Passing a dynamically allocated array from main to another function

Question

First off, I have already looked at the example presented here:

Passing dynamically allocated array as a parameter in C

I am trying to pass a dynamically allocated array as a parameter to another function.

void InputIterations(int *iterations);
void CalcE(long double *E, int iterations);

int main()
{
     int iterations;  
     long double *E;

     InputIterations(&iterations);

     E = (long double *) malloc(iterations * sizeof(long double));

     CalcE(&E, iterations);
}

void InputIterations(int *iterations)
{
     printf("Enter a number of iterations: ");
     scanf("%d", iterations);
}

void CalcE(long double *E, int iterations)
{
     long double sum = 0;
     int i;
     for(i=0; i<iterations; i++)
     {
          sum = sum + /*formula for calculating irrational constant e*/
          *E = sum;
          E++;
     }
 }

However, when I compile my code I get the following error:

error: cannot convert ‘long double**’ to ‘long double*’ for argument ‘1’ to ‘void CalcE(long double*, int)’ CalcE( &E, iterations );

Does anyone knows why I am getting this error ?

If you could please explain my mistake or point me to a source that explains it I would greatly appreciate the help.

Take a look at the error message again. It really tells you all you need to know. And if you still don't know, then go back to a book and read more about the address-of operator (&) and what it does. You need to do it anyway, because you really misuse it in more than one place (for example I suggest you take a closer look at that scanf call too). — Some programmer dude
– Some programmer dude, Commented Oct 5, 2016 at 8:15
You need to know that Pi is already a pointer and &Pi is a pointer to the pointer, a double poiner. So the call should be CalcPi(Pi, iterations); — The Apache
– The Apache, Commented Oct 5, 2016 at 8:17
Choose a language. In C++ the answer is simply, "use std::vector". — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Oct 5, 2016 at 8:27

marcinj · Accepted Answer · 2016-10-05 08:27:17Z

3

Here:

CalcE(&E, iterations);

you take address of E (of type long double *) and pass it as an argument to CalcE. CalcE accepts as first parameter a pointer to long double. But when you take an address of E you are given actually a pointer to pointer to long double (long double**), and that is not a pointer to long double. And this is what your error tells you:

error: cannot convert ‘long double**’ to ‘long double*’ for argument ‘1’ to ‘void CalcE(long double*, int)’ CalcE( &E, iterations );

So you should have:

CalcE(E, iterations);

edited Oct 5, 2016 at 8:27

answered Oct 5, 2016 at 8:25

marcinj

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Comments

sergej · Accepted Answer · 2016-10-05 08:37:14Z

3

CalcE(&E, iterations);

should be

CalcE(E, iterations);

Hope I helped

edited Oct 5, 2016 at 8:37

sergej

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answered Oct 5, 2016 at 8:20

Shay Gold

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2 Comments

Corey Over a year ago

Yes this worked. Thank you! Is it because E was already a pointer so I did not need to pass the address of the pointer?

George Over a year ago

@Corey Yes, in passing the address, you're trying to pass a double pointer where a pointer is expected, hence the error, error: cannot convert ‘long double**’ to ‘long double*’ and also your program leaks memory because you never free the memory that E is pointing to.

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