DynamoDB Span Pointers

[nhulston]

What does this PR do?

Adds span pointers to spans for Lambdas triggered by DynamoDB streams. This change will affect the downstream case for Universal Instrumentation Lambda runtimes (Java, .NET, Golang).

Span pointers are similar to Span Links, but for cases when it is impossible to pass the Trace ID and Span ID between the spans that need to be linked.

When the calculated hashes for the upstream and downstream lambdas match, the Datadog frontend will automatically link the two traces together.

Upstream case (requires tracer to be instrumented):

Downstream case (this code):

When clicking on the linked span, a new tab opens linking to the opposite Lambda function.

Describe how you validated your changes

Mostly manual testing, but I also added unit tests. There are also functional tests in https://github.com/DataDog/serverless-e2e-tests -- DynamoDB is TODO but will be implemented soon.

[duncanista]

Suggested change

	let (primary_key1, value1, primary_key2, value2) = if self.dynamodb.keys.len() == 1 {
	// Single primary key case
	let (key, attr_value) = self
	.dynamodb
	.keys
	.iter()
	.next()
	.expect("No DynamoDB keys found");
	let value = attr_value.to_string()?;
	(key.clone(), value, String::new(), String::new())
	} else {
	// Two keys case, sort lexicographically for consistent ordering
	let mut keys: Vec<(&String, &AttributeValue)> = self.dynamodb.keys.iter().collect();
	keys.sort_by(|a, b| a.0.cmp(b.0));
	let (k1, attr1) = keys[0];
	// If unable to get string value, just return None
	let v1 = attr1.to_string()?;
	let (k2, attr2) = keys[1];
	let v2 = attr2.to_string()?;
	(k1.clone(), v1, k2.clone(), v2)
	};
	let (primary_key1, value1, primary_key2, value2) = match self.dynamodb.keys.len() {
	1 => {
	let (key, attr_value) = self.dynamodb.keys.iter()
	.next()
	.expect("No DynamoDB keys found");
	(key.clone(), attr_value.to_string()?, String::new(), String::new())
	},
	_ => {
	let mut sorted_keys: Vec<_> = self.dynamodb.keys.iter()
	.collect::<Vec<_>>()
	.into_iter()
	.sorted_by_key(|k| k.0.clone())
	.collect();
	let (k1, attr1) = sorted_keys[0];
	let (k2, attr2) = sorted_keys[1];
	(k1.clone(), attr1.to_string()?, k2.clone(), attr2.to_string()?)
	}
	};

[duncanista]

Ideally, I'd not even return anything, but maybe push to the array that you're creating? Although if you push and someone modifies the code, they might break the order (in case the order matters), anyhow, looks good if you don't decide to change the return thing,

Nonetheless, I'd encourage the suggestion above

[duncanista]

LGTM – left a comment