How to use remove_if on an std::list of structs when you want to compare to a member variable of the-CodePudding

I have an std::list of structs and I would like to remove items from the list based on if a certain member variable matches a particular value.

My struct and list:

struct Foo
{
  uint64_t PID;
  uintptr_t addr;
};

std::list<Foo> FooList;

Code to remove entry:

uintptr_t Bar;
FooList.remove_if(???) // Remove when "Foo.addr == Bar";

I'm not sure how to reference to the struct instance inside of the remove_if() function, any help would be appreciated!

Thanks,

Naitzirch

CodePudding user response：

list::remove_if takes a function object as its argument. You can feed with an inline lambda function like this:

FooList.remove_if([=Bar] (auto &element) {
    return element.addr == Bar;
});

Edit: be advised that if Bar is a local variable declared outside if the lambda, you need to capture it via copy (=Bar) or reference (&Bar) within the lambda capture list (the leading square brackets)