英文:
The same logic on an array is producing different results
问题
以下是翻译好的部分：
这是该函数：
void print(int arr[], int size)
{
int* count = new int[size];
for(int i = 0; i < size; i++){
count[arr[i]]++;
cout << count[arr[i]] << endl;
}
}
当我在主函数中调用时，print(arr, sizeof(a)/sizeof(a[0]))，其中a = {4, 2, 4, 5, 2, 3}，我将获得数组中每个数字的更新计数。
现在，假设我想在我的int main()函数中做类似的事情：
int* my_arr = new int[10];
my_arr[1]++;
cout << "my_arr[1] = " << my_arr[1];
最后一条语句会打印类似于15799121的内容，我认为这是my_arr[1]的内存地址。为什么在print()函数中不是这样呢？难道cout << count[arr[i]] << endl;不应该产生相同的结果吗？
希望这有所帮助。如果您有其他问题，请随时提出。
英文:
This is the function:
void print(int arr[], int size)
{
int* count = new int [size];
for(int i = 0; i < size; i++){
count[arr[i]]++;
cout << count[arr[i]] << endl;
}
}
When I call (in my main) print(arr, sizeof(a)/sizeof(a[0])) where a = {4, 2, 4, 5, 2, 3}, I would get the updated count of each number in the array.
Now, let's say I want to do something very similar to this but in my int main():
int* my_arr = new int[10];
my_arr[1]++;
cout << "my_arr[1] = " << my_arr[1];
The last statement prints something like 15799121, which I assume is the memory address of my_arr[1].
How is this not the same in print()? Shouldn't cout << count[arr[i]] << endl; produce the same result?
答案1
得分: 4
对于默认初始化，
否则，什么都不会发生：具有自动存储期的对象（及其子对象）将被初始化为不定值。
这意味着给定new int [size]，数组的元素将被初始化为不定值，使用这些值会导致UB，意味着任何情况都有可能发生。
你可能希望使用值初始化，即new int [size]()；所有元素将被初始化为0。
英文:
> otherwise, nothing is done: the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.
That means given new int [size], the elements of the array are initialized to indeterminate values, use of these values leads to UB, means anything is possible.
You might want value initialization, i.e. new int [size](); all the elements would be initialized to 0 exactly.
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