问题

为了创建一个用随机数填充动态分配数组的方法，我使用了以下方法：

void autoFill(long int *fillArr[], int length) {
srand(1024);
for (int i = 0; i < length; i++) {
fillArr[i] = rand();
}
}

然而，`fillArr[i] = rand();` 不起作用，因为这行试图将 `int` 转换为 `long *`。您应该如何赋值呢？

<details>
<summary>英文:</summary>

To make a method for filling dynamically assigned array with random numbers, I used following method:

void autoFill(long int *fillArr[], int length) {
srand(1024);
for (int i = 0; i < length; i++) {
fillArr[i] = rand();
}
}

However, `fillArr[i] = rand();` doesn't work because this line tries to convert `int` to `long *`. How do I assign this?

</details>


# 答案1
**得分**: 0

I assume you have an 1D array that stores long int, you pass the argument in the wrong way
```cpp
#include <iostream>
#include <stdlib.h> 
#include <ctime>

void auto_fill(long int *arr, int length) {
    srand(1024);
    for (int i = 0; i < length; i++) {
        arr[i] = rand();
    }
}

int main()
{
    long int *myarray;
    int length = 20;
    myarray = new long int[length];
    auto_fill(myarray, length);
    for (int i = 0; i < length; i++)
        std::cout << myarray[i] << std::endl;
}

please note that rand() returns an int, not a long int

#include <iostream>
#include <stdlib.h> 
#include <ctime>

void auto_fill(long int *arr, int length) {
	srand(1024);
	for (int i = 0; i < length; i++) {
		arr [i]=rand();
	}
}

int main()
{
	long int *myarray;
	int length=20;
	myarray = new long int[length];
	auto_fill(myarray, length);
	for (int i=0;i<length;i++)
		std::cout << myarray [i]<<std::endl;
}

答案2

得分: -3

看起来我不得不写*(fillArr[i]) = rand();而不是fillArr[i] = rand();现在又出现了另一个问题。