Значения элементов массива, адреса элементов и приращение указателя
Это очень основная концепция. Но я запутался в общем. Помоги мне.
Случай 1:
Для приведенного ниже кода:
int oneDArray[] = {1,2,3};
cout<<&oneDArray<<endl;
cout<<oneDArray<<endl;
cout<<&oneDArray+1<<endl;
cout<<oneDArray+1<<endl;
Выход:
0x28fef4
0x28fef4
0x28ff00
0x28fef8
Почему есть разница в увеличенных значениях?
Случай 2:
int arr[2][3] = {{1,2,3}, {4,5,6}};
cout<<&arr<<endl;
cout<<arr<<endl;
cout<<*arr<<endl;
cout<<&arr+1<<endl;
cout<<arr+1<<endl;
cout<<*arr+1<<endl;
Выход:
0x28fee8
0x28fee8
0x28fee8
0x28ff00
0x28fef4
0x28feec
Почему вывод такой же для обр & * Обр? (Как это работает внутри)
Почему есть разница в увеличенных значениях?
Решение
Для вашего одномерного массива:
//starting at memory address: 0x28FEF4
int My_Array[3] = {1,2,3};
//memory dumping, each 'int' element takes 4 bytes,
//high bytes are at higher addresses
0x28FEF4 01 00 00 00
0x28FEF8 02 00 00 00
0x28FEFC 03 00 00 00
Это перевод с языка C на английский язык:
My_Array = (address) of-the (first-byte) of (My_Array)
My_Array[0] = (value) of-the (element-at-index-0)
My_Array[k] = (value) of-the (element-at-index-k)
&My_Array = (address) of-the (first-byte) of (My_Array)
&My_Array[0] = (address) of-the (first-byte) of (element-at-index-0)
&My_Array[k] = (address) of-the (first-byte) of (element-at-index-k)
*My_Array = (value) pointed-by (My_Array)
*My_Array[0] = (value) pointed-by (My_Array[0])
*My_Array[k] = (value) pointed-by (My_Array[k])
The values of 'My_Array' and '&My_Array' are always identical for array type
but there's a difference when using them with mathematical operators.
My_Array is a pointer to 'int' type, while &My_Array is a pointer
to 'int[3]' type. When incremented, the values added in are sizeof(int)
and sizeof(int[3]) respectively.
The last two notations: *My_Array[0] and *My_Array[k]
are valid in mathematical sense but not valid in C language when using
with single-dimension array because C langugae doesn't allow using
'int' value as pointer.
Пояснение к выходному результату вашего случая 1:
(&My_Array) is identical to (My_Array) =
0x28FEF4
(My_Array) is identical to (&My_Array) =
0x28FEF4
(&My_Array+1) = 0x28FEF4 + sizeof(int[3]) = 0x28FEF4 + 12 =
0x28FF00
(My_Array+1) = 0x28FEF4 + sizeof(int) = 0x28FEF4 + 4 =
0x28FEF8
Примените мои заметки выше к вашему второму делу, со следующим вниманием:
(1) Value of element in single dimension array is the base type. Eg:
int My_Array[3]; //base type is 'int'
(2) Value of element in multiple dimension array is either pointer
or base type. Eg:
int My_Array[2][3][4]; //base type is 'int'
My_Array is a pointer to 'int' type
My_Array[i] is a pointer to 'int' type
My_Array[i][j] is a pointer to 'int' type
My_Array[i][j][k] is an 'int' value
&My_Array is a pointer to 'int[2][3][4]'
&My_Array[i] is a pointer to 'int[3][4]'
&My_Array[i][j] is a pointer to 'int[4]'
&My_Array[i][j][k] is a pointer to 'int'
(3) Values of 'My_Array' and '&My_Array' are always equal,
however, when dealing with operators, they work differently.
Eg.
int My_Array[2][3][4]; //base type is 'int'
'&My_Array' here is a pointer to 'int[2][3][4]', but
'My_Array' is a pointer to base type 'int'.
(4) A pointer is always incremented by the size of the type
that it points to. 'int' has size of 4 bytes.
Eg.
int My_Array[2]; //type size = (2)*4
int My_Array[2][3]; //type size = (2*3)*4
int My_Array[2][3][4]; //type size = (2*3*4)*4
int My_Array[2][3][4][5]; //type size = (2*3*4*5)*4
Другие решения
Других решений пока нет …