Question

For code such as this:

std::list<int> a[3][3];

int myNumber = 0;

for(int i = 0; i < 3; i++)
{
    for(int j = 0; j < 3; j++)
    {
        a[i][j].push_back(myNumber);
        myNumber++;
    }
}

The Local window of the debugger shows:

There is no easy way to go through and see that:

list a[ 0 ][ 0 ] contains 0

list a[ 0 ][ 1 ] contains 1

list a[ 0 ][ 2 ] contains 2

list a[ 1 ][ 0 ] contains 3
etc.

I can only see what lists [ 0 ][ 0 ], [ 1 ][ 0 ], and [ 2 ][ 0 ] contain but I want to see what all the lists contain. How do I go about doing this in Visual Studio 2010?

Answer 1

“There is no easy way to go through and see that list a[0][2] contains 2″

You right click somewhere into code, chose Quick Watch, write there a[0][2] and with Add Watch you put into “Watch 1” so that you can see your list<int> at a[0][2] properly.

When you declare simple list<int> l;, Visual Studio shows it properly. You are able to see all elements.

But when you declare array of lists like this:

std::list<int> l[2];
l[1].push_back(3);
l[1].push_back(4);

then variable l is considered pointer to first std::list<int> so even if you push elements into list at index 1, Visual Studio just shows you that l is empty list: some address [0](). I can see l[1] only if its in Watch:

Possible solution is to replace simple c-style array ([]) with std::vector :

std::vector<std::list<int> > l;
l.resize(2);

l[1].push_back(3);
l[1].push_back(4);

so l is not considered a pointer to first list anymore. Since l is vector Visual Studio displays you all elements properly: