首先，SIZE被声明为一个整数

  final   int  SIZE =  5 ; 

但是你试图将它用作两者for循环行中的整数：

  for （ int  i =  0 ; i< = SIZE; i ++）; 

以分号结尾，所以 i 超出范围，以后不能使用 - 并且会尝试访问不存在的元素：当你声明一个 n $的数组时c $ c>元素，索引开始为零，因此您可以访问元素[0]，[1]，...，[n - 1]但不能访问[n]，因为它不存在。

并且还使用SIZE作为后面一行中的数组：



 SIZE [i] = i; 

它不能同时出现！

下一行没问题！

  int  A [] [] = {{ 1 ， 2 }，{ 3 ， 4 }}; 

然后再次出错，使用错误的语法：

 A [ 1 ， 1  ] =  5 ; 

试试这个：

  final   int  SIZE =  5 跨度>; 
 int  data [] =  new   int  [SIZE]; 
 for （ int  i =  0 ; i< SIZE; i ++）{
 data [i] = i; 
} 
 int  A [] [] = {{ 1 ， 2 }，{ 3 ， 4 }}; 
 A [ 1 ] [ 1 ] =  5 ; 

  for （ int  i =  0 ; i< = SIZE; i ++）;  //   由于此处有分号，  
 SIZE [i] = i;  //   此行不在循环中  

final int SIZE = 5;
for (int i = 0; i <= SIZE; i++);
SIZE [i] = i;
int A [] [] = {{1,2},{3,4}};
A [1,1] = 5;

What I have tried:

I have tried using a different variable

Firstly, SIZE is declared as an integer

final int SIZE = 5;

But you are trying to use it as both an integer in your for loop line:

for (int i = 0; i <= SIZE; i++);

Which ends at the semicolon, so i goes out of scope and can't be used later - and would try to access elements that don't exist: when you declare an array of n elements, the indexes start are zero, so you can access elements [0], [1], ..., [n - 1] but not [n] as it doesnt exist.
And also use SIZE as an array in the line after it:

SIZE [i] = i;

It can't be both at the same time!
The next line is fine!

int A [] [] = {{1,2},{3,4}};

But then you go wrong again, by using it with the wrong syntax:

A [1,1] = 5;

Try this:

final int SIZE = 5;
int data[] = new int[SIZE];
for (int i = 0; i < SIZE; i++){
    data[i] = i;
}
int A[][] = { {1, 2}, {3, 4} };
A[1][1] = 5;

for (int i = 0; i <= SIZE; i++);  // Because of semicolon here,
SIZE [i] = i;  // this line is not in the loop