‘sizeof’ on array function parameter ‘arr’ will return size of ‘int*’ [-Wsizeof-array-argument]

#include <iostream>
using namespace std;
int sum(int arr[]){
	int length = sizeof(arr)/sizeof(arr[0])
    int sum = 0;
    for (int i = 0 ; i < length; i++){
        sum += arr[i];
    }
    return sum;
}
int main(){
    int arr[] = {1,3,5,7,9};
    //cout<<sizeof(arr)/sizeof(int);
    int summary = sum(arr);
    cout << summary << endl;
}

原因是数组作为参数传给函数时，是传给数组的地址，而不是传给整个的数组空间，因而
sizeof(arr)这句话会报错
正确的用法是：不在函数内部使用sizeof

#include <iostream>
using namespace std;
int sum(int arr[], int length){
	int length = sizeof(arr)/sizeof(arr[0])
    int sum = 0;
    for (int i = 0 ; i < length; i++){
        sum += arr[i];
    }
    return sum;
}
int main(){
    int arr[] = {1,3,5,7,9};
    //cout<<sizeof(arr)/sizeof(int);
    int length = sizeof(arr)/sizeof(arr[0])
    int summary = sum(arr,length);
    cout << summary << endl;
}