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I am working on a simple softmodem program. The modem is designed to implement the Audio FSK of a dial-up modem and should be able to initiate a phone call by emitting DTMF tones. Currently I am having some problems with a function which will generate sine values. 

 double* generate_sine( int numberOfElements, double amplitude, 
 double phase_in_degrees, double numberOfCycles)
{
static int i;

double *sine_output;
sine_output = malloc( numberOfElements*sizeof(double) ); 

for( i=0; i<numberOfElements; i++ )
{

     sine_output[i] = (amplitude*sin(( (2.0*M_PI*i*numberOfCycles)/(double)numberOfElements )+
     ((M_PI*phase_in_degrees)/180 )));

   }
return sine_output; 
}

There is a segmentation error in the function. the variable "i" appears to become a pointer after the first iteration of the loop, its value is 4206692. sine_ptr also has a valid address until the second iteration of the loop at which point it become 0 (NULL). Here is where the function is called. DEFAULT_PHASE = 0.0

int main()
{
 int i;
     int numElements = 10;
 double* sine_ptr = generate_sine( numElements, 0.5, DEFAULT_PHASE, 440 );

  for( i=0; i<numElements; i++)
  {
    printf( "%e \n", *(sine_ptr + i ) );
  }
 free(sine_ptr);
 return 0;
}

After taking all of the edit suggested into consideration I was able to solve the problem in my code, thank you very much for any help that you gave me.

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2  
Undefined behaviour. You're passing a double when printf expects an int. – chris Jan 5 at 7:03
I tried specifying that I want printf() to display a 3.5%d number. The program froze at run time. "sine_output" somehow becomes a null pointer during the loop in the function "generate_sine()". – user1505399 Jan 5 at 7:15
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1 Answer

up vote 3 down vote

EDIT Added another point.

Issue Number 1

You are allocating enough memory for 5 elements (numberOfElements is 5). You are then trying to print 10 elements from the initial point. The last 5 elements will be accessing unallocated space and can result in undefined behavior. Including segmentation faults.

Issue Number 2

You are not freeing the pointer to the allocated space but a location 10 places later. This will also cause other problems if you solve the segmentation fault.

Issue Number 3

This is one of the minor ones. But sine_ptr is double, and trying to show it as int is undefined. will cause compiler warnings. Even with warnings, the numbers are downcasted. In your case the output will be all zeros. To see correct results, use %lf.

Use the following code.

int main()
{
  int i;
  int numElements = 5;
  double* sine_ptr = generate_sine( numElements, 0.5, DEFAULT_PHASE, 440 );
   for( i=0; i<numElements; i++)
   {
     printf( "%lf \n", *(sine_ptr + i) );
   }
   free(sine_ptr);
   return 0;
}
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1  
For completeness, include the printf type mismatch. You can never have too little UB. – chris Jan 5 at 7:20
@Potatoswatter The formatting will not cause a segmentation fault. May be an illegal conversion. But those are usually warnings and some compilers don't even complain unless you ask for it. – Pavan Jan 5 at 7:20
@chris done. Updated the code too. – Pavan Jan 5 at 7:23
1  
@user1505399 What do you mean it freezes ? Did the segmentation fault go away ? – Pavan Jan 5 at 7:25
1  
@Pavan, Using a double instead of an int is undefined, as per C11 7.16.1.1/2 (emphasis mine): If there is no actual next argument, or if type is not compatible with the type of the actual next argument (as promoted according to the default argument promotions), the behavior is undeļ¬ned ... (more cut out). Therefore, you could very well not see all zeros. – chris Jan 5 at 7:26
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