How to create an array of n rows and 5 columns in C++, but the n value is changing

I've this code, where I want to create a dynamic 2-d array with n x 5 values -

int** bump = new int*[b_bump_num];

for(int i = 0; i < b_bump_num; i++) {

    bump[i] = new int[5];

}



for(int i = 0; i < b_bump_num; i++) {

    delete bump[i];

}

delete bump;

But I'm facing this error -

*error:Expressions must contain pointer types-The third last line-bump[i]

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

which compiler are you using? Nothing wrong with your code. Please post the exact error message and a Minimal, Complete, and Verifiable example

– Alan Birtles
Nov 20 '18 at 7:53

I am very sorry, I use this software for the first time, many things do not understand, thank you for explaining to me

– dou
Nov 20 '18 at 7:54

An error is reported on the The third last line line: the expression must contain pointer type

– dou
Nov 20 '18 at 7:56

3

@dou Unrelated: In C++ you have a better option. Its called std::vector<>. And as usual, the question would be; is there any special reason that you are not using it?

– JeJo
Nov 20 '18 at 8:01

@JeJo:Can you give me an example of code usage?I do not have the programming foundation, a lot of basic knowledge do not understand, can only learn from the program example

– dou
Nov 20 '18 at 8:11

|
show 4 more comments

I've this code, where I want to create a dynamic 2-d array with n x 5 values -

int** bump = new int*[b_bump_num];

for(int i = 0; i < b_bump_num; i++) {

    bump[i] = new int[5];

}



for(int i = 0; i < b_bump_num; i++) {

    delete bump[i];

}

delete bump;

But I'm facing this error -

*error:Expressions must contain pointer types-The third last line-bump[i]

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

which compiler are you using? Nothing wrong with your code. Please post the exact error message and a Minimal, Complete, and Verifiable example

– Alan Birtles
Nov 20 '18 at 7:53

I am very sorry, I use this software for the first time, many things do not understand, thank you for explaining to me

– dou
Nov 20 '18 at 7:54

An error is reported on the The third last line line: the expression must contain pointer type

– dou
Nov 20 '18 at 7:56

3

@dou Unrelated: In C++ you have a better option. Its called std::vector<>. And as usual, the question would be; is there any special reason that you are not using it?

– JeJo
Nov 20 '18 at 8:01

@JeJo:Can you give me an example of code usage?I do not have the programming foundation, a lot of basic knowledge do not understand, can only learn from the program example

– dou
Nov 20 '18 at 8:11

|
show 4 more comments

I've this code, where I want to create a dynamic 2-d array with n x 5 values -

int** bump = new int*[b_bump_num];

for(int i = 0; i < b_bump_num; i++) {

    bump[i] = new int[5];

}



for(int i = 0; i < b_bump_num; i++) {

    delete bump[i];

}

delete bump;

But I'm facing this error -

*error:Expressions must contain pointer types-The third last line-bump[i]

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

I've this code, where I want to create a dynamic 2-d array with n x 5 values -

int** bump = new int*[b_bump_num];

for(int i = 0; i < b_bump_num; i++) {

    bump[i] = new int[5];

}



for(int i = 0; i < b_bump_num; i++) {

    delete bump[i];

}

delete bump;

But I'm facing this error -

*error:Expressions must contain pointer types-The third last line-bump[i]

c++ arrays

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

edited Nov 20 '18 at 8:00

asked Nov 20 '18 at 6:49

dou

134

asked Nov 20 '18 at 6:49

dou

134

asked Nov 20 '18 at 6:49

dou

134

which compiler are you using? Nothing wrong with your code. Please post the exact error message and a Minimal, Complete, and Verifiable example

– Alan Birtles
Nov 20 '18 at 7:53

I am very sorry, I use this software for the first time, many things do not understand, thank you for explaining to me

– dou
Nov 20 '18 at 7:54

An error is reported on the The third last line line: the expression must contain pointer type

– dou
Nov 20 '18 at 7:56

3

@dou Unrelated: In C++ you have a better option. Its called std::vector<>. And as usual, the question would be; is there any special reason that you are not using it?

– JeJo
Nov 20 '18 at 8:01

@JeJo:Can you give me an example of code usage?I do not have the programming foundation, a lot of basic knowledge do not understand, can only learn from the program example

– dou
Nov 20 '18 at 8:11

|
show 4 more comments

which compiler are you using? Nothing wrong with your code. Please post the exact error message and a Minimal, Complete, and Verifiable example

– Alan Birtles
Nov 20 '18 at 7:53

I am very sorry, I use this software for the first time, many things do not understand, thank you for explaining to me

– dou
Nov 20 '18 at 7:54

An error is reported on the The third last line line: the expression must contain pointer type

– dou
Nov 20 '18 at 7:56

3

@dou Unrelated: In C++ you have a better option. Its called std::vector<>. And as usual, the question would be; is there any special reason that you are not using it?

– JeJo
Nov 20 '18 at 8:01

@JeJo:Can you give me an example of code usage?I do not have the programming foundation, a lot of basic knowledge do not understand, can only learn from the program example

– dou
Nov 20 '18 at 8:11

which compiler are you using? Nothing wrong with your code. Please post the exact error message and a Minimal, Complete, and Verifiable example

– Alan Birtles
Nov 20 '18 at 7:53

I am very sorry, I use this software for the first time, many things do not understand, thank you for explaining to me

– dou
Nov 20 '18 at 7:54

An error is reported on the The third last line line: the expression must contain pointer type

– dou
Nov 20 '18 at 7:56

@dou Unrelated: In C++ you have a better option. Its called std::vector<>. And as usual, the question would be; is there any special reason that you are not using it?

– JeJo
Nov 20 '18 at 8:01

@JeJo:Can you give me an example of code usage?I do not have the programming foundation, a lot of basic knowledge do not understand, can only learn from the program example

– dou
Nov 20 '18 at 8:11

|
show 4 more comments

1 Answer
1

active

oldest

votes

This may be all you need -

Approach 1: Using a simple 2-D array

#include <iostream>



using namespace std;



int main()

{

    int dimension1, dimension2 = 5;

    cout << "Enter first dimension : " << endl;

    cin >> dimension1;



    // dynamic allocation

    int** array2d = new int*[dimension1];

    for(int i = 0; i < dimension1; ++i)

        array2d[i] = new int[dimension2];



    /* 

    // you may fill it with your values

    for(int i = 0; i < dimension1; ++i)

        for(int j = 0; j < dimension2; ++j)

            array2d[i][j] = i;

    */



    // print

    for(int i = 0; i < dimension1; ++i) {

        for(int j = 0; j < dimension2; ++j) {

            cout << array2d[i][j] << " ";

        }

        cout << endl;

    }



    // free

    for(int i = 0; i < dimension1; ++i)

        delete  array2d[i];



    delete  array2d;



    return 0;

}

Sample Output :

Enter first dimension : 

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Approach 2: Using Vectors -

#include <iostream>

#include <vector> 



using namespace std;



int main()

{

    // size of row 

    int rowSize;

    int colSize = 5;



    cout << "Enter row size : ";

    cin >> rowSize;



    // Create a vector of vector with size equal to rowSize; filled with 0.

    vector<vector<int>> vec(rowSize, vector<int>(colSize));



    for (int i = 0; i < rowSize; i++) { 

        for (int j = 0; j < vec[i].size(); j++) 

            cout << vec[i][j] << " "; 

        cout << endl; 

    }



    vec.clear();



    return 0;

}

Sample Output :

Enter row size : 3                  

0 0 0 0 0                           

0 0 0 0 0                           

0 0 0 0 0

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

2

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

This may be all you need -

Approach 1: Using a simple 2-D array

#include <iostream>



using namespace std;



int main()

{

    int dimension1, dimension2 = 5;

    cout << "Enter first dimension : " << endl;

    cin >> dimension1;



    // dynamic allocation

    int** array2d = new int*[dimension1];

    for(int i = 0; i < dimension1; ++i)

        array2d[i] = new int[dimension2];



    /* 

    // you may fill it with your values

    for(int i = 0; i < dimension1; ++i)

        for(int j = 0; j < dimension2; ++j)

            array2d[i][j] = i;

    */



    // print

    for(int i = 0; i < dimension1; ++i) {

        for(int j = 0; j < dimension2; ++j) {

            cout << array2d[i][j] << " ";

        }

        cout << endl;

    }



    // free

    for(int i = 0; i < dimension1; ++i)

        delete  array2d[i];



    delete  array2d;



    return 0;

}

Sample Output :

Enter first dimension : 

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Approach 2: Using Vectors -

#include <iostream>

#include <vector> 



using namespace std;



int main()

{

    // size of row 

    int rowSize;

    int colSize = 5;



    cout << "Enter row size : ";

    cin >> rowSize;



    // Create a vector of vector with size equal to rowSize; filled with 0.

    vector<vector<int>> vec(rowSize, vector<int>(colSize));



    for (int i = 0; i < rowSize; i++) { 

        for (int j = 0; j < vec[i].size(); j++) 

            cout << vec[i][j] << " "; 

        cout << endl; 

    }



    vec.clear();



    return 0;

}

Sample Output :

Enter row size : 3                  

0 0 0 0 0                           

0 0 0 0 0                           

0 0 0 0 0

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

2

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

add a comment |

This may be all you need -

Approach 1: Using a simple 2-D array

#include <iostream>



using namespace std;



int main()

{

    int dimension1, dimension2 = 5;

    cout << "Enter first dimension : " << endl;

    cin >> dimension1;



    // dynamic allocation

    int** array2d = new int*[dimension1];

    for(int i = 0; i < dimension1; ++i)

        array2d[i] = new int[dimension2];



    /* 

    // you may fill it with your values

    for(int i = 0; i < dimension1; ++i)

        for(int j = 0; j < dimension2; ++j)

            array2d[i][j] = i;

    */



    // print

    for(int i = 0; i < dimension1; ++i) {

        for(int j = 0; j < dimension2; ++j) {

            cout << array2d[i][j] << " ";

        }

        cout << endl;

    }



    // free

    for(int i = 0; i < dimension1; ++i)

        delete  array2d[i];



    delete  array2d;



    return 0;

}

Sample Output :

Enter first dimension : 

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Approach 2: Using Vectors -

#include <iostream>

#include <vector> 



using namespace std;



int main()

{

    // size of row 

    int rowSize;

    int colSize = 5;



    cout << "Enter row size : ";

    cin >> rowSize;



    // Create a vector of vector with size equal to rowSize; filled with 0.

    vector<vector<int>> vec(rowSize, vector<int>(colSize));



    for (int i = 0; i < rowSize; i++) { 

        for (int j = 0; j < vec[i].size(); j++) 

            cout << vec[i][j] << " "; 

        cout << endl; 

    }



    vec.clear();



    return 0;

}

Sample Output :

Enter row size : 3                  

0 0 0 0 0                           

0 0 0 0 0                           

0 0 0 0 0

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

2

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

add a comment |

This may be all you need -

Approach 1: Using a simple 2-D array

#include <iostream>



using namespace std;



int main()

{

    int dimension1, dimension2 = 5;

    cout << "Enter first dimension : " << endl;

    cin >> dimension1;



    // dynamic allocation

    int** array2d = new int*[dimension1];

    for(int i = 0; i < dimension1; ++i)

        array2d[i] = new int[dimension2];



    /* 

    // you may fill it with your values

    for(int i = 0; i < dimension1; ++i)

        for(int j = 0; j < dimension2; ++j)

            array2d[i][j] = i;

    */



    // print

    for(int i = 0; i < dimension1; ++i) {

        for(int j = 0; j < dimension2; ++j) {

            cout << array2d[i][j] << " ";

        }

        cout << endl;

    }



    // free

    for(int i = 0; i < dimension1; ++i)

        delete  array2d[i];



    delete  array2d;



    return 0;

}

Sample Output :

Enter first dimension : 

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Approach 2: Using Vectors -

#include <iostream>

#include <vector> 



using namespace std;



int main()

{

    // size of row 

    int rowSize;

    int colSize = 5;



    cout << "Enter row size : ";

    cin >> rowSize;



    // Create a vector of vector with size equal to rowSize; filled with 0.

    vector<vector<int>> vec(rowSize, vector<int>(colSize));



    for (int i = 0; i < rowSize; i++) { 

        for (int j = 0; j < vec[i].size(); j++) 

            cout << vec[i][j] << " "; 

        cout << endl; 

    }



    vec.clear();



    return 0;

}

Sample Output :

Enter row size : 3                  

0 0 0 0 0                           

0 0 0 0 0                           

0 0 0 0 0

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

This may be all you need -

Approach 1: Using a simple 2-D array

#include <iostream>



using namespace std;



int main()

{

    int dimension1, dimension2 = 5;

    cout << "Enter first dimension : " << endl;

    cin >> dimension1;



    // dynamic allocation

    int** array2d = new int*[dimension1];

    for(int i = 0; i < dimension1; ++i)

        array2d[i] = new int[dimension2];



    /* 

    // you may fill it with your values

    for(int i = 0; i < dimension1; ++i)

        for(int j = 0; j < dimension2; ++j)

            array2d[i][j] = i;

    */



    // print

    for(int i = 0; i < dimension1; ++i) {

        for(int j = 0; j < dimension2; ++j) {

            cout << array2d[i][j] << " ";

        }

        cout << endl;

    }



    // free

    for(int i = 0; i < dimension1; ++i)

        delete  array2d[i];



    delete  array2d;



    return 0;

}

Sample Output :

Enter first dimension : 

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

Approach 2: Using Vectors -

#include <iostream>

#include <vector> 



using namespace std;



int main()

{

    // size of row 

    int rowSize;

    int colSize = 5;



    cout << "Enter row size : ";

    cin >> rowSize;



    // Create a vector of vector with size equal to rowSize; filled with 0.

    vector<vector<int>> vec(rowSize, vector<int>(colSize));



    for (int i = 0; i < rowSize; i++) { 

        for (int j = 0; j < vec[i].size(); j++) 

            cout << vec[i][j] << " "; 

        cout << endl; 

    }



    vec.clear();



    return 0;

}

Sample Output :

Enter row size : 3                  

0 0 0 0 0                           

0 0 0 0 0                           

0 0 0 0 0

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

edited Nov 20 '18 at 8:50

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

answered Nov 20 '18 at 7:21

Tushar Walzade

1,82331630

2

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

add a comment |

2

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

Use std::vector.

– Fantastic Mr Fox
Nov 20 '18 at 7:36

if (b_bump_num == 0) { int bump[1]={0}; } else { int** bump=new int*[n]; for(int i=0;i<b_bump_num;++i) { bump[i]=new int[5]; } } for(int i=0;i<b_bump_num;++i) delete bump[i]; delete bump;(An error is reported on the penultimate line: the expression must contain pointer type)

– dou
Nov 20 '18 at 7:45

@Tushar Walzade:Thank you very much.

– dou
Nov 20 '18 at 8:43

add a comment |

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