atof: string to float

Hi i have just created some code to get the number out of a char array. Now i am not sure if it works. Below is the code:

Code:

---


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

int main(int argc, char ** argv)
{

FILE * output = stdout;
FILE * input = stdin;

char aperture[4];
float apertureNo

        fprintf(output,"Please enter the aperture value (f-stops): "); fflush(output);
        fscanf(input,"%s",&aperture);
       
        apertureNo = atof(aperture);   

        fprintf(output,"Your aperture is \n",aperture);
        fprintf(output,"Your aperture NUMBER is \n",apertureNo);

}

---

Could someone please tell me if it will work.

For example. I enter a1.5

aperture should be: a1.5
apertureNo should be: 1.5
Will the above code do it?

thanks

No that won't work "a1.5" is not a valid double.
Kurt

so how could i get the 1.5 out of it?

Same mistakes as previous posts....

> float apertureNo
This is missing a ;

> char aperture[4];
This isn't large enough to hold a 4-character string AND a \0

> apertureNo = atof(aperture);
If you're confident of the input format, then
apertureNo = atof( &aperture[1] );
will convert from the 2nd character onwards.

you could pass the start of the float tp atof

Code:

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apertureNo = atof(aperture+1);

---

Checking if aperture[0] really is an 'a' would be a good idea too.
Kurt

missing your format specifiers.

Code:

---

fprintf(output,"Your aperture is %s\n",aperture);
fprintf(output,"Your aperture NUMBER is %f\n",apertureNo);

---

Also, atof() is in <stdlib.h>.

But I'd suggest using strtof() instead, also in <stdlib.h>; it has better error checking. strtof() is just like strtod(), but with floats instead of doubles.

Code:

---

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

int main(int argc, char ** argv)
{

FILE * output = stdout;
FILE * input = stdin;

char aperture[5];
float apertureNo;

        fprintf(output,"Please enter the aperture value (f-stops): "); fflush(output);
        fscanf(input,"%s",&aperture);
       
        apertureNo = atof(&aperture[1]);   

        fprintf(output,"Your aperture is %s\n",aperture);
        fprintf(output,"Your aperture NUMBER %fis \n",apertureNo);

}

---

ok i changed it according to what you guys were saying. Now i use this now: apertureNo = atof(&aperture[1]);

Basically the input will always be in the form 'a1.5' or 'a4'

it will always start with a letter then the rest are numbers.

now will the new code work to get the numbers after the first letter?

Not quite.

Code:

---

fscanf(input,"%s",&aperture);

---

Drop the &, since aperture is a char[].

Code:

---

fprintf(output,"Your aperture NUMBER %fis \n",apertureNo);

---

That will generate output like "Your aperturn NUMBER 1.23is". You probably want

Code:

---

fprintf(output,"Your aperture NUMBER is %f\n",apertureNo);

---

The location of the format specifier like %f is where the number is printed.

[edit] And you're still not including <stdlib.h> for atof(). [/edit]

Code:

---

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

int main(int argc, char ** argv)
{

FILE * output = stdout;
FILE * input = stdin;

char aperture[5];
float apertureNo;

        fprintf(output,"Please enter the aperture value (f-stops): "); fflush(output);
        fscanf(input,"&#37;s",aperture);
       
        apertureNo = atof(&aperture[1]);   

        fprintf(output,"Your aperture is %s\n",aperture);
        fprintf(output,"Your aperture NUMBER is %f \n",apertureNo);
            return(0);

}

---

ok so basically now, will it take the number after the first character?

Not quite. ;)

Code:

---

apertureNo = atof(aperture[1]);

---

You need an & there. :)

Here's why. Both fscanf() and atof() expect a char* string. Since aperture is a string, you can pass it with no adornments to fscanf(). aperturn[1] is a character, so you need to take its address to make it a string as &aperture[1] to pass it to atof(). Or you could just use

Code:

---

apertureNo = atof(aperture + 1);

---

which is the same thing.

I might also suggest one or two other minor changes, like removing the redundant space after the &#37;f, moving the fflush() call onto its own line to make it more readable, and adding a return 0 at the end of main().

ah alright now hopefully its ok. i changed it in post10. seem good. so if i enter a1.4 it will take out the 1.4 and store in apertureNo?

Looks good to me.

By the way, you do know that you can use printf() and scanf() to always use stdout and stdin, right? I guess that you used fprintf() etc so that you can use other streams later, though.

yea thats right, we were taught like that so we what we learn later can be used easier

thanks for the help.

oh by the way: In pseudocode, whats the appropriate language to say get numbers out of char arrat like my case?

Quote:

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oh by the way: In pseudocode, whats the appropriate language to say get numbers out of char arrat like my case?

---

Do you mean, what is the pseudocode for your program? Or what's some pseudocode for extracting digits from array elements? Or pseudocode for searching though a string and getting numbers from it, where the numbers could be anywhere in the string? Or what?

[edit]

Quote:

---

yea thats right, we were taught like that so we what we learn later can be used easier

---

Personally, I don't think that's a very good way to do it. I use stdin and stdout all of the time. If I want to use files for input/output, I redirect the input/output of my program, or code it in specially.

And if you really did want to change all occurances of printing to go to a file, you could use freopen().

I realize that this is probably beyond your control, but I just thought I'd comment. [/edit]

hahaha yea man, thats how tutor teaches.

for the pseudocode i mean for this bit:
apertureNo = atof(&aperture[1]);
i guess to extract the digits from the array?

You already have C code, why do you need pseudo-code? . . .

Code:

---

number = string_to_number(string from element 1 onwards)

---

Are you wondering how atof() works? It probably does something like this:

Code:

---

#include <ctype.h>

float my_atof(const char *s) {
    float f = 0.0f, fraction = 0.0f;

    while(*s) {
        if(isdigit(*s)) {
            f *= 10;
            f += *s - '0';
            fraction *= 10;
        }
        else if(*s == '.') {
            fraction = 1;
        }
        else break;
    }

    if(fraction) f /= fraction;

    return f;
}

---

That's just one way to do it, of course. A way I haven't tested. I suppose you want its pseudo code. :)

Code:

---

for each character in the string
    if the character is a digit
        multiply the number by 10 to shift all of the other digits to the left
        add the digit to the number
    else if the character is a decimal
      take note of the position of the decimal point
    else the character is unrecognized, so end the loop

if a decimal point was noted, shift the number right \
enough times to put the decimal place where it belongs

return the number

---

Something like that.

ya well first i have to write pseudocode then program, i just wanted to see if there was a way to get the number from string and now there is.
theres no other simple pseudocode to describe getting the number from string? because yours sound too much like proper english??

Quote:

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theres no other simple pseudocode to describe getting the number from string? because yours sound too much like proper english??

---

I dislike very specific pseudo-code. In my opinion, if you've got the program nailed down to a high enough degree to write very specific pseudo-code, you might as well write C.

If you want really specific pseudo-code, just convert my C code into it. :) As long as you understand it, converting it to pseudo-code should be quite simple.

If you don't understand it, study the vague pseudo-code that I posted, or look elsewhere. :)

Quote:

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Originally Posted by dwks

I dislike very specific pseudo-code. In my opinion, if you've got the program nailed down to a high enough degree to write very specific pseudo-code, you might as well write C.

If you want really specific pseudo-code, just convert my C code into it. :) As long as you understand it, converting it to pseudo-code should be quite simple.

If you don't understand it, study the vague pseudo-code that I posted, or look elsewhere. :)

---

I agree. And pseudo-code isn't a standardized "language" (in the way that C/C++ is), you may do as you like in it. Different companies, universities or such may have definitions of their pseudo-code form. Where I worked previously, the pseudo-code looked pretty much like C, but it had some non-C "operations" that made it easier to write the pseudo-code [can't remember the exact details, as it was a while back, and I only read the pseudo-code, as it was written by the high-level architects].

If Taurus' school has a defined standard for pseudo-code, then the code dwks wrote should be fairly easy to translat inot that.

I'm also going to add that wriitng your own "atof" is probably not a good idea, but that's a different discussion.

--
Mats

i used this:

char aperture[4] (=f1.2)
int app

app = atof(aperture[1]);

Now i get an error for the atof?? whats wrong?
(error is: passing argument 1 of atof makes pointer from integer without cast)
what can i do?

ok correction . wen i use aperture + 1 it works. Now app is an int and i get my result of 1 from f1.2 but not the .2. But once app is double it gives a random number of 858993459

how can i get the .2???

thanks

use &#37;f format to print double

ah right thanks