Multidimensional variable length array in C++ [duplicate] - c++

Question

This question already has answers here:
invalid conversion from 'int' to 'int*' [-fpermissive] on passing array
(3 answers)
Closed 4 years ago.
void drawTable(int arg[], int length);
int main()
{
int length=0;
int counter=0;
int *pointer2ArrSize = new int[length];
cout << "Enter length of array: " << endl;
cin >> length;
do{
for(int j=0; j<length; j++){
cout << "Enter array elements: \n";
cin >> pointer2ArrSize[j];
cout << "You entered: " << pointer2ArrSize[j] << " in position: "
<< j+1
<< endl;
counter++;
}
cout << drawTable(pointer2ArrSize[j],length) << endl;
}while(!counter == length);
return 0;
}
void drawTable(int arg[], int length){
for(int i=0; i<length; i++){
cout << arg[i] << " ";
cout << '/n';
}
}
error: invalid conversion from 'int' to 'int*' [-fpermissive]
My goal is to display a 2D variable length array. I want the user to define the length of the array and the elements in it. However, am not too familiar with vectors yet. how could I approach this?

Answer 1

cout << drawTable(pointer2ArrSize[j],length) << endl;
is wrong since the argument type for drawTable is int* and you are passing pointer2ArrSize[j] to it, which is of type int.
You need to use
cout << drawTable(pointer2ArrSize, length) << endl;
More importantly, use of
int *pointer2ArrSize = new int[length];
cout << "Enter length of array: " << endl;
cin >> length;
is wrong.
The array will not be resized to length after you accept its value from the user. Consequently, pointer2ArrSize will continue to be an array of size 0. Since pointer2ArrSize continues to be an array of size 0, any attempt to access its elements will result in undefined behavior.
Move the line that allocates memory after you have accepted the input value for length.
cout << "Enter length of array: " << endl;
cin >> length;
int *pointer2ArrSize = new int[length];
It will be better still to use std::vector. Then, you don't have to worry about memory allocation and deallocation.
cout << "Enter length of array: " << endl;
cin >> length;
std::vector<int> pointer2ArrSize(length);
Of course, you should change drawTable to accept a std::vector<int>.
void drawTable(std::vector<int> const& arg);
There won't be the need for length as a second argument to drawTable since you can get the size of the array from the std::vector.