CODEWITHSUNDEEP
c++carrayspointers
Phorce
Phorce

Reputation: 4642

C++ passing reference to array to C function

I'm trying to call a function written in C which passes in a pointer to an array.

In C++ I have the following:

double* x = new double[10];

populateArray(x);

In C:

void populateArray(double* vars); 
{
    vars = (double*) malloc(10*sizeof(double));

    int i;

    for(i = 0; (i < 10); i++)
    {
        vars[i] = 1*20;
    }
}

However, when I try to output the contents of x inside the C++ code, the result is always 0?

Upvotes: 0

Views: 106

Answers (3)

HelloWorld123456789
HelloWorld123456789

Reputation: 5369

Problem arises because you are changing local variable vars.

Change to:

double* x; //you don't need to allocate here if you are allocating later

populateArray(&x);

And:

void populateArray(double** vars) 
{
    *vars = malloc(10*sizeof(double)); //also no need to cast malloc

    int i;

    for(i = 0; (i < 10); i++)
    {
        (*vars)[i] = 1*20;
    }
}

Upvotes: 4

Tim Čas
Tim Čas

Reputation: 10867

Short answer: Remove the malloc call inside the function.

Slightly longer answer:

You're changing the value of the vals pointer to another newly allocated memory -- thus, the argument you pass in gets unused (and thus is never populated).

The result being always 0 is a coincidence, it could be anything (or you could end up with nasal demons) because you're then reading uninitialized memory, which is undefined behavior.

This of it as this, if you remove the call:

vals = new double[10];
vals = malloc(10 * sizeof(double));

The first value is overwritten.

Upvotes: 0

mah
mah

Reputation: 39827

void populateArray(double* vars); 
{
    vars = (double*) malloc(10*sizeof(double)); // <---- remove this line completely

The problem is you're ignoring the vars you were given and creating a brand new buffer for it.

Worse, that buffer pointer gets lost when this function returns, causing a memory leak.

Just use the vars you were given -- delete the line indicated above.

Upvotes: 0

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