Find a pointer in a std::set based on a const pointer

Asked Nov 05 '17 at 14:10

Active Nov 05 '17 at 14:14

Viewed 39 times

The following code does not work because of the marked line, since s.find expects an int*, not const int*. Since nothing is being changed about the variable b, why is it not possible to call the method like that? Is there a nicer way than const_cast?

#include <set>

int main()
{
    int x = 5;
    int* a = &x;
    const int* b = a;

    std::set<int*> s;
    s.insert(a);
    s.find(b); // does not work
    s.find(const_cast<int*>(b)); // does work, but const_cast
}

The error is

error: no matching function for call to 'std::set<int*>::find(const int*&)'

after which it tries implicit conversion and fails with

error: invalid conversion from 'const int*' to 'std::set<int*>::key_type {aka int*}' [-fpermissive]

edited Nov 05 '17 at 14:14

asked Nov 05 '17 at 14:10

Post Self

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This is one of those times where I think using `const_cast` is fine, though if you never intend to modify the set elements you could always specialise for `const int*`. – George Nov 05 '17 at 14:13
Is the C++11 tag binding? – StoryTeller - Unslander Monica Nov 05 '17 at 14:14
@George Interesting. I'll wait for some other opinions, though. With specialize, you mean have the set itself be `std::set`? – Post Self Nov 05 '17 at 14:15
Can I mark my own question as a duplicate? – Post Self Nov 05 '17 at 14:17
@kim366 Yeah, though obviously that may not help here, but there's nothing unsafe or even especially dirty about using `const_cast`. – George Nov 05 '17 at 14:19
@kim366 Yes, on the web browser version there should be a link, `close` between the question and comments, hit that and it should give you an option to close off as a dupe :-) – George Nov 05 '17 at 14:21
@George Found it – Post Self Nov 05 '17 at 14:21

Find a pointer in a std::set based on a const pointer

0 Answers0