why i can't pass a char pointer to lambda. C Primer ex13.44-CodePudding

I am writing a simplified version of std::string. When i'm writing the free function, i use the for_each function like this:

void String::free()
{
    std::for_each(element, end, [this](char *c){ alloc.destroy(c); });
    alloc.deallocate(element, end-element);
}

This function will destory the char memory, and delete the memory space allocated by allocator. But it will be error when i compile it.

In file included from /usr/include/c  /9/algorithm:62,
                 from 13_44_String.cpp:2:
/usr/include/c  /9/bits/stl_algo.h: In instantiation of ‘_Funct std::for_each(_IIter, _IIter, _Funct) [with _IIter = char*; _Funct = String::free()::<lambda(char*)>]’:
13_44_String.cpp:20:69:   required from here
/usr/include/c  /9/bits/stl_algo.h:3876:5: error: no match for call to ‘(String::free()::<lambda(char*)>) (char&)’
 3876 |  __f(*__first);
      |  ~~~^~~~~~~~~~
13_44_String.cpp:20:33: note: candidate: ‘String::free()::<lambda(char*)>’ <near match>
   20 |     std::for_each(element, end, [this](char *c){ alloc.destroy(c); });
      |                                 ^
13_44_String.cpp:20:33: note:   conversion of argument 1 would be ill-formed:
In file included from /usr/include/c  /9/algorithm:62,
                 from 13_44_String.cpp:2:
/usr/include/c  /9/bits/stl_algo.h:3876:5: error: invalid conversion from ‘char’ to ‘char*’ [-fpermissive]
 3876 |  __f(*__first);
      |  ~~~^~~~~~~~~~
      |     |
      |     char

The right answer is change char * to char & like this:

void String::free()
{
    std::for_each(element, end, [this](char &c){ alloc.destroy(&c); });
    alloc.deallocate(element, end-element);
}

I do not know why i can't pass a char pointer to a lambda. Why i must use the &.

CodePudding user response：

From std::for_each documentation:

template< class InputIt, class UnaryFunction >
UnaryFunction for_each( InputIt first, InputIt last, UnaryFunction f );

Applies the given function object f to the result of dereferencing every iterator in the range [first, last), in order.

Note the emphasis on result of dereferencing in the above quote. Now lets apply this quote to your 1st code snippet. In your case:

f is equivalent to the lambda that you supplied

result of dereferencing the iterator is char

So the parameter should be a char type. But you're supplying/specifying a char* so you get the mentioned error.

Now in your 2nd code snippet, you fixed this by specifying the parameter type in the lambda to be char& and so the code works.