=1000, ES=2, S1=50, S2=40, S3=30, i1=40, i2=20, i3=10 Address of A[40][20][10]=1000+2*[40*40*30 + 20*30 + 10]=98220
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