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lambda expression error: expression must be a modifiable lvalue

Ok, the codes are:

vector<vector<double>> imageFiltered;

// some processing codes here

parallel_for( blocked_range<unsigned>(0, imageFiltered.size()),
    [=](const blocked_range<unsigned>& r) {
        for( unsigned i = r.begin(); i != r.end(); ++i ){
            for( unsigned j = 0; j != imageFiltered[i].size(); ++j ) {
                imageFiltered[i][j] = 0; // error here:expression must be a modifiable lvalue
            }
        }
});

And I've write another similar code block which works just fine. So, a little help here. PS: parallel_for is from Interl TBB.

like image 642
Adrian Yu Avatar asked Mar 22 '13 08:03

Adrian Yu


1 Answers

The [=] causes the lambda to capture by value, which means that it makes a copy of imageFiltered, and the copy is marked "const". Change the [=] to [&] to capture imageFiltered by reference, which should eliminate the problem.

like image 159
Arch D. Robison Avatar answered Nov 04 '22 13:11

Arch D. Robison