C++ lambda 按值语义捕获？

Question

class TestGetStr {
public:
    string a;
    string getString() {
        return a;
    }
};

void acceptConstString(const string & a) {
    std::cout << a << std::endl;
}

int main(int argc, const char * argv[]) {
    
    auto testGetStr = TestGetStr();

    acceptConstString(testGetStr.getString()); //compile ok

    auto testaaa = [testGetStr](){

//compile error 'this' argument to member function 'getString' has type 'const TestGetStr', but function is not marked const        
acceptConstString(testGetStr.getString());
    };
    

普通调用和 lambda 捕获之间的区别？

普通调用可以编译，但是lambda调用无法编译。

为什么？感谢您提供更多详细信息。

Answer 1

您可以添加mutable限定符。

auto testaaa = [testGetStr]() mutable {
//                            ^^^^^^^
    acceptConstString(testGetStr.getString());
};

否则lambda的operator()是 const 限定的，然后不能在捕获的对象上调用非常量成员函数。

• mutable: allows body to modify the objects captured by copy, and to call their non-const member functions

Unless the keyword mutable was used in the lambda-expression, the function-call operator is const-qualified and the objects that were captured by copy are non-modifiable from inside this operator().

PS:或者更好地使 TestGetStr::getString() const 成员函数。