atof的实现代码--atoi的补充_atof 补零-优快云博客

本文链接：https://blog.youkuaiyun.com/ministarler/article/details/12303861

本文介绍了如何使用C语言实现atoi和atof函数，将字符串转换为整数和浮点数。通过逐步解析代码，展示了如何处理空格、正负号及小数点等特殊情况。

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//atoi函数如下

#include<stdio.h>

#include<ctype.h>
int atoi(const char* str)
{
int i,sign ;
for(i = 0; isspace(str[i]);i++)
;
sign = (str[i]=='-')? -1:1 ; //三目表达式，不多解释
if(str[i]=='+'||str[i]=='-')
i++;
for(int n = 0 ;isdigit(str[i]);i++)
n = 10*n + (str[i] - '0');
return n*sign ;
}
int main()
{
char * s =" -13.23";
printf("%d \n",atoi(s));
return 0 ;
}

//下面就是对atoi函数小数点以后的数进行处理，以便于能够得到double类型的数据。

double atof(const char s[]) //把一个字符串转换成double型的数据

#include<stdio.h>

#include<ctype.h>

double atof(char s[])

{

double var,power;

int sign = 1;

int i ;

for( i = 0; isspace(s[i]) ;i++)

;

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

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

i++;

for(var = 0.0 ; isdigit(s[i]);i++)

var = 10.0 * var + (s[i] - '0');

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

i++;

for(power = 1.0;isdigit(s[i]);i++)

{

var = 10.0 * var +(s[i] - '0');

power *= 10;

}

return sign * var /power;

}

int main()

{

char *s = "12.23";

printf("%f",atof(s));

return 0 ;

}

//atof中两种格式的转化

//
//  main.c
//  StrToInt
//
//  Created by lichan on 13-10-16.
//  Copyright (c) 2013年 lichan. All rights reserved.
//

#include <stdio.h>
#include <ctype.h>
#include <math.h>

int atoi(const char * str)
{
    int i,sign=1,n;
    
    for (i  = 0; isspace(str[i]); i++)
        ;
    if (str[i]=='+'||str[i]=='-')
    {
        sign = (str[i]=='+')? 1:-1;
        i++;
    }
    for (n = 0; isdigit(str[i]); i++) 
        n = 10*n + (str[i]-'0');
    
    return n*sign;

}


float atof(const char * str)
{
    int i,sign = 1;
    float signP;
    int resultInt=0 ;
    float resultPoint=0.0;
    
    for (i  = 0; isspace(str[i]); i++)
        ;
    if (str[i]=='+'||str[i]=='-')
    {
        sign = (str[i]=='+')? 1:-1;
        i++;
    }
  
    while(isdigit(str[i]))
    {
        resultInt = 10*resultInt + (str[i] -'0');
        i++;
    }
    if(str[i]=='e'||str[i]=='E')
    {
        i++;
        printf("e出现了.\n");

         if (str[i]=='+'||str[i]=='-')
        {
           signP = (str[i]=='+')? 10:0.1;
           i++;
        }
         while(isdigit(str[i]))
        {
           resultPoint= 10*resultPoint + (str[i] -'0');
           printf("%f\n",resultPoint);
           i++;
       }

       return sign*resultInt*pow(signP, resultPoint);
    }
    else if(str[i++]=='.')
    {
        
        for (signP = 0.1; isdigit(str[i]); i++) 
        {
            resultPoint += signP*(str[i] -'0');
            signP *=0.1;
            printf("%f\n",resultPoint);
           
        }
    
        return sign*resultInt+resultPoint;

    }
    return 0 ;

}
int main(int argc, const char * argv[])
{
    char *str ="123e+5";
    char *str1="1234.23";
    
    printf("%f\n",atof(str1));
    return 0;
}