How to pass size of an array from command line

Lost User

Is there any way to pass the size of an array using a command prompt.

#include using namespace std;

int main(int argc, char *argv[])
{
//const int SIZE = 5;

int arr\[ argv\[1\] \];

return 0;

}

Rick York

Arguments come as an array of char * - that's the argv variable. One way to convert them is :

int size = atoi( argv[1] );

Lost User

Now consider this code

#include

using namespace std;

class cls
{
private:
int arr[len];

public:
	int len;
	void print\_arr();

};

int main(int argc, char *argv[])
{
cls obj;
obj.len = atoi( argv[1] );
obj.print_arr();

return 0;

}

void cls::print_arr()
{
for (int i = 0; i < len; i++)
{
arr[i] = i;
cout << arr[i] << " ";
}
}

Getting errors error: ‘len’ was not declared in this scope error: ‘arr’ was not declared in this scope

Rick York

The array needs to be dynamically allocated. This is done using the new operator. I usually prepend an "m_" to member variable names so I know that is what they are. The array will be allocated in the class constructor and released in the destructor.

class cls
{
private:
int * m_array;

public:
int m_length;

cls( int length );   // constructor
~cls();              // destructor
void print\_array();

};

cls::cls( int length ) // constructor
{
m_length = length;
m_array = new int[length];
}

cls::~cls() // destructor
{
delete [] m_array;
m_array = nullptr;
}

int main(int argc, char *argv[])
{
int length = atoi( argv[1] );
cls obj( length );
obj.print_array();
return 0;
}

Lost User

Thanks :)

CPallini

Consider also using a vector[^], instead of a C-like array.