How to waste CPU cycles in a program?

I want the program to spend some time running but 'for ( ; ; )' is not suitable because I do not want it to run forever. I want something that I can kind of control how long it runs. That is, I want something like

Code:

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for (i; i<1000; i++)
  x = x + i;

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and then I can adjust the maximum loop index (1000, 10000, 100000, ..) depending on the actual run. But apparently the compiler optimizes away the loop though I use the -O0 flag with gcc.
What is an effective and efficient way to do this?

The method you're trying to use for adding delays is a horrible one for 2 reasons: 1) The speed of the computer will affect the amount of delay, 2) You're chewing up valuable CPU time that could be better used for other applications while yours is sleeping.

A better method is to use your platform-specific function for putting your program to sleep for a certain amount of time. In Windows try the Sleep() function.

I know I am chewing up CPU resources, that's why the title is "How to WASTE CPU cycles in a program?" I do not mind that. I also know the speed of the computer will affect the amount of delay, that's why I need to adjust the max loop index from one run to another, just so that the time the program spends in the loop will approximately be what I want.
I do not want sleep() because I am testing a program involving Unix signals. sleep() sets the SIGALRM signal and thus affects the testing of the signal handler, at least on Solaris.

Quote:

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Originally Posted by hzmonte

I want the program to spend some time running but 'for ( ; ; )' is not suitable because I do not want it to run forever. I want something that I can kind of control how long it runs. That is, I want something like

Code:

---

for (i; i<1000; i++)
  x = x + i;

---

and then I can adjust the maximum loop index (1000, 10000, 100000, ..) depending on the actual run. But apparently the compiler optimizes away the loop though I use the -O0 flag with gcc.
What is an effective and efficient way to do this?

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Call gettimeofday() in a tight loop, breaking the loop when the time has reached whatever time you wanted. This meets the "wasting CPU" requirement and doesn't require sleep() or signals.

> What is an effective and efficient way to do this?
volatile int i;
should do the trick.

Notwithstanding all the other problems with the technique.

Quote:

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Originally Posted by Salem

Notwithstanding all the other problems with the technique.

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What problems? Why is "the solution must busy loop the CPU" an invalid requirement?

Quote:

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Originally Posted by brewbuck

What problems? Why is "the solution must busy loop the CPU" an invalid requirement?

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Nothing per se, except I don't feel like using that program unless there's a serious reason for it. :p

A for loop will not run the same speeds on all systems. You need a timer and the timer resolution to do this correctly.

1. Set a variable to 0 outside of the loop.
2. Increase the variable by the timer difference inside the loop.
3. When the variable reaches a certain delay value (as chosen by you), exit the delay loop and continue with the program then goto 1.

Code:

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UINT uTotalTime=0;
UINT uLastTimer=<some_func_to_get_timer_value>;

do
{
  uTotalTime+=<some_func_to_get_timer_value>-uLastTimer;
  if (uTotalTime><some_delay_value>)
  {
      uTotalTime=0;
      uLastTimer=<some_func_to_get_timer_value>;
      .....continue with code
  }
} while (<some_condition>);

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> I do not want sleep() because I am testing a program involving Unix signals.
An alternative is using select() with empty FD_SETs and the timeout set to what you want to sleep for.

How about putting the solution(s) in the FAQ "How to ...?"

> What problems? Why is "the solution must busy loop the CPU" an invalid requirement?
But that wasn't the underlying question, which was how to introduce a delay without using a signal generating sleep() call.

Besides, if you used a busy delay loop on a mobile device, you'd get your ears chewed off by the battery life brigade.