C++ Lambda Expression (chatGpt Discussions!)

Q: cpp lanbda expression

auto val = []() -> <type>{

};

I can provide variables inside [] n () also. But what is difference?

chatGpt:

In C++, lambda expressions can capture variables in two ways: through the capture list (`[]`) and the parameter list (`()`). Understanding the difference between these two is crucial for effectively using lambdas in C++.

### 1. **Capture List (`[]`)**

The capture list is used to capture variables from the surrounding scope (i.e., the scope in which the lambda is defined). The captured variables can be accessed inside the lambda's body.

- **Syntax:** `[]`

- **Purpose:** To specify which variables from the enclosing scope are accessible inside the lambda.

#### Types of Captures:

- **Capture by Value:** `[x]` captures the variable `x` by value. The lambda will create a copy of `x`.

- **Capture by Reference:** `[&x]` captures the variable `x` by reference. The lambda will use the original `x`.

- **Capture All by Value:** `[=]` captures all variables used inside the lambda by value.

- **Capture All by Reference:** `[&]` captures all variables used inside the lambda by reference.

  

#### Example:

```cpp

int a = 10, b = 20;

auto lambda = [a, &b]() {

    // a is captured by value, b is captured by reference

    std::cout << "a: " << a << ", b: " << b << std::endl;

};

lambda();

b = 30;

lambda(); // Note that a remains 10, while b changes to 30

```

### 2. **Parameter List (`()`)**

The parameter list is used to pass arguments to the lambda function, just like in a regular function. These are not captured variables from the enclosing scope but rather arguments passed when the lambda is invoked.

- **Syntax:** `()`

- **Purpose:** To define parameters that the lambda can accept when it is called.

#### Example:

```cpp

auto lambda = [](int x, int y) -> int {

    return x + y;

};

int result = lambda(5, 10); // Passes 5 and 10 as arguments

```

### **Differences:**

- **Scope of Variables:**

  - **Capture List (`[]`)**: Captures variables from the surrounding scope of where the lambda is defined.

  - **Parameter List (`()`)**: Accepts variables as input parameters when the lambda is invoked.

- **When to Use:**

  - **Capture List (`[]`)**: Use this when you need to access or modify variables from the enclosing scope inside the lambda.

  - **Parameter List (`()`)**: Use this when the lambda needs to take arguments at the time of invocation, similar to a regular function.

### Combined Example:

```cpp

int a = 5;

auto lambda = [a](int x) -> int {

    return a + x; // 'a' is captured from the surrounding scope, 'x' is passed as an argument

};

int result = lambda(10); // result is 15 (5 from 'a' and 10 from 'x')

```

### Summary:

- **Capture List (`[]`)**: Captures variables from the surrounding context for use within the lambda.

- **Parameter List (`()`)**: Defines parameters that must be provided when calling the lambda. 

Both the capture list and parameter list can be used together to create powerful and flexible lambda expressions in C++.

Q: 

Capture All by Value: [=] captures all variables used inside the lambda by value.

Capture All by Reference: [&] captures all variables used inside the lambda by reference.

how to do? give example

chatGpt: 

Sure! Let's break down the examples using **capture by value** (`[=]`) and **capture by reference** (`[&]`).

### **Capture All by Value `[=]`**

When you use `[=]` in the capture list of a lambda, it means that all variables from the enclosing scope that are used inside the lambda will be captured by value. This means that a copy of each variable is made, and any changes made inside the lambda do not affect the original variables outside the lambda.

#### Example:

```cpp

#include <iostream>

int main() {

    int a = 10;

    int b = 20;

    auto lambdaByValue = [=]() {

        // 'a' and 'b' are captured by value

        std::cout << "Inside lambda (by value): a = " << a << ", b = " << b << std::endl;

    };

    lambdaByValue();

    // Change original variables

    a = 15;

    b = 25;

    lambdaByValue();  // a and b inside lambda remain 10 and 20 because they were captured by value

    std::cout << "Outside lambda: a = " << a << ", b = " << b << std::endl;

    return 0;

}

```

**Output:**

```

Inside lambda (by value): a = 10, b = 20

Inside lambda (by value): a = 10, b = 20

Outside lambda: a = 15, b = 25

```

### **Capture All by Reference `[&]`**

When you use `[&]` in the capture list of a lambda, it means that all variables from the enclosing scope that are used inside the lambda will be captured by reference. This means that the lambda accesses and modifies the actual variables, and any changes inside the lambda will reflect on the original variables.

#### Example:

```cpp

#include <iostream>

int main() {

    int a = 10;

    int b = 20;

    auto lambdaByReference = [&]() {

        // 'a' and 'b' are captured by reference

        std::cout << "Inside lambda (by reference): a = " << a << ", b = " << b << std::endl;

        a += 5; // Modify 'a'

        b += 5; // Modify 'b'

    };

    lambdaByReference();

    // a and b have been modified by the lambda

    std::cout << "Outside lambda: a = " << a << ", b = " << b << std::endl;

    return 0;

}

```

**Output:**

```

Inside lambda (by reference): a = 10, b = 20

Outside lambda: a = 15, b = 25

```

### **Summary of Differences:**

- **Capture by Value (`[=]`)**: The lambda creates a copy of the variables. Changes made inside the lambda do not affect the original variables.

- **Capture by Reference (`[&]`)**: The lambda works directly with the original variables. Changes made inside the lambda are reflected outside.

These examples illustrate how you can use capture by value and capture by reference in lambda expressions, depending on whether you want to work with copies or directly modify the original variables.