运营商[] [英] Operator [ ]
问题描述
直到最近我掌握了指针和数组之间的关系和
我终于明白[]是一个运算符。但是我还是不能确定这一点：在一个声明中真正做到了[]：
int array [5];
int matrix [5] [5];
dArray = new int [];
dMatrix = new int * [];
此外，这是否有效（有用？）：
int x;
int * array =& x ;
for（int i = 0; i< some_number; ++ i）
array [i] = //一些东西
Until recently I grasp the relationship between pointers and arrays and
I have finally understand that [ ] is an operator. But I still can''t
quite figure it out: What does really does [ ] in a statement such as:
int array[5];
int matrix[5][5];
dArray = new int[];
dMatrix = new int*[];
Also, is this valid (useful?):
int x;
int *array=&x;
for(int i=0; i<some_number; ++i)
array[i] = //some stuff
推荐答案
Gaijinco写道：
Gaijinco wrote:
直到最近我才掌握指针和数组之间的关系和
我终于明白[]是一个运营商。但是我仍然无法弄明白：在一个声明中真正做到了[]：
int array [5];
int matrix [5 ] [5];
dArray = new int [];
dMatrix = new int * [];
这里的括号是语法的一部分。但是，它呈现的形式
它不会编译，括号内必须有一个整体表达式
。
此外，这是否有效（有用？）：
int x;
int * array =& x;
for（int i = 0; i< some_number; ++ i）
array [i] = //一些东西
Until recently I grasp the relationship between pointers and arrays and
I have finally understand that [ ] is an operator. But I still can''t
quite figure it out: What does really does [ ] in a statement such as:
int array[5];
int matrix[5][5];
dArray = new int[];
dMatrix = new int*[];
The brackets here are part of the syntax. However, it the presented form
it isn''t going to compile, there has to be an integral expression inside
the brackets.
Also, is this valid (useful?):
int x;
int *array=&x;
for(int i=0; i<some_number; ++i)
array[i] = //some stuff
不，它对于除'0'以外的''i'的值无效（它有未定义的
行为），因此在正常编程中没用。
V
No, it''s not valid for values of ''i'' other than 0 (it has undefined
behaviour), and therefore not useful in normal programming.
V
>这里的括号是语法的一部分。但是，它呈现的形式是
> The brackets here are part of the syntax. However, it the presented form
它不会编译，括号内必须有一个整体表达式。
it isn''t going to compile, there has to be an integral expression inside
the brackets.
我很少迷路！
当我使用类似的东西时：
int array [5];
我说'寻找5块空闲内存来分配int'和
创建一个指针数组到第一个街区是真还是假？
所以，当我说'
int matrix [5] [5];
它看起来是25块而不是5块，但是我不明白[5] [5]
的确意味着什么？
如果这就是它的工作方式那么它意味着什么：
那么
int array [5]
与
相同
int * array = new int [5]
？
I''m a little lost!
When I use something like:
int array[5];
I''m saying "look for 5 block of free memory to allocate int''s and
create a pointer "array" to the first block". True or false?
So when I''m saying
int matrix[5][5];
It looks for 25 blocks instead of 5, but I cn''t understand what [5][5]
really means?
And if that''s the way it works then what it means:
Then
int array[5]
is the same as
int* array = new int[5]
?
>这里的括号是语法的一部分。但是，它呈现的形式是
> The brackets here are part of the syntax. However, it the presented form
它不会编译，括号内必须有一个整体表达式。
it isn''t going to compile, there has to be an integral expression inside
the brackets.
我很少迷路！
当我使用类似的东西时：
int array [5];
我说'寻找5块空闲内存来分配int'和
创建一个指针数组到第一个街区是真还是假？
所以，当我说'
int matrix [5] [5];
它看起来是25块而不是5块，但是我不明白[5] [5]
的确意味着什么？
如果这就是它的工作方式那么
int array [5]
与
相同
int * array = new int [5]
？
I''m a little lost!
When I use something like:
int array[5];
I''m saying "look for 5 block of free memory to allocate int''s and
create a pointer "array" to the first block". True or false?
So when I''m saying
int matrix[5][5];
It looks for 25 blocks instead of 5, but I cn''t understand what [5][5]
really means?
And if that''s the way it works then
int array[5]
is the same as
int* array = new int[5]
?
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