Using lambda to return a boolean inside if else statement

Question

Here is my code:

int main(int argc, char** argv) {
    bool gg;
    if( [&]()->decltype(gg){

    return false;  //try changing this to true or false and you'll get the same result.

    } ){

    std::cout<<"all even"<<std::endl;   
    }else {
    std::cout<<"all odd"<<std::endl;
    }


    return 0;
}

It's just simple, i have an if else statement and a lambda function inside it that checks the condition. I don't know if it's the code or compiler but even if i change false to true and vice versa, i get the same result. I am using Dev CPP. What's wrong with my code?

well if you wrote the same code with a named function: if (returnFalse) { …} what would you expect? If you don't call the function, how do you expect a return value out of it? — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Jul 24, 2015 at 5:45
... I don't quite understand why this even compiles. Must be another gcc extension. — dyp
– dyp, Commented Jul 24, 2015 at 5:45
I'm voting to close this question as off-topic because it shows lack of research and of understanding of fundamentals of the language. — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Jul 24, 2015 at 5:47
@dyp not necessarily, you can do the same thing with function pointers… — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Jul 24, 2015 at 5:48
It does implicit cast from function pointer to bool, which is pretty dangerous. VS actually refuses to compile that code. — Ari0nhh
– Ari0nhh, Commented Jul 24, 2015 at 5:48

Community · Accepted Answer · 2020-06-20 09:12:55Z

6

5.1.2 Lambda expressions

6 The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator.

That is exactly what happens in your case. You forgot to invoke your closure object's () operator. Instead you used the closure object itself as a condition in if.

Since your closure object does not capture anything, per 5.1.2/6 it is implicitly convertible to plain function pointer type bool (*)(). And so your object got implicitly converted to a function pointer. Since the pointer is not null, it acts as true under if.

On other words, your code is interpreted by the compiler in the following way

bool gg;
auto lf = [&]() -> decltype(gg) { return false; };

bool (*f)() = lf;
if (f) {
  ...

If you make your lambda function actually capture something (e.g. replace return false; with return gg;), your original code will immediately fail to compile.

P.S. As you can read in the comments below, the very presence of [&] apparently should have disabled 5.1.2/6 for your lambda. But apparently your compiler treats 5.1.2/6 more loosely and looks for actual captures instead of simply checking for presence of non-[] capture clause.

edited Jun 20, 2020 at 9:12

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answered Jul 24, 2015 at 5:49

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7 Comments

Carlos Miguel Colanta Over a year ago

OOoo alright. no wonder it got evaluated and i didnt get any error.

M.M Over a year ago

@dyp this paragraph doesn't exclude capturing lambdas from having the conversion; it only defines the behaviour for non-capturing lambdas.

dyp Over a year ago

Ok, checked again: I was right, this lambda is not convertible to a function pointer. It does have a capture, and clang++ complains about it. g++ accepts it, which is an extension.

dyp Over a year ago

That is, a capture-default grammatically is a lambda-capture. Hence, this lambda does have a lambda-capture and is not convertible to a function pointer. coliru.stacked-crooked.com/a/5b938c774611398a

dyp Over a year ago

@MattMcNabb See p3. An implementation is not allowed to add such a conversion. You can observe whether or not the conversion exists via SFINAE-reflection.

|

Thomas Moulard · Accepted Answer · 2015-07-24 05:43:26Z

5

You don't evaluate the lambda, you just check the object:

int main(int argc, char** argv) {
    bool gg;
    if( [&]()->decltype(gg){

    return false;  //try changing this to true or false and you'll get the same result.

    }() == true){

    std::cout<<"all even"<<std::endl;   
    }else {
    std::cout<<"all odd"<<std::endl;
    }


    return 0;
}

edit: Ideone code https://ideone.com/yxloQf

answered Jul 24, 2015 at 5:43

Thomas Moulard

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5 Comments

Carlos Miguel Colanta Over a year ago

Doesn't it get evaluated already since it returned a true or false?

M.M Over a year ago

@CarloBrew evaluating a function is different to calling it. It's the same error as: bool f(); if ( f ) { instead of if ( f() ) {

Carlos Miguel Colanta Over a year ago

Oh alright, i forgot lambda's aren't automatically called unless it's something like std transform.

Thomas Moulard Over a year ago

Just to be clear: I added == true because I feel this is clearer but what was really missing was the parenthesis after the brackets. This evaluates the lambda and ask "is the return value true?" without it it means "does this lambda object is convertible to boolean and is this boolean true?" which is different and does not evaluate the lambda.

The Paramagnetic Croissant Over a year ago

@CarloBrew they are never "called automatically". Higher-order functions (such as std::transform()) call the supplied callbacks/transformation functions explicitly as well.

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