Thread: problem with atof

>for(ctr=1;ctr<=argc;ctr++)
This is suspicious.

>converted[ctr-1]=atof(*argv[ctr]);
And this is broken. argv[argc] is a null pointer, and because your loop continues through ctr having a value of argc, you're dereferencing a null pointer. Dereferencing argv[ctr] is also incorrect as it's already a pointer to char and that's what atof expects. Perhaps something more like this is what you want:

Code:

for ( i = 1; i < argc; i++ ) /* Skip program name */
  converted[i - 1] = atof ( argv[i] );

>any examples of atof error will be appriciated
If the string doesn't represent a valid floating point value?

Originally Posted by ssharish

Code:

  float converted[argc];

and output will look like this

Code:

[d4088019@outcast ~/other]$ gcc ex.c
[d4088019@outcast ~/other]$ a.out 1.25 + 3.25
The result is:0.000000

This line is definitely an error and I'm surprised that you didn't get warning or error while compiling
Anyway you should use

Code:

float *converted = malloc(argc*sizeof(float));
  float result;
  int i;
  for ( i = 1; i < argc; i++ ) /* Skip program name */
  converted[i - 1] = atof ( argv[i] );

Note that atof returns double so expect to get a compiler warning about loosing precision.

And one thing more:
PRELUDE IS ALWAYS RIGHT, AND HER EXAMPLES ARE ALWAYS CORRECT.

Can't say I see your problem, ssharish.

Code:

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

int main(int argc, char *argv[])
{
	double *converted = malloc(argc*sizeof(double));
	int i;

	for (i = 1; i < argc; i++ ) /* Skip program name */
	{
		converted[i - 1] = atof ( argv[i] );
		printf("%f ",converted[i-1]);
	}

	//and here some mathematical function which could perform basic operation will be called
	//by sending the converted[], &result
	free(converted);
	return 0;
}

With input:

Code:

1.25 + 3.25

Output:

Code:

1.250000 0.000000 3.250000

And one thing more:
PRELUDE IS ALWAYS RIGHT, AND HER EXAMPLES ARE ALWAYS CORRECT.

Quoted for emphasis.

>printf("%d",converted[i-1]);
%d is for signed integers, you want %f to print a double.

Come on now. You've posted here enough to know better.

Code:

int main(int argc, char *argv[])
 {
  double *converted=(double *)malloc(argc*sizeof(double));

1) Stop casting malloc.

Originally Posted by ssharish

still it gives me an error Prelude

2) Go read the forum announcements again, because you've obviously skipped the part about posting the error you're getting.

3) Your indentation is absolutely horrible.

Code:

  calcs(ch,converted,&result);

4) calcs? What the hell is calcs? You want us to find the error in a function you don't include? We're psychics now?


Quzah.

That's funny, it works for me:

Code:

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

int main( int argc, char *argv[] )
{
    if( argc > 1 )
    {
        int i;
        for( i = 1; i < argc; i++ )
            printf("%s is %f\n", argv[ i ], atof( argv[ i ] ) );
    }

    return 0;
}


/*
$ atofproggy 10.2 11.3 14.4
10.2 is 10.200000
11.3 is 11.300000
14.4 is 14.400000
*/

Quzah.

Because your compiler feels like it. Or doesn't, as the case may be. There is nothing in regards to C which says how your compiler has to work as far as being nice enough to correct you when you do something silly like leave out headers.

What it should do is give you error messages saying it can't find the function you're trying to use. If it does work, well, you're lucky. If not, you end up with something like this. I suspect you were getting some warnings that you were ignoring.

Quzah.