我很快写了这个代码：

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

float binary_f(char fraction[]){
  char *pointp; //pointer to the point
  long integer=strtol(fraction, &pointp, 2); //Gives the integer and the pointer to the point, at the same time!
  int point=(int)(pointp-fraction);
  
  long div = 2;
  int lenght=strlen(fraction);
  float result=0;
  
  for (int i=point+1;i<lenght;i++){
    if (fraction[i]=='1'){
      result+=(float)1/div; 
    }
    div <<= 1; //Multiply the divisor by 2
  }
  return result+(float)integer;
}

int main() {
  char fractionary[]="10.001";
  float a=binary_f(fractionary);
  printf("%f",a);
  return 0;
}

图纸：

2.125000

正如您可能注意到的，它并不是真正健壮的，但它可以完成预期参数的工作。