No matching member function error inside lambda expression?

Question

This is quite a strange error to me. Check the code below:

void test(void){
    vector<string> v;
    v.push_back("hello");
    auto fn=[=](){
        v.push_back("world");
    };
}

The first push_back method passed the compilation but the second failed, yielding the error:

Error:no matching member function for call to 'push_back'

The compiler note is:

**Note:(687, 36) candidate function not viable: 'this' argument has type 'const vector' (aka 'const vector, allocator > >')

But the method is not marked const**.
Well I am not using any const argument and I cannot figure what the compiler is trying to tell me. Could anybody help me?

Kerrek SB · Accepted Answer · 2016-04-19 14:56:23Z

10

Lambda call operator member functions are const by default. If you want a mutable call operator, say mutable:

auto fn = [=]() mutable {
//              ^^^^^^^
    v.push_back("world");
};

Having const be the default forces you to be explicit about the fact that you mean to capture a copy of the vector and mutate that copy, rather than the original vector v.

By contrast, variables that are captured by reference can be mutated by const-qualified member functions:

auto fn = [&]() {
//        ^^^
    v.push_back("world");  // modifies original "V"!
};

(This is essentially because const T is the same as T when T = U &; there are no "constant references" in C++.)

edited Apr 19, 2016 at 14:56

answered Apr 19, 2016 at 14:50

Kerrek SB

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5 Comments

NathanOliver Over a year ago

Probably want to capture by reference as well.

Kerrek SB Over a year ago

@NathanOliver: Not "as well", but rather "instead".

NathanOliver Over a year ago

Ah yes. Instead indeed

Heranort Over a year ago

Well what is it mean that you want to "capture a copy and mutate that copy, rather than the original v"? I tested this using mutate, add a line of code that each time i call that lambda function, i print &v out, and seems that the address of that two "copies" are the same?

Kerrek SB Over a year ago

@Heranort: Compare it with the address of the very first vector v in your code.

Andreas DM · Accepted Answer · 2016-04-19 15:54:49Z

4

capture by value is const use the keyword mutable (doesn't change the original vector):

auto fn = [=]() mutable {
    v.push_back("world");
};

or by reference (changes the original vector):

auto fn = [&]() {
    v.push_back("world");
};

edited Apr 19, 2016 at 15:54

answered Apr 19, 2016 at 14:53

Andreas DM

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Comments

skypjack · Accepted Answer · 2016-04-19 17:33:08Z

0

Because of the C++14 tag and for the sake of completeness, I'd cite also the initializer list as an alternative solution:

[&vec = v](){ vec.push_back("world") };

If you want to capture by copy instead of by reference:

[vec = v]() mutable { vec.push_back("world") };

Initializer lists as a (let me say) capture method are available since C++14, as already mentioned.

answered Apr 19, 2016 at 17:33

skypjack

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