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I think you need to understand previous answer and it is not difficult to do it with pointer access. I haven't test following lines :

int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
    p = I.ptr<uchar>(i)+i;
    for ( j = i; j < nCols; ++j,p++)
    {
        *p = table[*p];
    }
}

int i,j;
uchar* p;
for( i = 0; i < nCols; ++i)
{
    p = I.ptr<uchar>(i)+i+1;
    for ( j = i+1; j < nRows; ++j,p+=nCols)
    {
        *p = table[*p];
    }
}


int i,j;
uchar* p=I.ptr<uchar>(i);
for( i = 0; i < min(nCols,nRows); ++i,p+=nCols+1)
{
        *p = table[*p];
}

I think you need to understand previous answer and it is not difficult to do it with pointer access. I haven't test following lines :(Mat i must be continuous):

int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
    p = I.ptr<uchar>(i)+i;
    for ( j = i; j < nCols; ++j,p++)
    {
        *p = table[*p];
    }
}

int i,j;
uchar* p;
for( i = 0; i < nCols; ++i)
{
    p = I.ptr<uchar>(i)+i+1;
    for ( j = i+1; j < nRows; ++j,p+=nCols)
    {
        *p = table[*p];
    }
}


int i,j;
uchar* p=I.ptr<uchar>(i);
for( i = 0; i < min(nCols,nRows); ++i,p+=nCols+1)
{
        *p = table[*p];
}