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For example, I can define a lambda function as
auto f = [](double value) {double ratio = 100; return value * ratio;}
Now I want to generate a function with the ratio an argument and return the lambda function like
auto makeLambda(double ratio) { return [=](double value) {return value * ratio;}; }
How to do it?
640
Returning a valueIf you use the RequestResponse invocation type, such as Synchronous invocation, AWS Lambda returns the result of the Python function call to the client invoking the Lambda function (in the HTTP response to the invocation request, serialized into JSON).
The return type for a lambda is specified using a C++ feature named 'trailing return type'. This specification is optional. Without the trailing return type, the return type of the underlying function is effectively 'auto', and it is deduced from the type of the expressions in the body's return statements.
Returning a value from a lambda expressionYou can explicitly return a value from the lambda using the qualified return syntax. Otherwise, the value of the last expression is implicitly returned.
The lambda must contain the same number of parameters as the delegate type. Each input parameter in the lambda must be implicitly convertible to its corresponding delegate parameter. The return value of the lambda (if any) must be implicitly convertible to the delegate's return type.
With C++14 function return type deduction, that should work.
In C++11, you could define another lambda (which can deduce the return type), rather than a function (which can't):
auto makeLambda = [](double ratio) {
return [=](double value) {return value * ratio;};
};
As noted in the comments, this could be further wrapped, if you particularly want a function:
auto makeLambdaFn(double ratio) -> decltype(makeLambda(ratio)) {
return makeLambda(ratio);
}
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