C++使用unique_ptr的字符數組

Question

首先，我知道這不是最好的方式來做到這一點，我只是看它應該如何完成。我創建了一個名爲BORD類，其中包含一個成員

 std::unique_ptr<std::unique_ptr<char>[] > char_bord; 

這應該是正確的語法，然後我嘗試在構造函數初始化此：

bord::bord():char_bord(new std::unique_ptr<char>[10]) 
{ 
    //char_bord=new std::unique_ptr<char>[10]; //This did not seem to work aswell. 
    for(int i=0;i<10;i++) 
     char_bord[i]=new std::unique_ptr<char>[](new char[10]); 
    return; 
} 

這將導致錯誤的下面堆，我沒有設法破譯。

jelmer@jelmer-N56JN:~/Git/Board/lib$ g++ -std=c++0x bord.c 
In file included from bord.c:1:0: 
bord.h:20:1: error: new types may not be defined in a return type 
class bord 
^ 
bord.h:20:1: note: (perhaps a semicolon is missing after the definition of ‘bord’) 
bord.c:3:12: error: return type specification for constructor invalid 
bord::bord():char_bord(new std::unique_ptr<char>[10]) 
      ^
bord.c: In constructor ‘bord::bord()’: 
bord.c:7:46: error: expected primary-expression before ‘]’ token 
     char_bord[i]=new std::unique_ptr<char>[](new char[10]); 
              ^
bord.c:7:60: error: parenthesized initializer in array new [-fpermissive] 
     char_bord[i]=new std::unique_ptr<char>[](new char[10]); 
                  ^
bord.c:7:19: error: no match for ‘operator=’ (operand types are ‘std::unique_ptr<char>’ and ‘std::unique_ptr<char>*’) 
     char_bord[i]=new std::unique_ptr<char>[](new char[10]); 
       ^
bord.c:7:19: note: candidates are: 
In file included from /usr/include/c++/4.9/memory:81:0, 
       from bord.h:19, 
       from bord.c:1: 
/usr/include/c++/4.9/bits/unique_ptr.h:249:7: note: std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(std::unique_ptr<_Tp, _Dp>&&) [with _Tp = char; _Dp = std::default_delete<char>] 
     operator=(unique_ptr&& __u) noexcept 
    ^
/usr/include/c++/4.9/bits/unique_ptr.h:249:7: note: no known conversion for argument 1 from ‘std::unique_ptr<char>*’ to ‘std::unique_ptr<char>&&’ 
/usr/include/c++/4.9/bits/unique_ptr.h:269:2: note: template<class _Up, class _Ep> typename std::enable_if<std::__and_<std::is_convertible<typename std::unique_ptr<_Up, _Ep>::pointer, typename std::unique_ptr<_Tp, _Dp>::_Pointer::type>, std::__not_<std::is_array<_Up> > >::value, std::unique_ptr<_Tp, _Dp>&>::type std::unique_ptr<_Tp, _Dp>::operator=(std::unique_ptr<_Up, _Ep>&&) [with _Up = _Up; _Ep = _Ep; _Tp = char; _Dp = std::default_delete<char>] 
    operator=(unique_ptr<_Up, _Ep>&& __u) noexcept 
^
/usr/include/c++/4.9/bits/unique_ptr.h:269:2: note: template argument deduction/substitution failed: 
bord.c:7:19: note: mismatched types ‘std::unique_ptr<_Tp, _Dp>’ and ‘std::unique_ptr<char>*’ 
     char_bord[i]=new std::unique_ptr<char>[](new char[10]); 
       ^
In file included from /usr/include/c++/4.9/memory:81:0, 
       from bord.h:19, 
       from bord.c:1: 
/usr/include/c++/4.9/bits/unique_ptr.h:278:7: note: std::unique_ptr<_Tp, _Dp>& std::unique_ptr<_Tp, _Dp>::operator=(std::nullptr_t) [with _Tp = char; _Dp = std::default_delete<char>; std::nullptr_t = std::nullptr_t] 
     operator=(nullptr_t) noexcept 
    ^
/usr/include/c++/4.9/bits/unique_ptr.h:278:7: note: no known conversion for argument 1 from ‘std::unique_ptr<char>*’ to ‘std::nullptr_t’ 

我做錯了什麼，假設我做錯了什麼。

Answer 1

下面是一些代碼，演示了什麼，我想你想：

#include <iostream> 
#include <algorithm> 
#include <iterator> 
#include <memory> 

using namespace std; 

using array_ptr_type = std::unique_ptr<char[]>; 
using array_of_arrays_type = std::unique_ptr<array_ptr_type[]>; 

auto main() -> int 
{ 
    auto x = array_ptr_type(new char[10]); 
    auto y = array_ptr_type(new char[10]); 
    for (int i = 0 ; i < 10 ; ++i) 
    { 
     x[i] = char('0' + i); 
     y[i] = char('0' + 9 - i); 
    } 

    auto pxy = array_of_arrays_type(new array_ptr_type[2]); 
    pxy[0] = move(x); 
    pxy[1] = move(y); 

    for (int i = 0 ; i < 2 ; ++i) { 
     copy(&pxy[i][0], &pxy[i][10], ostream_iterator<char>(cout, ", ")); 
     cout << endl; 
    } 
    return 0; 
} 

預期輸出：

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 

當然，如你所知，這一切都不建議 - vector<vector<char>>會很多更清潔，更易於維護。

Answer 2

你必須使用專門：

std::unique_ptr<char[]> chars(new char[1024]); 

這是因爲std::unique_ptr不支持定製刪除作爲std::shared_ptr確實（在此寫作風格）。

std::unique_ptr使用std::default_delete作爲刪除者。不久，如果您指定參數類型爲class T它將使用默認delete，但如果您編寫class T[]（在此專業化）std::unique_ptr將使用delete[]。

但是最好使用一些容器而不是c風格的數組。