在這個二維數組中可以存儲多少個整數？

Question

我正在爲我的考試而學習，而且我遇到了一個我無法解決的問題。
這裏的問題是「在運行代碼之後，可以在二維數組」arr「中存儲多少個不同的int值？」

int[][] arr = new int[3][]; 
arr[0] = new int[5]; 

for (int i = 1; i < arr.length; i++) 
{ 
arr[i] = arr[i-1]; 
} 

我認爲答案是如圖7所示，這裏的原因：
您創建2-d陣「改編」，並立即宣佈，它在1名3個維開值。
然後你說第一個打開的值本身就是一個數組，它包含5個開放值。
最後，for-loop指出原始數組的第二個值成爲第一個值，原始數組的第三個值也是相同的。 （這些值沒有初始化，所以它歸結爲0成爲0，因爲這是一個整數的標準值），這也是7

儘管答案應該根據我的書是5

，和我找不到原因。

在此先感謝！

Answer 1

這是5，因爲每個外陣列的3個位置的包含到相同的5元件陣列的引用。所以執行代碼在你的問題，如果一個人要做到這一點後：

arr[0][0] = 5; 

那麼下面將舉行：如果你說the for-loop says that the second value of the original array becomes the first value and the same goes for the third value of the original array. (These values weren't initialized, so it boils down to 0 becoming 0 since that's the standard value for an integer)

arr[0][0] == 5; // true 
arr[1][0] == 5; // true 
arr[2][0] == 5; // true 

你是正確的大多隻是arr[0]被初始化。它被設置爲new int[5]。

在您的文章的代碼運行在此之後是數組的樣子：

即使數組是int[][]型，外陣列中的3個元素不持有int小號，他們持有對int[] s的引用。

Answer 2

說明直列，帶註釋

/* Declare a two dimensional array, but only specify the first dimension 
* of '3'. This effectively leaves three arrays of undefined length 
* to which 'arr' can point. 
*/ 
int[][] arr = new int[3][]; 

/* Declare that arr[0] - the first element in the outer dimension 
* is a array of length 5. This creates 5 locations in an array of 
* length 5 into which integers can be stored. 
*/ 
arr[0] = new int[5]; 

/* No further declarations creating new space to store data in arr. 
* With no additional operations, below, to allocate memory, there 
* may never be more than 5 locations in arr to store anything. 
* This means that the rest of the question, as written, is like the 
* extra garbage you sometimes get in word problems, to try and confuse 
* you 
* 
* "A train, carrying 1000 apples, is traveling from Des Moines to Boston 
*  at 90mph. B train, carrying 1 apple, is traveling from Boston to Des Moines 
*  at 30mph on exactly teh same track. At what point on the track do they 
*  crash?" 
* 
* The apples are unnecessary to the problem at hand. 
*/ 

/* Iterate through arr - the outer array of length 3. */ 
for (int i = 1; i < arr.length; i++) 
{ 
    /* Set the current value of arr[i] to the value stored at arr[i-1]. 
    * Remember what each value of arr[] is before entering the loop... 
    * arr[0] = an array of length 5, whose values are not yet explicitly set 
    * arr[1] = null 
    * arr[2] = null 
    */ 
    arr[i] = arr[i-1]; 
    /* The first time we get this far, arr[1] will have been set to arr[0]... 
    * arr[0] = array length 5 
    * arr[1] = the same length 5 array pointed to by arr[0] 
    * arr[2] = null 
    * 
    * The second time we get this far, arr[1] will have been set to arr[0]... 
    * arr[0] = array length 5 
    * arr[1] = the same length 5 array pointed to by arr[0] 
    * arr[2] = the same length 5 array pointed to by arr[0] 
    */ 
}