将字符串转换为C中的双变量

I have written the following code....It should convert a string like "88" to double value 88 and print it

我编写了以下代码....它应该将像“88”这样的字符串转换为double值88并打印出来

void convertType(char* value)
{
   int i = 0;
   char ch;
   double ret = 0;
   while((ch = value[i] )!= '\0')
   {
      ret = ret*10 +(ch - '0');
      ++i;
   }
   printf("%d",ret);//or %f..what is the control string for double?
}



//input string :88

But it always prints 0...But when i change type of ret to int ...it works fine...when the type is float or double,it prints zero...so why am i getting this ambiguous results?

但它总是打印0 ...但是当我将ret的类型更改为int ...它工作正常...当类型为float或double时,它打印为零...所以为什么我得到这个模棱两可的结果?

7 个解决方案

#1

Use sscanf (header stdio.h or cstdio in C++):

使用sscanf(C ++中的头文件stdio.h或cstdio):

char str[] = "12345.56";
double d;

sscanf(str, "%lf", &d);

printf("%lf", d);

#2

But it always prints 0...But when i change type of ret to int ...it works fine...when the type is float or double,it prints zero.

但它总是打印0 ...但是当我将ret的类型更改为int ...它工作正常...当类型为float或double时,它打印为零。

Logic is fine. Just your format specifier is wrong. Change it to %f and all is well!

逻辑很好。只是你的格式说明符是错误的。把它改成%f,一切都很好!

#3

You might be able to use atof() it returns a double.

你可能能够使用atof()它返回一个double。

source

#4

You should use function "atof" if you want to parse a char* to double.

如果要将char *解析为double,则应使用函数“atof”。

You should also use the delimiter "%f" to print the double:

您还应该使用分隔符“%f”来打印双精度:

More information and example of use can be found here.

可以在此处找到更多信息和使用示例。

Example of use:

使用示例:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
   float val;
   char str[20];

   strcpy(str, "98993489");
   val = atof(str);
   printf("String value = %s, Float value = %f\n", str, val);

   strcpy(str, "tutorialspoint.com");
   val = atof(str);
   printf("String value = %s, Float value = %f\n", str, val);

   return(0);
}

To print it you must print it as a float:

要打印它,您必须将其打印为浮动:

printf("This is the value in float: %f\n", yourFloatValue);

#5

converting string to a double variable in C

将字符串转换为C中的双变量

If overflow is not a concern, yet code wants to detect extra non-white-space text after the numeric text:

如果溢出不是问题,但代码想要在数字文本后检测额外的非空白文本:

// return 1 on success
int convertType(const char* value, double *destination) {
  char sentinel;
  return sscanf(value,"%f %c", destination, &sentinel) == 1;
}

If the sscanf() fails to find a double, the return value of sscanf() will be EOF or 0.

如果sscanf()找不到double,则sscanf()的返回值将为EOF或0。

If the sscanf() finds non-white-space text after numeric text, it will return 2.

如果sscanf()在数字文本后找到非空白文本,它将返回2。

If only a double is scanned, without extra, sscanf() returns 1. Leading and trailing white-spaces are OK.

如果只扫描一个double,没有额外的,sscanf()返回1.前导和尾随空格都可以。

Example:

double x;
if (convertType(some_string, &x)) {
  printf("%.17e\n", x);  // or whatever FP format you like
} else {
  puts("Failed");
}

#6

#define ZERO 48
#define NINE 57
#define MINUS 45
#define DECPNT 46

long strtolng_n(char* str, int n)
{
    int sign = 1;
    int place = 1;
    long ret = 0;

    int i;
    for (i = n-1; i >= 0; i--, place *= 10)
    {
        int c = str[i];
        switch (c)
        {
            case MINUS:
                if (i == 0) sign = -1;
                else return -1;
                break;
            default:
                if (c >= ZERO && c <= NINE) ret += (c - ZERO) * place;
                else return -1;
        }
    }

    return sign * ret;
}

double _double_fraction(char* str, int n)
{
    double place = 0.1;
    double ret = 0.0;

    int i;
    for (i = 0; i < n; i++, place /= 10)
    {
        int c = str[i];
        ret += (c - ZERO) * place;
    }
    return ret;
}
double strtodbl(char* str)
{
    int n = 0;
    int sign = 1;
    int d = -1;
    long ret = 0;

    char* temp = str;
    while (*temp != '\0')
    {
        switch (*temp)
        {
            case MINUS:
                if (n == 0) sign = -1;
                else return -1;
                break;
            case DECPNT:
                if (d == -1) d = n;
                else return -1;
                break;
            default:
                if (*temp < ZERO && *temp > NINE) return -1;
        }
        n++;
        temp++;
    }

    if (d == -1)
    {
        return (double)(strtolng_n(str, n));
    }
    else if (d == 0)
    {
        return _double_fraction((str+d+1), (n-d-1));
    }
    else if (sign == -1 && d == 1)
    {
        return (-1)*_double_fraction((str+d+1), (n-d-1));
    }
    else if (sign == -1)
    {
        ret = strtolng_n(str+1, d-1);
        return (-1) * (ret + _double_fraction((str+d+1), (n-d-1)));
    }
    else
    {
        ret = strtolng_n(str, d);
        return ret + _double_fraction((str+d+1), (n-d-1));
    }
}

#7

The following code works for me.

以下代码适用于我。

#include <stdio.h>

void convertType(char* value);

int main(int argc, char *argv[]) {
    char *str="0929";
    convertType(str);

    return  0;
}

void convertType(char* value) {
    double ret = 0;

    while(*value != '\0') {
        ret = ret*10 +(*value - '0');
        value++;
    }

    fprintf(stdout, "value: %f\n", ret);
}