How can i print once per second

Question

1.00/5 (1 vote)

See more:

Hi,

I have the code snippet below:

C++

#include "TimeUtils.hpp"

FILE *output;                   // internal filename

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

//…………………………….
//More code
//…………………………….

time01 = time_now(); // Returns the time now
    while(1)
    {
        time02 = time_now(); // Returns the time now
        static long count  = 0;
        count  = (time02-time01)/1000; // Number of seconds since "time01"

        Rangek.runge4(count, y, dist); // Method executed in real time [Is this the way to do this?]

/*
My problem:
	I want to print to file "output" only once per second.
	How do I do that?
*/
        if((count%2) != 0)
        {
            fprintf(output, "%f\t%f\n", count, y[0]);
            continue;
        }

        cout << "Counting: " << count << " seconds\n";

        if( ((time02-time01)/1000) >= 10) break; // Stop after 10 second
    }


    fclose(output);

return 0;

}

I am supposed to see only 10 lines on my file, but I have a few hundred lines.
I can I make it print only once per second?

Thank you for your help.

Regards,

Herve

Posted 7-Feb-12 4:14am

The_Real_Chubaka

Add a Solution

2 solutions

Solution 1

It looks like you write to the file with every iteration that (count%2) is true.

Try adding a variable called lastcount and set it equal to -1.

Then if count and lastcount are different, print to the file and make lastcount = count.

C++

if(lastcount != count)
{
    fprintf(output, "%f\t%f\n", count, y[0]);
    lastcount = count;
    continue;
}

Posted 7-Feb-12 4:20am

JackDingler

Comments

Chuck O'Toole 7-Feb-12 10:47am

This is correct, "time02 = time_now();" could return the exact same value for the time a thousand times before the clock ticks to the next second. You are not doing "timing based output". As noted, you are doing "iteration based output".

The_Real_Chubaka 7-Feb-12 10:56am

Do you mean like this:

while(1)
{
time02 = time_now(); // Returns the time now
static long count = 0;
count = (time02-time01)/1000; // Number of seconds since “time01”
Rangek.runge4(count, y, dist); // Method executed in real time [Is this the way to do this?]

static long lastCount = -1; // 'static' so it does not get re-initialised
cout << "Counting: " << lastCount << " seconds\n";
//if((count%2) == 0)
if(lastCount == count)
{
fprintf(output, "%f\t%f\n", count, y[0]);
lastCount = count;
continue;
}

cout << "Counting: " << count << " seconds\n";

if( ((time02-time01)/1000) >= MAX) break;// Stop after 10 second

}

I have two problems:

I made "lastCount" static, but it is always getting re-initialized!!! Why is that?

The "countinue" statement will make this line "Rangek.runge4(count, y, dist);" not be executed in real-time.

Richard MacCutchan 7-Feb-12 12:29pm

I found the following comment on your profile page: I am very proficient at C, C++, Python and MATLAB.
?

The_Real_Chubaka 7-Feb-12 12:52pm

Yes but, the last line of code i wrote was a few years ago.

The_Real_Chubaka 7-Feb-12 10:59am

Since, it is always getting re-initialized, it will always be equal to -1 after the first iteration.
That mean lastCount will always be != from count after the first iteration.
That is also why it is only printing the first line.

Jochen Arndt 7-Feb-12 11:11am

Move 'static long lastCount = -1;' out of the while loop to the top of your function. Than it is not re-initialized.

Richard MacCutchan 7-Feb-12 11:32am

You don't need statics, just local variables that are outside the while loop. See my suggestion below.

JackDingler 7-Feb-12 11:33am

Further it should be

if (lastCount != count)

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Richard MacCutchan · Accepted Answer · 2012-02-07T04:36:00

C++

Rangek.runge4(count, y, dist); // Method executed in real time [Is this the way to do this?]

What is it supposed to be doing, and what are the types and values of y and dist?

C++

if((count%2) != 0)
{
    fprintf(output, "%f\t%f\n", count, y[0]);
    continue;
}

cout << "Counting: " << count << " seconds\n";

count % 2 will yield non-zero for odd numbers so that will be after 1 second, 3 seconds etc; so every two seconds.
The fprintf() statement uses the %f format control character to print a floating point value, but the variable count is a long integer. I don't know what y[0] might be.
The continue statement goes back to the top of the loop and gets the time again. This will (probably) be fairly close to the last iteration so it will call the fprintf() statement a second time within the current second.

You need to rethink your logic and figure out exactly what you are trying to do. It would be better to save the difference between your times in a second variable and wait till that changes before doing the print; something like:

C++

timeDifference = 0;
lastDifference = 0;
while (true)
    timeDifference = time2 - time1
    if (timeDifference != lastDifference)
        lastDifference = timeDifference
        print details
    end if
    if (lastDifference > 10)
        return
end while