Multiplication of thousands of small values

[SoFaraway]

Dear all,

I need to do multiplication of thousands of floating points in C, here are some values:
4.71678396160628e-09
0.072222226257693
4.71678396160628e-09
0.0111111157230777

I need to read these values from a file, then do multiplication of them. I use

long double *value;
value = (long double *)malloc(n * sizeof(long double));
for(i = 1...n)
  fscanf(fp, "%lf", value[i]);

However, I always get some very funny values. When I change long double to double, it also fails.

Anyone can help me?

Thank you very much!

[JosAH]

It could be an 'ill conditioned' system you're working with. First read all the numbers,
sort them and multiply the largest numbers first if they're all |x| < 1

kind regards,

Jos

[SoFaraway]

I forgot to mention that there is error when I use the following code to read in values:

3 values to be read from the file:
# 0.000152823802513597 476

code to read and check the values:

long double d1, d2;
fscanf(fp, "%s %lf %d", tmp, &d1, &d2);
printf("%s %.30lf %d ", tmp, d1, d2);

The values printed are:
# 0.000152823802513597008798443277 0

Any idea about this?

Thank you very much!

[JosAH]

I forgot to mention that there is error when I use the following code to read in values:

3 values to be read from the file:
# 0.000152823802513597 476

code to read and check the values:

long double d1, d2;
fscanf(fp, "%s %lf %d", tmp, &d1, &d2);
printf("%s %.30lf %d ", tmp, d1, d2);

The values printed are:
# 0.000152823802513597008798443277 0

Any idea about this?

Thank you very much!

You're trying to read an int (%d) and store the four bytes in a long double (d2).
First make sure that you can read those numbers correctly before you attempt
to multiply certain values. As far as I can see now this has nothing to do with
ill conditioned numerical systems, just IO bugs.

kind regards,

Jos

[SoFaraway]

You're trying to read an int (%d) and store the four bytes in a long double (d2).
First make sure that you can read those numbers correctly before you attempt
to multiply certain values. As far as I can see now this has nothing to do with
ill conditioned numerical systems, just IO bugs.

kind regards,

Jos

Hi Jos,

Thank you for your reply.

Sorry that I typed wrongly in the previous post. My original code is

long doube d1;
int d2;

which is in correct format.

Do you have any suggestion about the error?

Thanks again!

[JosAH]

Hi Jos,

Thank you for your reply.

Sorry that I typed wrongly in the previous post. My original code is

long doube d1;
int d2;

which is in correct format.

Do you have any suggestion about the error?

Nope, I'm sorry; after my first answer you come up with something completely
different; I answer again and then you twist the problem to something else
again. I can't follow you anymore. Please make up your mind first, state the
problem you're really experiencing and only *then* raise a question.

The way this thread works out now is not the way questions are stated and
possibly answered. You're just playing hide an' seek and I don't play those games.


kind regards,

Jos

[pbmods]

Heya, SoFaraway.

What code are you using right now (copy and paste are your friends ~_^)?

What is an example of an input file that your script is supposed to read?

What is the output that you are getting when you run this file through your code?

[weaknessforcats]

OK. Let's start again:

This data:

4.71678396160628e-09
0.072222226257693
4.71678396160628e-09
0.0111111157230777

was put into a text file and that file was read by this code:

int main()
{
int n = 1000;
FILE* fp;
fp = fopen("C:\\scratch\\instructor\\data.txt", "r");
if (!fp)
{
   cout << "Failed to open" << endl;
   return 0;
}
long double *value;
int numread = 0;
value = (long double *)malloc(n * sizeof(long double));
while (fscanf(fp, "%lf", &value[numread]) != EOF)
{
    ++numread;
}
//Playback
for (int i = 0; i < numread; ++i)
{
     printf("%e\n", value[i]);}
}
 

These are the results:

4.71678e-009
0.0722222
4.71678e-009
0.0111111

You can't reall fscanf() into a long double since the %f means to scan to a float (6 significant figures). The excess is rounded. The %lf really does nothing more than %f and is Microsoft specific.


There were too many errors in the sample to address each one.

[ravenspoint]

The trick is to specify the precision you want on output.

If you use

printf("%.20e\n", value[i]);

you should get

4.71678396160628030000e-009
7.22222262576930060000e-002
4.71678396160628030000e-009
1.11111157230777000000e-002

James