atof exercise（K＆R第4.2節，練習4.2）

Question

K & R ex。 4.2要求您修改缺少指數處理機制來處理指數（如123e6或456e-7）的給定（非標準）atof函數。我添加了一個小改動來處理正確輸入的無空格的單位指數。爲了檢查它是否工作，我添加了樣本輸入數據和一個printf函數給main。返回值是完全關閉的（有些是零，沒有符號或小數，沒有明顯的關係）。有人可以幫我改進嗎？代碼：

#include <ctype.h> 

double antof(char[]); /* name changed to protect the innocent 
         and avoid references to stdlib functions */ 

int main() 
{ 
    char putin1[] = "12345"; 
    char putin2[] = "987.65"; 
    char putin3[] = " -2468"; 
    char putin4[] = "12e2"; 
    char putin5[] = " -34E-3"; 
    printf ("%s \t %s \t %s \t %s \t %s \n\n", putin1, putin2, putin3, putin4, putin5); 

    double converted1 = antof(putin1); 
    double converted2 = antof(putin2); 
    double converted3 = antof(putin3); 
    double converted4 = antof(putin4); 
    double converted5 = antof(putin5); 
    printf ("%d \t %d \t %d \t %d \t %d", converted1, converted2, converted3, converted4, converted5); 

    return 0; 
} 

/* atof: convert string s to double */ 
double antof(char s[]) 

{ 
    double val, power; 
    int i, sign; 

    for (i = 0; isspace(s[i]); i++) /* skip white space */ 
     ; 
    sign = (s[i] == '-') ? -1 : 1; 

    if (s[i] == '+' || s[i] == '-') 
     i++; 

    for (val = 0.0; isdigit(s[i]); i++) 
     val = 10.0 * val + (s[i] - '0'); 

    if (s[i] == '.') 
     i++; 

    for (power = 1.0; isdigit(s[i]); i++) { /*tracks right side of decimal, keeps adding to val */ 
     val = 10.0 * val + (s[i] - '0');  /* but keeps multiplying power by 10 to keep track of decimal place */ 
     power *= 10; 
    } 

    /* added from here to handle scientific notation */ 
    int exponenty; 
    int exponentysign; 

    if (s[i] == "e" || s[i] == "E") 
     i++; 

    if (s[i] == '-') 
     exponentysign == -1; 
     i++; 

    exponenty = (s[i] - '0'); 
     /* full functionality would require storing data like val and power 
     but here I assume single digit exponent as given in the example */ 

    if (exponentysign == -1) 
     exponenty = (1/exponenty); 

    return (sign * val/power) * (10^exponenty); 
} 

一如既往的感謝。

Answer 1

修改的功能：antof

double antof(char s[]) 
{ 
    double val = 0.0, power = 1; 
    int i, sign; 

    /* skip white space */ 
    for (i = 0; isspace(s[i]); i++); 

    sign = (s[i] == '-') ? -1 : 1; 
    if (s[i] == '+' || s[i] == '-') i++; 

    for (val = 0.0; isdigit(s[i]); i++) 
     val = 10.0 * val + (s[i] - '0'); 

    if (s[i] == '.') 
    { 
     i++; 

     for (power = 10.0; isdigit(s[i]); i++) { /*tracks right side of decimal, keeps adding to val */ 
      val = 10.0 * val + (s[i] - '0');  /* but keeps multiplying power by 10 to keep track of decimal place */ 
      power *= 10; 
     } 
    } 

    /* added from here to handle scientific notation */ 
    double exponenty = 0; 
    int exponentysign = 1; 

    if (s[i] == 'e' || s[i] == 'E') 
    { 
     i++; 

     if (s[i] == '-') 
     { 
      exponentysign = -1; 
      i++; 
     } 

     exponenty = (s[i] - '0'); 
     /* full functionality would require storing data like val and power 
     but here I assume single digit exponent as given in the example */ 

     exponenty *= exponentysign; 
    } 

    return (sign * val/power) * pow(10.0, exponenty); 
} 

而且，你要注意的是^執行bitwise xor，而不是power。您必須使用math.h中的pow（或者如果您不想使用pow，則必須執行重複的乘法或除法）。