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c++ - 识别字符串中的 C 11 关键字(string.at() 问题)

转载 作者:行者123 更新时间:2023-11-28 03:07:56 24 4
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在教授提供的代码片段中,我在main函数中搜索c++11关键字成功。尽管在打印关键字时,如果它们包含在其他单词中,则打印关键字,例如当字符串片段存在 donut 时打印 do 。所以我创建了两个 if 语句来识别字符串中的下一个字符以及它是否是空格字符,如果是则继续打印它。但是我在编译但不运行的 x.at() 语句中遇到问题。有人知道我做错了什么吗?

// 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;
}

最佳答案

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

一旦 found > 0last_char >= x.length() 和您的 at() 调用就会尝试读出的界限。

另请注意,空格不是唯一可能的标记分隔符。一个人经常写这样的东西 break;,或 return;,或 while(condition),或 f(int[]),或 this->member。所有这些都包含一个没有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”

关于c++ - 识别字符串中的 C 11 关键字(string.at() 问题),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19234180/

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