我试图用C语言做一个用户输入的程序：＆＃34;一百九十九＆＃34;对于三位数字（直到一百），输出将为199 但我只能将其转换为1-9，而且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;
}
答案 0 :(得分：0)
以下函数分析下一个标记，将其转换为int并跳过它。通过反复调用该函数，您可以解析任何数字。它是荷兰语，但意思应该是明确的：
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：＆＃34; driehonderdeenendertigduizendzeshondertweeentwintig＆＃34;
或者：＆＃34;六个人，两个人，两个人，二十个人。 （需要用英语管理空格：在较小的数字后面的任何较大的数字意味着乘以，似乎？）
答案 1 :(得分：0)
解决此问题的一种方法是，将输入拆分为标记，并按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);
}
}
这是一般性的想法，您可以进一步改进，并根据您的输入，您需要处理所有角落案件。