Question

这是检查给定字符串的代码是identifier还是keyword。这是代码：

#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<ctype.h>


int main(){

    int i = 0, flag = 0;
    char a[10][10] = {"int", "float", "break", "long", "char", "for", "if", "switch", "else", "while"}, string[10];

    //clrscr();

    printf("Enter a string :");
    gets(string);

    /*----Checking whether the string is in array a[][]----*/

    for(i = 0 ; i < 10; i++){
        if( (strcmp( a[i], string) == 0) )
            flag = 1;
    }

    /*----If it is in the array then it is a keyword----*/

    if( flag == 1)
        printf("\n%s is a keyword ", string);

    /*----Otherwise check whether the string is an identifier----*/
    else{
        flag = 0;
        /*----Checking the 1st character*----*/

        if( (string[0] == '_') || ( isalpha(string[0]) != 0 ) ){
            /*---Checking rest of the characters*---*/
            for(i = 1; string[i] != '\0'; i++)
            if( (isalnum(string[i]) == 0 ) && (string[i]!='_') )
                flag = 1;
        }
        else
            flag = 1;
        if( flag == 0)
            printf("\n%s is an identifier ", string);
        else
            printf("\n%s is neither a keyword nor an identifier ", string);
    }
        getch();
}

我想更轻松地执行此代码。是否有可能得到或确定所有关键字都没有在char中声明？怎么做？

S.O可以为我提供该代码吗？

Answer 1

这是一个简单的方法：

static const std::string keywords[] =
{
  "char", "class", 
  "struct",
  /* ... */
};
static const size_t keyword_quantity =
  sizeof(keywords) / sizeof(keywords[0]);

std::string search_word;
cin >> search_word;

std::string const * const iterator = 
    std::find(&keywords[0], &keywords[keyword_quantity],
              search_word);
if (iterator != &keywords[keyword_quantity])
{
  cout << "Word is a keyword!\n";
}

std::string数据类型使文本或字符串处理更容易 std::find功能很简单，因此您不必编写它（并且已经过测试）。

给定的字符串是c ++中的有效标识符或关键字

1 个答案: