要转换的字符串 [英] String to float conversion

问题描述

你好，



你能告诉我如何在不使用标准功能的情况下将字符串转换为浮点数，例如 atoi ， atof ， strtod ， sscanf 等

我的代码是：

Hello,

Could you please let me know how to convert string to float without using standard functions, like atoi, atof, strtod, sscanf etc.
My code is:

main()
{
    char inputstr[]="123.45";
    float result= 0.0F;
    float temp;
    int ndecimaldigit,i;

    ndecimaldigit = 1;
    for(i=0;i<=4;i++)
    {
        temp=inputstr[i] - '0';
        ndecimaldigit *= 10;
        temp /= ndecimaldigit;
        result += temp;
    }

    printf("%f %f %d \n",result,temp,ndecimaldigit);
}

但我得到的输出为：0.122671 -0.000129 -31072

这是不正确。我需要输出为123.45

But i am getting output as: 0.122671 -0.000129 -31072
which is not correct. I need output as 123.45

推荐答案

尝试

Try

int  main()
{
  char inputstr[] = "123.45";
  float result= 0.0f;
  size_t len = sizeof(inputstr)-1;
  size_t dotpos = 0;
  for (size_t n = 0; n < len; n++)
  {
    if (inputstr[n] == '.')
    {
      dotpos = len - n  - 1;
    }
    else
    {
      result = result * 10.0f + (inputstr[n]-'0');
    }
  }
  while ( dotpos--)
  {
    result /= 10.0f;
  }
  printf("%f\n",result);
}

Quote：

你也可以更改/添加下面的注释行，这样你也可以支持负值：

You may also change/add the commented lines below,so that you could support also negative values:

这个想法只是不触及CPallini为正值做的优秀短代码，只是将负值视为正值并乘以最后为-1使其为负。

The idea is simply by not touching the great short code CPallini made for positive values and just treat the negative value as positive and multiplying by -1 at the end to make it negative.

#include <stdio.h>
int  main()
{
    char inputstr[] = "-.45";
    float result= 0.0f;
    int len = sizeof(inputstr)-1;
    int dotpos=0;
    int n;
    if (inputstr[0]=='-'||inputstr[0]=='+')   //Added line to check sign.If the number is signed,
        n=1;                 //set n to position 1.
    else                     //(number is not signed)
        n=0;                 //set n to position 0.
/*If the number was signed,then we set n to 1,so that we start with inputstr[1],and at the end if the number was negative we will multiply by -1.*/
    for (; n < len; ++n)         //n is already set to the position of the fisrt number.
        if (inputstr[n] == '.')
            dotpos = len - n  - 1;
        else
            result = result * 10.0f + (inputstr[n]-'0');
    while (dotpos--)
        result /= 10.0f;
    if (inputstr[0]=='-')  //If inputstr[] is "negative",
        result*=(-1);      //multiply the number by -1,making it also negative.
    printf("%f\n",result);
    return 0;
}

我已将size_t更改为整数，因为我不知道如何使用它们。

I have changed size_t to integer,because I don't know how to use them.

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