如何在 C 中解析美元和美分

Question

假设我有一个 3.50（字符串）的输入，我该如何解析它以便将它存储为 3 美元和 50 美分。美元和美分都是整数，不允许使用atoi。

我有这个想法，但显然它在 C 中不起作用（假设令牌是 3.50）：

dollars = int(token); /* dollars is 3 */
cents = atoi(token) - dollars; /* atoi is not allowed but I can't think of anything else */

谢谢！

Answer 1

您是否想将自己的string 滚动到int 解析器...

int parse_int(char **s) {
  char sign = **s;
  if (sign == '-' || sign == '+') {
    (*s)++;
  }
  int result = 0;
  unsigned char ch;
  while (isdigit(ch = **s)) {
    (*s)++;
     // By accumulating the int on the negative side, we can parse INT_MIN
    result = result * 10 + '0' - ch;
  }

  if (sign != '-') result = -result;
  return result;
}


void parse_dollars_cents(char **s, int *dollar, int *cent) {
  *dollar = parse_int(s);
  if (**s == '.') {
    (*s)++;
    *cent = parse_int(s);
  } else {
    *cent = 0;
  }
}


int main(void) {
  char buf[100];
  fgets(buf, sizeof buf, stdin);
  int dollar, cent;
  char *s = buf;
  parse_dollars_cents(&s, &dollar, &cent);
  printf("$%d.%02d\n", dollar, cent);
  return 0;
}

Answer 2

你可以使用sscanf

sscanf(stringWhereDollarCentValueIsStored , "%d.%d" , &dollars , &cents);

Answer 3

您需要修改这些以优雅地处理无效输入，但这些函数将适用于有效输入：

int get_dollars(char *str)
{
    int i = 0;
    int accum = 0;

    // While the current character is neither the null-terminator nor a '.' character
    while(str[i]&&str[i]!='.')
    {
        // Shift any previously accumulated values up one place-value by multiplying it by 10
        accum *= 10;
        // Add the numeric value of the character representation of the digit in str[i] to accum
        // (('0'-'9')-0x30=integer value 0-9)
        accum+=str[i]-0x30;
        // Increment the loop counter
        i++;
    }
    // Return the result
    return accum;
}

int get_cents(char *str)
{
    int i = 0; 
    int accum = 0;
    int start_pos = 0;
    // Get the location of the '.' character so we know where the cent portion begins at
    while(str[i]&&str[i]!='.') {i++; start_pos++; }
    i = start_pos+1;
    // Same algorithm as get_dollars, except starting after the '.' rather than before
    while(str[i])
    {
        accum *= 10;
        accum += str[i]-0x30;
        i++;
    }
    return accum;
}

Answer 4

假设您所追求的美分长度仅为小数点后 2 位且令牌为 double，则从令牌中减去美元并乘以 100。

dollars=int(token);
token-=dollars;
cents=int(token*100.);

编辑：根据评论，token 是char *：

int dollars, cents;
sscanf(token,"%d.%d",&dollars,&cents);

编辑 2：从评论开始，没有 itoa() 和 sscanf()：

按照here 的描述实现以下功能。然后使用它：

double amount=parseFloat(token);
dollars=int(amount);
cents=int((amount-dollars)*100.);