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识别字符串中的C 11关键字(string.at()问题)

[英]Identifying C 11 keywords in Strings (string.at() issues)

我已经成功地在教授提供的主要功能的代码片段中搜索了c ++ 11关键字。 尽管当打印的关键字包含在其他单词中时,将在打印这些关键字时进行打印,例如,当字符串片段中存在甜甜圈时,将打印do。 因此,我创建了两个if语句,这些语句标识字符串中的下一个char以及是否为空格字符,如果是则继续打印它。 但是我在x.at()语句中遇到了一个问题,该问题可以编译但无法运行。 有人知道我在做什么错吗?

// C++11 Keyword Search Program

#include <iostream>
using std::cout;
using std::endl;

#include <string>
using std::string;

#include <vector>
using std::vector;
typedef vector<string> string_vector;

// return the C++11 keywords that appear in the code snippet
string_vector find_keywords(const string snippet)
{
  const string x=snippet;
  int len=x.length();
  int back=x.back();
  int last_char, first_char;
  vector<string> answer;
  string keywords[]= {"alignas", "alignof", "and", "and_eq", "asm", "auto", "bitand", 
      "bitor", "bool", "break", "case", "catch", "char", "char16_t", "char32_t", "class", "compl", "const", "constexpr",
      "const_cast", "continue", "decltype", "default", "delete", "do", "double", "dynamic_cast", "else", "enum", 
      "explicit", "export", "extern", "false", "float", "for","friend", "goto", "if", "inline", "int", "long", "mutable",
      "namespace", "new", "noexcept", "not", "not_eq", "nullptr", "operator", "or", "or_eq", "private", "protected public", 
      "register", "reinterpret_cast","return","short","signed","sizeof","static","static_assert","static_cast","struct",
      "switch", "template", "this","thread_local","throw","true","try","typedef","typeid","typename","union","unsigned",
      "using","virtual","void", "volatile","wchar_t","while","xor","xor_eq"};



     for(int i=0;i<83;++i)
  {
    string test=keywords[i];
    size_t found = x.find(test);
    unsigned int testsize=test.size();
    int test1=0,test2=0;

    if(found!=string::npos)
    {
        char a=x.at(found-1);
        char b=x.at(found+testsize);
        int c=isalpha(a);
        int d=isalpha(b);
        if (c>=1)
            test1=1;
        if (d>=1)
            test2=1;
    }   
    if(found!=string::npos && test1==0 &&test2==0)
        cout<<test<<" ";        
  }
return answer;
}

int main (int argc, char* const argv[])
{
  const string snips[] =
  {
    "int i = 0;",
    "i += 1;",
    "return (double) x / y;",
    "char foo(char bar);",
    "double bar(void) const",
    "garbage = fuzz + lint;"
  };

  const string_vector snippets(snips,snips + 6);

  for (auto snippet : snippets)
  {
    cout << "**Snippet**  " << snippet << endl << "**Keywords** ";
    for (auto keyword : find_keywords(snippet))
    {
      cout << ' ' << keyword;
    }
    cout << endl << endl;
  }

  return EXIT_SUCCESS;
}
 last_char=found+len-1; char2=x.at(last_char); 

一旦found > 0 ,则last_char >= x.length()并且您的at()调用尝试越界读取。

还要注意,空格不是唯一可能的令牌定界符。 一个人通常会写诸如break;东西break; ,或return; while(condition)f(int[])this->member 所有这些都包含一个不被空格包围的关键字。

并且,看起来像关键字的字符串可能会出现在字符串文字中"like this"

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