需要幫助理解為什么這個無效的C 代碼可以成功編譯

以下代碼在某些環境中為我正確編譯（例如，在帶有 GCC 9.3.0 的編譯器資源管理器上）并在其他環境中抱怨（CentOS 7.9.2009 (Core)、GCC 9.3.1）。

#include <iostream>
#include <string>
using namespace std;

int main() {
    std::string name="aakash";
    name.erase(name.begin() 2, name.cend());
    return 0;
}

當我收到錯誤時，錯誤是：

test.cpp:6:40: error: no matching function for call to 'std::basic_string<char>::erase(__gnu_cxx::__normal_iterator<char*, std::basic_string<char> >, std::basic_string<char>::const_iterator)'
    6 |  name.erase(name.begin() 2, name.cend());
      |                                        ^
In file included from /opt/rh/devtoolset-9/root/usr/include/c  /9/string:55,
                 from /opt/rh/devtoolset-9/root/usr/include/c  /9/bits/locale_classes.h:40,
                 from /opt/rh/devtoolset-9/root/usr/include/c  /9/bits/ios_base.h:41,
                 from /opt/rh/devtoolset-9/root/usr/include/c  /9/ios:42,
                 from /opt/rh/devtoolset-9/root/usr/include/c  /9/ostream:38,
                 from /opt/rh/devtoolset-9/root/usr/include/c  /9/iostream:39,
                 from test.cpp:1:
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4698:7: note: candidate: 'std::basic_string<_CharT, _Traits, _Alloc>& std::basic_string<_CharT, _Traits, _Alloc>::erase(std::basic_string<_CharT, _Traits, _Alloc>::size_type, std::basic_string<_CharT, _Traits, _Alloc>::size_type) [with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>; std::basic_string<_CharT, _Traits, _Alloc>::size_type = long unsigned int]'
 4698 |       erase(size_type __pos = 0, size_type __n = npos)
      |       ^~~~~
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4698:23: note:   no known conversion for argument 1 from '__gnu_cxx::__normal_iterator<char*, std::basic_string<char> >' to 'std::basic_string<char>::size_type' {aka 'long unsigned int'}
 4698 |       erase(size_type __pos = 0, size_type __n = npos)
      |             ~~~~~~~~~~^~~~~~~~~
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4714:7: note: candidate: 'std::basic_string<_CharT, _Traits, _Alloc>::iterator std::basic_string<_CharT, _Traits, _Alloc>::erase(std::basic_string<_CharT, _Traits, _Alloc>::iterator) [with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>; std::basic_string<_CharT, _Traits, _Alloc>::iterator = __gnu_cxx::__normal_iterator<char*, std::basic_string<char> >; typename _Alloc::rebind<_CharT>::other::pointer = char*]'
 4714 |       erase(iterator __position)
      |       ^~~~~
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4714:7: note:   candidate expects 1 argument, 2 provided
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4734:7: note: candidate: 'std::basic_string<_CharT, _Traits, _Alloc>::iterator std::basic_string<_CharT, _Traits, _Alloc>::erase(std::basic_string<_CharT, _Traits, _Alloc>::iterator, std::basic_string<_CharT, _Traits, _Alloc>::iterator) [with _CharT = char; _Traits = std::char_traits<char>; _Alloc = std::allocator<char>; std::basic_string<_CharT, _Traits, _Alloc>::iterator = __gnu_cxx::__normal_iterator<char*, std::basic_string<char> >; typename _Alloc::rebind<_CharT>::other::pointer = char*]'
 4734 |       erase(iterator __first, iterator __last);
      |       ^~~~~
/opt/rh/devtoolset-9/root/usr/include/c  /9/bits/basic_string.h:4734:40: note:   no known conversion for argument 2 from '__normal_iterator<const char*,[...]>' to '__normal_iterator<char*,[...]>'
 4734 |       erase(iterator __first, iterator __last);
      |                               ~~~~~~~~~^~~~~~
      |                               ~~~~~~~~~^~~~~~

這個錯誤對我來說看起來很合理，因為根據C 檔案，兩個引數都std::basic_string::erase應該是 typeiterator或const_iterator.

那么，當它作業時（例如這里），是什么讓它作業？

uj5u.com熱心網友回復：

從 C 11 開始，雙迭代器多載的兩個std::string::erase()引數都是const_iterator. 請參閱此 cppreference 頁面上的表格#3 （您在問題中鏈接的那個）。

此外，C 11 標準要求容器的iterator型別可以轉換為等價的const_iterator.

從這個 C 11 標準草案中，在§23.2.1 [container.requirements.general]中，表 96具有以下條目（粗體強調我的）：

因此，在您對 的呼叫中.erase()，第一個引數被轉換為string::const_iterator.

因此，假設您已將它們設定為使用 C 11（或更高版本）標準，則錯誤出現在那些不接受您的呼叫的編譯器中。

標籤：C c 11

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