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?
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;
}
}
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.
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.
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