c - 在没有外部函数的情况下复制 atoi()

Question

我希望在不使用任何其他函数(如 strtonum())的情况下重现 atoi() 函数的行为。我真的不明白如何将 char 或 char * 转换为 int 而不是将其转换为 ASCII 值。我在 C 中这样做，有什么建议吗？

Answer 1

哇，我很慢。很多人回答...顺便说一句，这是我的标志检测的简单示例。如果遇到无效字符，它将返回字符串的第一个解析部分:

#include <stdio.h>

int my_atoi(const char* input) {
  int accumulator = 0, // will contain the absolute value of the parsed number
      curr_val = 0, // the current digit parsed 
      sign = 1; // the sign of the returned number
  size_t i = 0; // an index for the iteration over the char array 

  // Let's check if there is a '-' in front of the number,
  // and change the final sign if it is '-'
  if (input[0] == '-') {
    sign = -1;
    i++;
  }
  // A '+' is also valid, but it will not change the value of
  // the sign. It is already +1!
  if (input[0] == '+')
    i++;

  // I think it is fair enough to iterate until we reach 
  // the null char...
  while (input[i] != '\0') {
    // The char variable has already a "numeric"
    // representation, and it is known that '0'-'9'                                 
    // are consecutive. Thus by subtracting the
    // "offset" '0' we are reconstructing a 0-9
    // number that is then casted to int.
    curr_val = (int)(input[i] - '0'); 

    // If atoi finds a char that cannot be converted to a numeric 0-9
    // it returns the value parsed in the first portion of the string.
    // (thanks to Iharob Al Asimi for pointing this out)
    if (curr_val < 0 || curr_val > 9)
      return accumulator;

    // Let's store the last value parsed in the accumulator,
    // after a shift of the accumulator itself.
    accumulator = accumulator * 10 + curr_val;
    i++;
  } 

  return sign * accumulator;
}


int main () {
   char test1[] = "234";
   char test2[] = "+6543";
   char test3[] = "-1234";
   char test4[] = "9B123";

   int a = my_atoi(test1);
   int b = my_atoi(test2);
   int c = my_atoi(test3);
   int d = my_atoi(test4);

   printf("%d, %d, %d, %d\n", a, b, c, d);
   return 0;
}

它打印:

234, 6543, -1234, 9