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tainted flag not propagated in a function defined with eval() and its toSource() value #11

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dmitris opened this Issue · 1 comment

2 participants

Dmitri S Stefano Di Paola
Dmitri S
Collaborator

Consider the following example:

var s = String.newTainted("foo"); // "foo" is the potential attack payload
s.tainted; // true, of course
eval("function myfun() {return '" + s + "';}"); // define function f() that returns the tainted string
var x = myfun();  // invoke newly define function
x.tainted; // currently false, should be true

Additionally, consider the return value of toSource() call that also contains the attack payload:

var src = myfun.toSource(); "function myfun() {return "foo";}"
src.tainted; // currently false, should be true
Stefano Di Paola
Owner
wisec commented

The eval case is the same as the Function issue #10
About the toSource() it was actually left untainted in the C/C++ code.
If you want you can still wrap it and return a tainted string.

Stefano Di Paola wisec closed this
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