I have seen that the array values will change if the function parameter is "int arr[]" or "int * arr". Where is the difference?
int array[]:
void myFunction(int arr[], int size) {
for (int i = 0; i < size; ++i)
arr[i] = 1;
}
int * array:
void myFunction(int * arr, int size) {
for (int i = 0; i < size; ++i)
arr[i] = 1;
}
Both functions change the array values.
int main(){
int array[3];
array[0] = 0;
array[1] = 0;
array[2] = 0;
myFunction(array, 3);
return 0;
}
There is no difference. Both functions types (after adjustment) are "function taking a pointer to int
and an int
, returning void
." This is just a syntactic quirk of C++: the outermost []
in a function parameter of non-reference type is synonymous with *
.
表达同一事物的不同方式。您只是通过指针将数组传递给函数。
when you pass an array to functions it implicitly passes it by pointer. because if there is many elements in array pass by value copying it is a Huge overhead. even-though it looks different its same.
you can't use both functions at same time. if you do its compile time error
error: redefinition of 'void myFunction(int*, int)'
it is already defined.
There are no difference. In fact, a declaration of an array return allawys a pointer. Only, when you specify in the declaration that the variable is an array (ie. with []), it is not necessary to specify that it will be a pointer, because this is made implicitly. But when you do not specify this, you must declare it as a pointer if you went affect it a table variable, or a result of "new []". The use of pointers of table has an interest only for a dynamic allocation, when the array size is not known on design-time.
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