Can I define a C++ lambda function without auto?

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I have plenty of C# experience before but I am new to C++. I have seen this problem when trying to use lambda as I used to do.

For example:

auto compare = [] (int i1, int i2) { return i1*2 > i2; }

Is there any way to define the lambda with a specific type, rather than auto deduction?

I am asking this because I want to define a common lambda for my class. This lambada will be used in multiple places so I don't want to define them multiple times. However, 'auto' can only be used on static members, while on the other hand, I want to access non-static fields in the lambda.

You use std::function, which can glob any lambda or function pointer.

std::function< bool(int, int) > myFunc = []( int x, int y ){ return x > y; };

See C++ Reference.

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You could use std::function, but if that's not going to be efficient enough, you could write a functor object which resembles what lambdas do behind the scenes:

auto compare = [] (int i1, int i2) { return i1*2 > i2; }

is almost the same as

struct Functor {
    bool operator()(int i1, int i2) const { return i1*2 > i2; }
Functor compare;

If the functor should capture some variable in the context (e.g. the "this" pointer), you need to add members inside the functor and initialize them in the constructor:

auto foo = [this] (int i) { return this->bar(i); }

is almost the same as

struct Functor {
    Object *that;
    Functor(Object *that) : that(that) {}
    void operator()(int i) const { return that->bar(i); }
Functor foo(this);

What's the difference between declaring and defining in C and C++ , define is a C preprocessor directive used to define macros. well it defines a name against a value (this value can be anything) . so before compiling source � Air conditioning, something that people in hot places love and worship.

You can use as std::function<Signature> as mentioned, but at the cost of type-erasing the lambda. This will add an indirection (basically a virtual function call) when you call your lambda. So keep in mind it is less efficient if you will use it in a context where this matters.

#define directive (C/C++), This is a function declaration; it does not provide the body of the function, but it does tell the compiler that it can use this function and expect that it will be defined � Define a/c. a/c synonyms, a/c pronunciation, a/c translation, English dictionary definition of a/c. abbr. account current abbreviation for 1. account 2. account

The way to do this without the overhead of std::function, if you feel a functor is too cumbersome, is to use function pointers, a typedef, and optionally a macro.

In your example, this function pointer would look like this:

bool (*compare)(int, int) = [] (int i1, int i2) { return i1*2 > i2; };

Note how compare, the pointer's identifier, is tucked into the parenthesis, this trips people up sometimes, but it's the proper way to declare a function pointer.

You can write a typedef in the same way:

typedef bool (*Compare)(int, int);

The typedef name is where the identifier would ordinarily be. This allows you to now write the more concise:

Compare func = [] (int i1, int i2) { return i1*2 > i2; };

If you allow yourself to use macros, you can go even further in terms of ergonomics:

#define CompareLambda [](int i1, int i2)
typedef bool (*Compare)(int, int);
Compare func = CompareLambda{ return i1*2 > i2;};

Macros and typedefs exist as tools to make your work as a programmer easier by helping you avoid repeating yourself, don't be afraid to use them.

Does the position of `#define` in C code matter?, After the macro is defined, the compiler can substitute the token string for each occurrence of the identifier in the source file. Syntax. #define� This C program would print the following: is over 10 years old. Expression. You can use the #define directive to define a constant using an expression. For example: #define AGE (20 / 2) In this example, the constant named AGE would also contain the value of 10. Below is an example C program where we use an expression to define

typedef versus #define in C, C files are parsed top to bottom at both the preprocessing stage and the compilation stage. (Note: As MSalters points out, each stage starts� Microsoft C/C++ lets you redefine a macro if the new definition is syntactically identical to the original definition. In other words, the two definitions can have different parameter names. This behavior differs from ANSI C, which requires that the two definitions be lexically identical.

Difference between const and #define in C, C++ programming , #define will just copy-paste the definition values at the point of use, while typedef is the actual definition of a new type. typedef follows the scope� You can also define a symbol with the -define compiler option. You can undefine a symbol with #undef. A symbol that you define with -define or with #define does not conflict with a variable of the same name. That is, a variable name should not be passed to a preprocessor directive and a symbol can only be evaluated by a preprocessor directive.

C preprocessor, Macro can be used in anywhere in the program or in other files to by including the related header file, thus macros are not scope controlled, but the constant can � Can definition is - be physically or mentally able to. How to use can in a sentence. can vs. may

  • Try using std::function.
  • You can access non-static fields by capturing this.
  • Note that std::function is less efficient because of the dynamicism. Usually doesn't matter.
  • @siyu less efficient than lambdas, but more efficient than pretty much any other language's equivalent abstraction. 1.5-3x the call overhead of a function pointer last I checked (vtable lookup then call). There are std::function implememtations that rivsl function pointers, but not in major libraries that I know of.
  • If you're so worried about the performance hit of std::function then there's probably better questions to ask like why you're using lambdas in a performance sensitive environment.
  • @morp what? lambdas should be used in performance sensitive environments: they let you use really complex harnesses to deliver code packages with next to zero overhead. Why do you think they should not be used?
  • Lambda is not slow at all.