将一系列字符转换为它们各自的 int答案

【问题标题】：Convert a sequence of chars into their respective int将一系列字符转换为它们各自的 int
【发布时间】：2023-03-08 15:10:01
【问题描述】：

我正在尝试用 C 语言编写一个程序，用户输入如下：“一百九十九”，对于三位数字（直到一百），输出将是 199。
但我只有这个可以从 1 到 9 转换，我不知道如何实现更大的数字。

#include <stdio.h>
#include <string.h>
 int main() {
     char numbers[10][10] = {"zero", "one" , "two", "three","four",  "five", "six", "seven","eight", "nine"};
        char input[100], word[10], *ptr, *tmp;
        int i, len, value;
        int values[9];
        /* get the number in words from user */
        printf("Enter number in word:\n");
        fgets(input, 100, stdin);
        input[strlen(input) - 1] = '\0';
        tmp = input;
        while (1) {
        /* move pointer to the space to extract word by word */
        ptr = strchr(tmp, ' ');
        if (ptr != NULL) {
            len = ptr - tmp;
            strncpy(word, tmp, len);
            word[len] = '\0';
            tmp = ptr + 1;
            } else {
                    /* last word in the given string */
                    len = strlen(tmp);
                    strncpy(word, tmp, len);
                    word[len] = '\0';
                        }
                     for (i = 0; i < 10; i++) {
                     /* word to number conversion */
                    if (strcasecmp(word, numbers[i]) == 0) {
                        value = (value * 10) + i;
                        break;
                        }
                    }

                    if (!ptr)
                        break;
                     }
        /* print the number */
        printf("Value in number is %d\n", value);
        int iN = 9;
        int arr[iN];
        while (iN--) {
         arr[iN]=value%10;
         value/=10;
        }
        printf("Value: %i", arr[0]);
        return 0;

}

【问题讨论】：

我猜你必须编写一个编译器来解决这个问题，我想你可以用一个 c 程序来做这个，但这需要你一些时间。
您的输入总是三位数还是您也需要支持两位数？您需要支持的最小和最大数字是多少？
@JVMATL Off-by-1。你的评论是关于input[strlen(input) - 0] = '\0';。 OP 正在使用input[strlen(input) - 1] = '\0'; 删除'\n'，它是黑客可利用的，因为input[0] 可能为0。
len = ptr - tmp; strncpy(word, tmp, len); word[len] = '\0'; 是 UB，输入类似于 "abcdefghij"。希望您的用户只输入预期的输入。

标签： c string

【解决方案1】：

以下函数分析下一个标记，将其转换为 int 并跳过它。通过重复调用该函数，您可以解析任何数字。它是荷兰语，但意思应该很清楚：

static int lookupnum(const char *s, const char **s2, long *num)
{
    long t;

    if (strncmp(s, "nul", 3)==0)
    {
        *num= 0; *s2= s+3; return(TRUE);
    }
    else if (t=(strncmp(s, "en",      2)==0?s+=2,0:     // "and"
                strncmp(s, "één",     3)==0?s+=3,1:     // numbers
                strncmp(s, "eén",     3)==0?s+=3,1:
                strncmp(s, "een",     3)==0?s+=3,1:
                strncmp(s, "twee",    4)==0?s+=4,2:
                strncmp(s, "drie",    4)==0?s+=4,3:
                strncmp(s, "vier",    4)==0?s+=4,4:
                strncmp(s, "vijf",    4)==0?s+=4,5:
                strncmp(s, "zes",     3)==0?s+=3,6:
                strncmp(s, "zeven",   5)==0?s+=5,7:
                strncmp(s, "acht",    4)==0?s+=4,8:
                strncmp(s, "negen",   5)==0?s+=5,9:
                strncmp(s, "tien",    4)==0?s+=4,10:
                strncmp(s, "elf",     3)==0?s+=3,11:
                strncmp(s, "twaalf",  6)==0?s+=6,12:
                strncmp(s, "dertien", 7)==0?s+=7,13:
                strncmp(s, "veertien",8)==0?s+=8,14:
                strncmp(s, "twintig", 7)==0?s+=7,20:
                strncmp(s, "dertig",  6)==0?s+=6,30:
                strncmp(s, "veertig", 7)==0?s+=7,40:
                strncmp(s, "vijftig", 7)==0?s+=7,50:
                strncmp(s, "zestig",  6)==0?s+=6,60:
                strncmp(s, "zeventig",8)==0?s+=8,70:
                strncmp(s, "tachtig", 7)==0?s+=7,80:
                strncmp(s, "negentig",8)==0?s+=8,90:
                strncmp(s, "honderd", 7)==0?s+=7,100:
                strncmp(s, "duizend", 7)==0?s+=7,1000:
                strncmp(s, "miljoen", 7)==0?s+=7,1000000:
                strncmp(s, "miljard", 7)==0?s+=7,1000000000:0))
    {
        *num= t;
        *s2 = s;
        return(TRUE);    // return success
    }
    else
        return(FALSE);   // return error
}

示例：“driehonderdeenendertigduizendzeshondertweeentwintig”

或者：“六百二万四百二十”（英语中需要管理空格：任何较大的数字跟在较小的数字后面似乎都意味着乘？）

【讨论】：

【解决方案2】：

解决此问题的一种方法是将您的输入拆分为标记，并以相反的顺序解析每个标记并累积结果。

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

#define MAX_NUM_WORDS   10
#define MAX_INPUT_LEN   128
#define ARRAY_SIZE(a)   (sizeof(a) / sizeof(a[0]))

enum digit_state {
    UNITS,
    TENS,
    HUNDREDS,
    MAX_STATE
};

int get_numerical_value(const char **array, unsigned int length, char *token)
{
    int i;
    int val = -1;

    for (i = 0; i < length; ++i) {
        if (!strcmp(array[i], token)) {
            val = i;
            break;
        }
    }
    return val;
}

unsigned int string2num(char *str)
{
    int i;
    int n;
    int num_words;
    unsigned int result;
    unsigned int state;
    char *token;
    char *words[MAX_NUM_WORDS];
    char input[MAX_INPUT_LEN];
    const char *units[] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
    const char *tens[] = {"zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};

    // save input
    memset(input, 0, MAX_INPUT_LEN);
    strncpy(input, str, strlen(str));

    // get each token, allocate memory and save them in an array
    num_words = 0;
    token = strtok(input, " ");
    while (token) {
        int len = strlen(token) + 1;

        words[num_words] = malloc(len);
        memset(words[num_words], 0, len);

        strncpy(words[num_words], token, len);
        token = strtok(NULL, " ");
        num_words++;
    }

    result = 0;
    state = UNITS;
    n = num_words;

    // process each token in the reverse order
    while (n--) {
        int val;

        if (state == UNITS) {
            val = get_numerical_value(units, ARRAY_SIZE(units), words[n]);
            if (val > 0)
                result += val;
            else
                // units digit missing so ignore this word
                ++n;

            state++;
        } else if (state == TENS) {
            val = get_numerical_value(tens, ARRAY_SIZE(tens), words[n]);
            if (val > 0)
                result += val*10;
            else
                ++n;    // tens digit missing so ignore this word

            state++;
        } else if (state == HUNDREDS) {
            val = 0;

            if (!strcmp(words[n], "hundred"))
                val = get_numerical_value(units, ARRAY_SIZE(units), words[--n]);

            result += val*100;
            state++;
        } else if (state == MAX_STATE) {
            break;
        }
    }

    while (num_words--)
        free(words[num_words]);

    return result;
}


int main(void)
{
    int i;
    char *input[] = {
        "one hundred fifty four",
        "two hundred thirty",
        "eighty one",
        "five",
        "zero",
        "forty",
    };

    for (i = 0; i < ARRAY_SIZE(input); ++i) {
        unsigned int val = string2num(input[i]);
        printf("%s: %u\n", input[i], val);
    }
}

这是一般的想法，您可以进一步改进它，并根据您需要处理所有极端情况的输入。

【讨论】：