Pointer issue in C - c

Question

Greetings,
I am trying to learn pointers in C, I simply want my "addtwo" function to add 2 to every element of the input integer array, yet I get odd compilation errors, here is the non-pointer version which indeed won't properly compile.
addtwo(int *arr[]) {
int i=0;
for(;i< sizeof(arr)/sizeof(int);i++) {
arr[i] = arr[i] + 2;
}
}
main() {
int myarray[] = {1,2,3,4};
addtwo(myarray);
}
Regards

Answer 1

You've some problems. First, you try to pass a int* to a parameter that's type int**. That won't work. Give it type int*:
void addtwo(int *arr){
int i=0;
for(;i< sizeof(arr)/sizeof(int);i++){
arr[i] = arr[i] + 2;
}
}
Then, you need to pass the size in an additional argument. The problem is, that when you pass arrays, you really pass just a pointer (the compiler will make up a temporary pointer that points to the array's first element). So you need to keep track of the size yourself:
void addtwo(int *arr, int size){
int i=0;
for(;i<size;i++){
arr[i] = arr[i] + 2;
}
}
int main(void) {
int myarray[] = {1,2,3,4};
addtwo(myarray, sizeof myarray / sizeof myarray[0]);
}
Now it will work. Also put the return type before them. Some compilers may reject your code, since it doesn't comply to the most recent C Standard anymore, and has long been deprecated (omitting the return type was the way you coded with the old K&R C).

Answer 2

addtwo(int *arr[]) should be addtwo(int *arr)
You cannot use sizeof to get the size of an array from a pointer. Typically you would either pass the size of the array as a separate arg or have some special value marking the last element.

Answer 3

Not to do with the compile error, but...
You have to pass sizeof(arr) to the function instead of calling it in the function. When an array is passed to a function, C no longer sees it as an array, but as a single pointer to memory, so that sizeof(arr) as you are calling it now, will return the size of the pointer arr, which is most likely 4.
Here's what I mean in code:
void addtwo(int *arr, int size){
int i=0;
for(;i< size;i++){
arr[i] = arr[i] + 2;
}
}
int main(){
int myarray[] = {1,2,3,4};
addtwo(myarray, sizeof(arr)/sizeof(int));
return 0;
}

Answer 4

In C a notation int *arr[] is the same as int** arr.

Answer 5

You need to pass a pointer to the first element of the array and the array size. Array types decay to pointers in the context of function parameters. Try:
void addtwo(int *arr, size_t size){
for(size_t i = 0; i < size; i++){
arr[i] = arr[i] + 2;
}
}
int main() {
int v[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 };
addtwo(v, sizeof v / sizeof v[ 0 ]);
return 0;
}

Answer 6

Though others already gave the correct response, basically you have an array of pointers when you have
int *arr[]
I doubt that is what you want. If you have
int arr[]
then that will also be equivalent to
int *arr

Answer 7

addtwo argument declaration really reads:
arr is an array of pointers to integer
when you probably really want
a pointer to an array of integers
"How to Read C Declarations" has really helped me to grok the topic a while ago, maybe it will do the same for you.