Converting string to a double and back to a string

Question

I'm having issues converting a string to a double and not exactly sure what's wrong. My add function:

int add(const char *a,const char *b,char* str,int length)
{
  printf("\n*you are in add function %s,%s*\n",a,b);

  //double aa = strtod(a,NULL);
  double aa =atof(a);
  //double bb = strtod(b,NULL);
  double bb = atof(b);
  printf("\n after converting you get %f ,%f \n",aa,bb);
  double c;
  c= aa+bb;
  //snprintf(str,length,"%.2f\n",c);
  sprintf(str,"%.2f\n",c);
  printf("\nthis is your new char %s\n",str);
  return 0;
}

Here is my open fuse function portion:

else if((strcmp(mytokens.toks[0],"add")) ==0 && (strcmp(mytokens.toks[1],"doc") != 0))
   {
     printf("\n You are in the ADD function and are trying to pass in %s,%s \n",mytokens.toks[1],mytokens.toks[2]);
     char str[1024];
     add(mytokens.toks[1],mytokens.toks[2],str,1024);
     printf("\n This is the str after add %s \n",str);
     len = strlen(str);
     if (offset < len) {
       if (offset + size > len)
     size = len - offset;
     printf("\nthis is for memcpy str %s",str);
       memcpy(buf, str + offset, size);
       }
   }

so I tried cat test/add/1/2 and get a result of 0.00 and in my debugger I'm getting:

 You are in the ADD function and are trying to pass in 1,2 

*you are in add function 1,2*

after converting you get 0.000000 ,0.000000 

this is your new char 0.00

This is the str after add 0.00

this is for memcpy str 0.00

So inside the add function the first printf is showing the string "1" and "2" correctly but when I try to convert it using strtod() or atof() , it converts my strings into 0.000000 and 0.000000. Any ideas?

Did you tried: printf("\n after converting you get %lf ,%lf \n",aa,bb); ? a and b are double, not float. However %f usually works with double. snprintf(str,length, "%.2lf\n",c); — nopsoft, Apr 25 '14 at 20:21
There is no difference between "%f" and "%lf" in a *printf* format (float is converted to double in a variable argument list). — Martin R, Apr 25 '14 at 20:23
You're right http://stackoverflow.com/questions/4264127/correct-format-specifier-for-double-in-printf — nopsoft, Apr 25 '14 at 20:25
Rather than use a buffer of 1024 (which might not be enough for `%f`) use a `char str[3+DECIMAL_DIG+8]; sprintf(str,"%.*e\n",DECIMAL_DIG,c);` to print to sufficient precision without way over doing it. http://stackoverflow.com/questions/16839658/printf-width-specificer-to-maintain-precision-of-floating-point-value/19897395#19897395 — chux - Reinstate Monica, Apr 25 '14 at 20:46

score 2 · Answer 1 · edited May 23 '17 at 10:26

2

Probably you missed the includes:

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

What is wrong with usage of atof function?

edited May 23 '17 at 10:26

Community

1
1

answered Apr 25 '14 at 20:36

nopsoft

922
7
10

BINGO! I commented out the #include and got the same set of zeros. Great call! – Mahonri Moriancumer Apr 25 '14 at 20:39
I already have both but here is a list of all my includes: #include #include #include #include #include #include – dspaces1 Apr 25 '14 at 20:47

score 0 · Accepted Answer · answered Apr 25 '14 at 21:08

0

Something was wrong with the buffer so I used a string copy to fix the issue.

int add(char *a,char *b,char* str,int length)
{
  char a1[100];
  char b1[100];
  strcpy(a1,a);
  strcpy(b1,b);
  printf("\n*you are in add function %s,%s*\n",a1,b1);
  double aa =atof(a1);
  double bb = atof(b1);
  printf("\n after converting you get %f ,%f \n",aa,bb);
  double c;
  c= aa+bb;
  snprintf(str,length,"%.2f\n",c);
  printf("\nthis is your new char %s\n",str);
  return 0;
}

answered Apr 25 '14 at 21:08

dspaces1

193
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15

Make sure mytokens.toks[1] and mytokens.toks[2] are not modified during add(). It is very similar to "magic" that function starts to work when you change one of variables to static but real problem is overwriting. – nopsoft Apr 25 '14 at 21:15

Converting string to a double and back to a string

2 Answers2