how to capture a pointer in CPP

In C++, you can capture a pointer by dereferencing it using the `*` operator. For example, if you have a pointer `int* ptr` that points to an integer, you can capture the value of the integer by dereferencing the pointer like this:

int value = *ptr;

This will assign the value of the integer that `ptr` points to, to the variable `value`. You can then use `value` in your code as needed.

It's worth noting that capturing a pointer in this way can be dangerous if the pointer is not valid (i.e. it points to memory that has been deallocated or is otherwise inaccessible). It's important to make sure that the pointer is valid before dereferencing it.

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how to capture a pointer in lambda function in CPP

In C++, you can capture a pointer in a lambda function by specifying the pointer in the lambda's capture list.

The capture list is a comma-separated list of variables or expressions that you want to capture. To capture a pointer, you can use the syntax `[&ptr]` to capture the pointer by reference, or `[=ptr]` to capture the pointer by value.

Here's an example:

int* ptr = new int(42);

auto lambda = [&ptr]() {
    int value = *ptr;
    std::cout << "The value of the pointer is: " << value << std::endl;
};

lambda(); // prints "The value of the pointer is: 42"

delete ptr; // make sure to delete the pointer when you're done with it

In this example, we create a lambda function that captures the `ptr` pointer by reference using the `&` symbol in the capture list. Inside the lambda function, we dereference the pointer to get the value it points to, and print it to the console.

Note that if you capture a pointer by reference, you need to make sure that the pointer remains valid for the lifetime of the lambda function. If the pointer goes out of scope or is deleted before the lambda is executed, you may get undefined behavior.