Problem in declaring array size inside a function

I'm passing an array (pointer) as an argument of a function, but I know the size of the array only inside the function, so I'd need something like this:

Code:

---

#include <stdio.h>

void foo(int *vec)
{
    int n = 3;
    vec = malloc(n*sizeof(int));
    vec[0] = 41;
    vec[1] = 10;
    vec[2] = 47;
    printf("vec[1]: %d", vec[1]); 
}


int main()
{
    int *vec;
    foo(vec);
    printf("vec[1]: %d", vec[1]);
    getch();
    return 0; 
}

---

The first printf gives me the right output, but the second one seems to crash the program. If I knew the size of the array before executing the function I would use int vec[n] to declare the array and delete the malloc line, but unfortunately that's not the case.

Any ideas?

Thanks.

You are changing vec inside that function through malloc() so you need to pass a int ** if you want to see the changes back in main.

It's like saying why doesn't:

void foo(int x) change my int I passed from main.

You can have foo return the size or have another int * passed as parameter which keeps track of the size.

Also compile with your warnings on, you are probably getting implicit declaration of getch(), which is a ........ function from an old crappy library for DOS.

Also free the memory you allocate

Thank you claudiu, but I think I didn't get it... I tried this:

Code:

---

#include <stdio.h>

void foo(int **vec)
{
    int n = 3;
    vec = malloc(n*sizeof(int**));
    *vec[0] = 41;
    *vec[1] = 10;
    *vec[2] = 47;
    printf("vec[1]: %d", vec[1]);
    free(vec);
}
int main()
{
    int *vec;
    foo(&vec);
    printf("vec[1]: %d", vec[1]);
    _getch();
    return 0; 
}

---

But it's not working, even though I don't get any errors or warnings. What's wrong?
I didn't know that getch() was deprecated, and my compiler (Dev-C++) doesn't show any warning when I use it. Is it ok to use _getch() instead?

Merci!

Dev-C++ is another obsolete crap IDE that hasn't been maintained in the past 8 years or so. Get rid of it. Use getchar() instead of all the getch() varieties.

Your function doesn't work because you are freeing the memory as soon as you allocate it. You need to free it at the end of main after you are done using it.

Oops, my bad.
But it's not working yet =P. Actually the black screen closes just after it opens. You think the code is wrong or it's a problem with the compiler?

Did you put in getchar() before return 0? What's your latest code?

I did. Here it's:

Code:

---

#include <stdio.h>

void foo(int **vec)
{
    int n = 3;
    vec = malloc(n*sizeof(int*));
    *vec[0] = 41;
    *vec[1] = 10;
    *vec[2] = 47;
    printf("vec[1]: %d", vec[1]);
}
int main()
{
    int *vec;
    foo(&vec);
    printf("vec[1]: %d", vec[1]);
    free(vec);
    getchar();
    return 0; 
}

---

In foo, it should be

Code:

---

*vec = malloc(n*sizeof(int*));

---

Code:

---

*vec = malloc(n*sizeof(**vec));
(*vec)[0] = 41;
(*vec)[1] = 10;
(*vec)[2] = 47;

---

Also need to be mindful of operator precedence here.

I prefer:

Code:

---

*vec = malloc(n * sizeof(**vec));

---

Also, this:

Code:

---

*vec[0] = 41;
*vec[1] = 10;
*vec[2] = 47;

---

should be:

Code:

---

(*vec)[0] = 41;
(*vec)[1] = 10;
(*vec)[2] = 47;

---

It may be easier to just write:

Code:

---

int *temp = malloc(n * sizeof(*temp));
/* ... access temp[0] etc */
*vec = temp;

---

My example was wrong, it should have been:

Code:

---

*vec = malloc(n*sizeof(int));  // sizeof(int) not sizeof(int*)

---

But, like laserlight, I prefer

Code:

---

*vec = malloc(n * sizeof **vec);

---

since that will be the right type even if the base type is changed.

Thank you guys, that's exactly what I need!