Comparing AST Nodes

Question

For calculating duplication, I am parsing a java project into the AST.

But when trying to compare the AST nodes, Declaration attributes such as src are compared too, which leads to the comparison failing.

This change was introduced < 1 year ago where Declaration annotations were removed.

To negate this I have to do the following:

myAst = visit(myAst) {
    case node n : {
        n.src = |unknown:///|;
    }
}

And then compare the nodes.

Is there any other way to make comparing nodes easier?

score 3 · Accepted Answer · answered Nov 29 '17 at 15:16

Pattern matching ignores keyword parameters if they arent part of the pattern.

rascal>data  X = z(int y = 0);
ok
rascal>x1 = z(y=3);
X: z(y=3)
rascal>x2 = z(y=4);
X: z(y=4)
rascal>x1 == x2
bool: false
rascal>x1 := x2
bool: true
rascal>z() := x1
bool: true
rascal>z(y=5) := x1
bool: false
rascal>

So if you want to test for equality, use pattern matching. If you are putting them in a map or a set, you will indeed have to "unset" them. There is a rascal function that does this in the most efficient way possible. The function unset in Node does it for the current node, and the unsetRec function does it recursively using a similar visitor as in your code.

Comparing AST Nodes

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