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