Lambdas

No-Capture Lambdas

I assume that you have experience with lambdas from C#, so what we will do here is cover the syntax that C++ has adopted. All code snippets are from the same file in the same sample.

Sample: LambdaSample\LambdaSample.cpp

	// Create a lambda-expression closure.
	auto lm1 = []()
	{
		wcout << L"No capture, parameterless lambda." << endl;
	};
	
	// Invoke the lambda.
	lm1();

Lambdas with Parameters

	// Create a lambda closure with parameters.
	auto lm2 = [](int a, int b)
	{
		wcout << a << L" + " << b << " = " << (a + b) << endl;
	};
	
	lm2(3,4);

Specifying a Lambda’s Return Type

The trailing return type here is -> int after the parameter specification.

	// Create a lambda closure with a trailing return type.
	auto lm3 = [](int a, int b) -> int
	{
		wcout << a << L" % " << b << " = ";
		return a % b;
	};
	
	wcout << lm3(7, 5) << endl;

Capturing Outside Variables

	int a = 5;
		int b = 6;
		// Capture by copy all variables that are currently in the scope. 
		// Note also that we do not need to capture the closure; 
		// here we simply invoke the anonymous lambda with the 
		// () after the closing brace.
		[=]()
		{
			wcout << a << L" + " << b << " = " << (a + b) << endl;
			//// It's illegal to modify a here because we have 
			//// captured by value and have not specified that 
			//// this lambda should be treated as mutable.
			//a = 10;
		}();
	
		[=]() mutable -> void
		{
			wcout << a << L" + " << b << " = " << (a + b) << endl;
			// By marking this lambda as mutable, we can now modify a. 
			// Since we are capturing by value, the modifications 
			// will not propagate outside.
			a = 10;
		}();
	
		wcout << L"The value of a is " << a << L"." << endl;
	
		[&]()
		{
			wcout << a << L" + " << b << " = " << (a + b) << endl;
			// By capturing by reference, we now do not need 
	// to mark this as mutable. 
			// Because it is a reference, though, changes now 
			// propagate out.
			a = 10;
		}();
	
		wcout << L"The value of a is " << a << L"." << endl;
	
		// Here we specify explicitly that we are capturing a by 
		// value and b as a reference.
		[a,&b]()
		{
			b = 12;
			wcout << a << L" + " << b << " = " << (a + b) << endl;
		}();
	
		// Here we specify explicitly that we are capturing b as 
		// a reference and that all other captures should be by 
		// value.
		[=,&b]()
		{
			b = 15;
			wcout << a << L" + " << b << " = " << (a + b) << endl;
		}();
	
		// Here we specify explicitly that we are capturing a by 
		// value and that all other captures should be by reference.
		[&,a]()
		{
			b = 18;
			wcout << a << L" + " << b << " = " << (a + b) << endl;
		}();

Lambdas in Class-Member Functions

When you use a lambda in a class-member function, you cannot use a default capture by reference. This is because the lambda will be provided with a this pointer, and it must be copied. Also, when dealing with reference-counted smart pointers, it’s common to run into problems with the lambda holding a reference to the class. Usually you will never get back to a reference count of zero, causing a memory leak in your program.

Conclusion

If you're familiar with lambda expressions in C#, then you shouldn't have a hard time getting used to the syntax in C++. In the next article, we take a look at the C++ Standard Library.

This lesson represents a chapter from C++ Succinctly, a free eBook from the team at Syncfusion.