Wrong behaviour of atof()

Question

Sample.c:

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

main()
{
    char str[10] = "111.1";
    float f = (float)atof(str);
    printf ("\n (%s  ,  %f) \n",str,f);
}

Output of above code is : (111.1 , 111.099998)

Please suggest whether i am missing something. Thanks in advance.

Try to use it with a `double`. I often had some random issues by using `float` — Guillaume Munsch, Jul 13 '15 at 09:59
its because of typecasting from double(return type of `atof`) to float, resulting into closest possible float. — Dayal rai, Jul 13 '15 at 10:03
Not related to the problem, but 1) you don't need to hardcode the size of an array if you're initializing it, so here `char str[] = "111.1";` would've been enough 2) implicit `int` return type for functions is pretty old C; whatever you're learning from, I suggest you use something newer. — user4520, Jul 13 '15 at 10:22

score 2 · Answer 1 · answered Jul 13 '15 at 15:10

Changing from float to double helps a bit, but doesn't "solve" the "problem." First, there is no problem, just a perception that there is one. As Dayal Rai said, 111.099998 is the closest you can get to 111.1 using a float. But even a double cannot represent 111.1, as you will find if you print its value using more decimal places. See the reference quoted by Sohil Omer.

score 0 · Answer 2 · edited May 23 '17 at 10:26

0

Use this , as atof returns double , Reference Man Page of atof()

double f = atof(str);

For more info, Double vs Float in C

edited May 23 '17 at 10:26

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answered Jul 13 '15 at 10:10

Sohil Omer

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Wrong behaviour of atof()

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