std :: less中的分段錯誤<char>

Question

我有以下代碼（C ++ 0x）：

const set<char> s_special_characters =  { '(', ')', '{', '}', ':' };

void nectar_loader::tokenize( string &line, const set<char> &special_characters )
{
    auto it = line.begin();
    const auto not_found = special_characters.end();

    // first character special case
    if( it != line.end() && special_characters.find( *it ) != not_found )
        it = line.insert( it+1, ' ' ) + 1;

    while( it != line.end() )
    {
        // check if we're dealing with a special character
        if( special_characters.find(*it) != not_found ) // <----------
        {
            // ensure a space before
            if( *(it-1) != ' ' )
                it = line.insert( it, ' ' ) + 1;
            // ensure a space after
            if( (it+1) != line.end() && *(it+1) != ' ' )
                it = line.insert( it+1, ' ');
            else
                line.append(" ");
        }
        ++it;
    }
}

崩潰指向指示的線。 這導致與此gdb回溯的段錯誤：

#0  0x000000000040f043 in std::less<char>::operator() (this=0x622a40, __x=@0x623610, __y=@0x644000)
    at /usr/lib/gcc/x86_64-unknown-linux-gnu/4.5.2/../../../../include/c++/4.5.2/bits/stl_function.h:230
#1  0x000000000040efa6 in std::_Rb_tree<char, char, std::_Identity<char>, std::less<char>, std::allocator<char> >::_M_lower_bound (this=0x622a40, __x=0x6235f0, __y=0x622a48, __k=@0x644000)
    at /usr/lib/gcc/x86_64-unknown-linux-gnu/4.5.2/../../../../include/c++/4.5.2/bits/stl_tree.h:1020
#2  0x000000000040e840 in std::_Rb_tree<char, char, std::_Identity<char>, std::less<char>, std::allocator<char> >::find (this=0x622a40, __k=@0x644000)
    at /usr/lib/gcc/x86_64-unknown-linux-gnu/4.5.2/../../../../include/c++/4.5.2/bits/stl_tree.h:1532
#3  0x000000000040e4fd in std::set<char, std::less<char>, std::allocator<char> >::find (this=0x622a40, __x=@0x644000)
    at /usr/lib/gcc/x86_64-unknown-linux-gnu/4.5.2/../../../../include/c++/4.5.2/bits/stl_set.h:589
#4  0x000000000040de51 in ambrosia::nectar_loader::tokenize (this=0x7fffffffe3b0, line=..., special_characters=...)
    at ../../ambrosia/Library/Source/Ambrosia/nectar_loader.cpp:146
#5  0x000000000040dbf5 in ambrosia::nectar_loader::fetch_line (this=0x7fffffffe3b0)
    at ../../ambrosia/Library/Source/Ambrosia/nectar_loader.cpp:112
#6  0x000000000040dd11 in ambrosia::nectar_loader::fetch_token (this=0x7fffffffe3b0, token=...)
    at ../../ambrosia/Library/Source/Ambrosia/nectar_loader.cpp:121
#7  0x000000000040d9c4 in ambrosia::nectar_loader::next_token (this=0x7fffffffe3b0)
    at ../../ambrosia/Library/Source/Ambrosia/nectar_loader.cpp:72
#8  0x000000000040e472 in ambrosia::nectar_loader::extract_nectar<std::back_insert_iterator<std::vector<ambrosia::target> > > (this=0x7fffffffe3b0, it=...)
    at ../../ambrosia/Library/Source/Ambrosia/nectar_loader.cpp:43
#9  0x000000000040d46d in ambrosia::drink_nectar<std::back_insert_iterator<std::vector<ambrosia::target> > > (filename=..., it=...)
    at ../../ambrosia/Library/Source/Ambrosia/nectar.cpp:75
#10 0x00000000004072ae in ambrosia::reader::event (this=0x623770)

我很茫然，不知道我做錯了什么。 任何幫助深表感謝。

編輯：崩潰時的字符串是

sub Ambrosia：lib libAmbrosia

更新：

我根據評論/答案中的建議替換了上述功能。 結果如下。

const string tokenize( const string &line, const set<char> &special_characters )
{
    const auto not_found = special_characters.end();
    const auto end = line.end();
    string result;

    if( !line.empty() )
    {
        // copy first character
        result += line[0];

        char previous = line[0];
        for( auto it = line.begin()+1; it != end; ++it )
        {
            const char current = *it;

            if( special_characters.find(previous) != not_found )
                result += ' ';

            result += current;
            previous = current;
        }
    }
    return result;
}

Answer 1

另一個猜測是line.append(" ")有時會使it無效，具體取決於線路的原始容量。

Answer 2

在第一次取消引用it之前，不要檢查it != line.end() 。

Answer 3

我無法發現錯誤，我建議您使用調試器慢慢迭代，因為您已經確定了問題。

我只是這樣，修改你正在迭代的內容非常容易失敗。

我建議使用Boost Tokenizer ，更准確地說： boost::token_iterator與boost::char_separator結合使用（包括代碼示例）。

然后，您可以簡單地從第一個構建一個新string ，並從該函數返回新字符串。 計算速度應該包括內存分配。