c - 如何在 c 中创建灵活的占位符表达式 (%d %s %d)？

Question

该程序可让您输入数学问题，例如:

Bob has 1 apple and Mike 2. What is the sum?

程序然后理解单词“sum”并将两个数字相加。至少那是我想要实现的目标。它不起作用，因为我得到了非常大的数字。我怀疑表达式 %d %s %d 不够灵活，只有在问题是:

1 sum 2

顺序似乎很重要。那么我该如何让它变得灵活，以便顺序无关紧要呢？

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

int main() {
   char question[100];
   char buffer[100];
   int result;
   int n1, n2;
   int operation;

   printf ("Your mathematical question: ");

   fgets(question, sizeof(question), stdin);
   fflush(stdin);

   if(strstr(question, "sum") || strstr(question, "add")){
        operation = 0;
   }

   sscanf(question, "%d %s %d", &n1, buffer, &n2);

   printf ("%d %d \n", n1, n2);

    switch(operation) {
        case 0: result = n1 + n2;
        break;
    }

   printf ("%d", result);
   return(0);
}

Answer 1

尝试使用 strtok 将您的输入字符串拆分为运算符和操作数列表。问题在于，在表达数学短语时，英语等自然语言没有运算符和操作数的正式顺序，而您的程序需要这样。

将输入拆分为运算符和操作数可让您按所需顺序匹配它们。

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

typedef enum
{
    NO_OP,
    ADDITION,
    /* Add new operators here. */
}   operator_t;

/* One operator plus two operands. */
#define NUMBER_OF_ITEMS 3

int main()
{
    char question[100];
    char buffer[100];
    char *token;

    int result = 0;
    /* Extend this to allow more complex sentences with multiple operations. */
    int operands[2];
    int* nextOperand = &operands[0];
    int itemCount = 0;
    /* Turn this into an array to extend the parser. */
    operator_t operator = NO_OP;

    printf ("Your mathematical question: \r\n");

    fgets(question, sizeof(question), stdin);

    /* Tokens are seperated by SPACES in this example. */
    const char delims[] = " ";

    /* Get the first token. */
    token = strtok(question, delims);

    /* Walk through all tokens in the question string. */
    while((token != NULL)&&(NUMBER_OF_ITEMS > itemCount)) 
    {
        /* use strcmp to avoid matches on words like 'summertime'. */
        if((0 == strcmp(token, "sum")) || (0 == strcmp(token, "add")))
        {
            operator = ADDITION;
            itemCount++;
        }
        else
        {
            /* Check that one integer can be parsed from the token. */
            if(1 == sscanf(token, "%d", nextOperand))
            {
                nextOperand++;
                itemCount++;
            }
        }

        /* Find the next token. */
        token = strtok(NULL, delims);
    }

    /* Perform the operation, if possible. */
    if((NUMBER_OF_ITEMS == itemCount) && (NO_OP != operator))
    {
        switch(operator)
        {
            case ADDITION:
                result = operands[0] + operands[1];
                printf("The answer to your question (%d + %d) is %d.\r\n",
                    operands[0], operands[1], result);
                break;
            default:
                printf("Oops, please ask my programmer to do some debugging.");
                break;
        }

    }
    else
    {
        printf ("I did not understand your question.\r\n");    
    }
}

请注意，您需要添加更多的错误检查以获得良好的稳健性，但上面的代码应该有助于说明如何解决问题。

然后您可以使用类似 Reverse Polish Notation 的东西扩展解析器的功能。