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Signal Problem

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# 1  
Old 10-10-2006
Signal Problem
I am using the signal function, and passing it a function named quit procedure...I get the following warning....

passing arg2 of signal from incompatible pointer type...

void quit_procedure(void); //this is the way i define my prototype...

signal(SIGINT, quit_procedure);


Please guide me..How can i eliminate this warning?
# 2  
Old 10-10-2006
This is the prototype for signal:
Code:
 void (*signal(int sig, void (*func)(int)))(int);

The second argument is a function pointer for a function that takes an int as an argument.
Code:
void quit_procedure(int sig_number)

will work.
# 3  
Old 10-10-2006
Hi,

I am not trying to pass any argument to my quit_procedure method...In that case, what do i need to do?
# 4  
Old 10-10-2006
You are not calling the quit_procedure function - signal() is.
Just make
Code:
void quit_procedure(int something)
{
  // ignore the "something" variable"
}

If you call the code yourself, then you are not responding to a signal. You can do that if you want, but it will confuse you later on unless you document what you did.
# 5  
Old 10-10-2006
thanks jim
# 6  
Old 11-17-2006
Lightbulb
Quote:
Originally Posted by jacques83
I am using the signal function, and passing it a function named quit procedure...I get the following warning....

passing arg2 of signal from incompatible pointer type...

void quit_procedure(void); //this is the way i define my prototype...

signal(SIGINT, quit_procedure);


Please guide me..How can i eliminate this warning?

Here as told .. the function is of type
signal(int sig,<func-name>)

Here func-name(int sig){ }
Here the signal handler function, takes the signal number as argument ...
If u try to print the variable,it ll give the signal number //Try!
So whether you use that variable or not ,u shd declare dut variable
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