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So i am new to programming and i have this program that i have to write, it has an array of integers and i have to pass it to a function with a pointer and then with a double pointer. After i wrote the code i had this error that i couldn't find a solution to. The error is :

initialization of 'int **' from incompatible pointer type 'int *' [-Wincompatible-pointer-type]

The code that i wrote is this:

int main(){
    int i;
    int arr[]={3,-15,19,0,-984};
    int *p=arr;
    funzione(p,sizeof(arr)/sizeof(arr[0]));      //The first function
    for(i=0;i<(sizeof(arr)/sizeof(arr[0]));i++)  //print of the single pointer (p)
        printf("%d\n",p[i]);
    int **g=p;                                   //the error results to be coming from p in this row
    for(i=0;i<(sizeof(arr)/sizeof(arr[0]));i++)  //print of the double pointer (g)
        printf("%d\n",g[i]);
    funzione2(g,sizeof(arr)/sizeof(arr[0]));     //The second function
    return 0;
}

My question is how can i remove this error.

Cromocon
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    "how can i remove this error." Don't try to assign a pointer to `int` to a pointer to `int *`. What are you *trying* to do here? You can't just convert pointers like that and expect everything to work. `funzione2` would be expecting the first argument to point to one or more `int *`, not to just be a cast pointer to `int`. – Thomas Jager Sep 27 '22 at 15:39

2 Answers2

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The variable p has the type int * while the initialized variable g has the type int **.

int **g=p;                                   //the error results to be

These pointer types are not compatible and there is no implicit conversion from one pointer type to another.

You need to write

int **g = &p;
for(i=0;i<(sizeof(arr)/sizeof(arr[0]));i++)  //print of the double pointer (g)
    printf("%d\n",( *g )[i]);

Pay attention to that the for loop can look the same way if you will declare the pointer g also the following way

int ( *g )[5] = &arr;
for(i=0;i<(sizeof(arr)/sizeof(arr[0]));i++)  //print of the double pointer (g)
    printf("%d\n",( *g )[i]);

Also the expression sizeof(arr)/sizeof(arr[0]) has the unsigned integer type size_t. So it would be better to declare the variable i also as having the unsigned type size_t instead of the signed type int.

size_t i;

Though it is even much better to declare the variable i in the minimal scope where it is used as for example

for( size_t i=0;i<(sizeof(arr)/sizeof(arr[0]));i++)  //print of the double pointer (g)
    printf("%d\n",( *g )[i]);

And this called function

funzione2(g,sizeof(arr)/sizeof(arr[0]));     //The first function

is not the first function as it is written in the comment.:)

Vlad from Moscow
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Note: This is not a rigorous answer but it is enough for you to continue and solve your compiler issue.

See, p has type int* while g is of type int**. They are incompatible as the error clearly states. p is a pointer to int while g is a point to a pointer to int. If you do int **g=&p; the compiler will stop complaining because the types now match.

The & operator takes the address of its argument (think of it like getting a pointer out of a variable). If you apply & to a pointer than you are taking a pointer out p which is a pointer already and this is how you end up with a pointer to pointer.

I suggest you to get a good C book and study pointers properly.

Davide Spataro
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