[英]C programming 2D array memory layout
Consider the c code below.考虑下面的 c 代码。
#include <stdio.h>
int main(){
unsigned int x[4][3] = {{1, 2, 3}, {4, 5, 6},
{7, 8, 9}, {10, 11, 12}};
printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
printf("\n%u, %u, %u", x,&x,*x);
return 0;
}
Now each of the printf statement prints the same values as mentioned below.现在,每个 printf 语句都打印出与下面提到的相同的值。
6356724, 6356724, 6356724
6356688, 6356688, 6356688
Process returned 0 (0x0) execution time : 0.128 s
Press any key to continue.
I wanted to know how is the outcome same in each printf statements.我想知道每个 printf 语句的结果如何相同。 This is my understanding of memory layout of 2-D array.
这是我对二维数组内存布局的理解。
Now I believe x+3 refers to base address of( x + 3 * size require for pointer arithmetic ) = 24 and *(x +3 ) = *(24) = 2036 but its not the case the first printf statement prints 2036, 2036 2036 .现在我相信 x+3 是指 ( x + 3 * 指针运算所需的大小 ) = 24 和 *(x +3 ) = *(24) = 2036 的基地址,但不是第一个 printf 语句打印 2036 的情况, 2036 2036。 I wanted to visually understand how exactly 2-D array is organized.
我想直观地了解二维数组是如何组织的。
Whenever an array is mentioned, it decays to a pointer to its first element (except in a few cases that are not interesting right now).每当提到一个数组时,它都会衰减到指向其第一个元素的指针(除了一些现在不感兴趣的情况)。
Why x
, &x
and *x
are all printed the same?为什么
x
, &x
和*x
都打印相同? That's because the array itself, its first element, (and if it's a 2D array, the first element of its first element, which is by itself a 1D array) all have the same address.这是因为数组本身,它的第一个元素(如果它是一个二维数组,它的第一个元素的第一个元素,它本身就是一个一维数组)都具有相同的地址。
So:所以:
&x
is the pointer to the array &x
是指向数组的指针x
is the array itself, which decays to the pointer of its first element x
是数组本身,它衰减到其第一个元素的指针*x
is the first element of the array, which is itself an array (1, 2, 3}
, so it decays to the pointer of its first element *x
是数组的第一个元素,它本身就是一个数组(1, 2, 3}
,所以它衰减到它的第一个元素的指针All these pointers have different types, but point to the same address.所有这些指针都有不同的类型,但指向相同的地址。
The same thing happens with x+3
and *(x+3)
.同样的事情发生在
x+3
和*(x+3)
。 First, x
decays to a pointer, then pointer arithmetic is performed.首先,
x
衰减为一个指针,然后执行指针运算。 Say y = x + 3
, so y
and *y
point to the same address just as x
and *x` do (see above).假设
y = x + 3
,因此y
和*y
指向与x
和 *x` 相同的地址(见上文)。
What about *(x+2)+3
? *(x+2)+3
呢? Well, *(x + 2)
is itself is an array of 3 elements.好吧,
*(x + 2)
本身就是一个包含 3 个元素的数组。 So *(x+2)+3
points to one element past the end of the *(x + 2)
array.所以
*(x+2)+3
指向超过*(x + 2)
数组末尾的一个元素。 This non-existing element has an address nevertheless , and it is the same as the address of the first element of the *(x + 3)
array --- because *(x + 2)
and *(x + 3)
are next to each other in the x
array.这个不存在的元素仍然有一个地址,它与
*(x + 3)
数组的第一个元素的地址相同 --- 因为*(x + 2)
和*(x + 3)
是下一个在x
数组中相互关联。
I'll try to explain from a layman's point of view.我将尝试从外行的角度进行解释。
Let us assume the base address is 3000, and assume that integer takes 4 bytes.让我们假设基地址是 3000,并假设整数占用 4 个字节。
Considering your 2-D array, this is how WE visualise a 2-D array in the memory :考虑到您的二维数组,这就是我们在内存中可视化二维数组的方式:
0 1 2
--------------------------------------------------
0 | 1 | 2 | 3 |
| 3000/3001/3002/3003| 3004/05/06/07 | 3008/09/10/11|
|___________________ |________________|______________|
1 | 4 | 5 | 6 |
| 3012 | 3016 | 3020 |
|____________________|________________|______________|
2 | 7 | 8 | 9 |
| 3024 | 3028 | 3032 |
|____________________|________________|______________|
3 | 10 | 11 | 12 |
| 3036 | 3040 | 3044 |
| | | |
------------------------------------------------------
Now, you should know how a 2-D array is actually stored just like a 1-D array in the memory:现在,您应该知道二维数组实际上是如何像一维数组一样存储在内存中的:
index 0 1 2 3 4 5 6 7 8 9 10 11
---------------------------------------------------------------------------------
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| | | | | | | | | | | | |
add. |3000__|_3004_|_3008_|_3012_|_3016_|_3020|3024__|_3028_|3032__|_3036_|_3040|3044_ |
Now, your code says:现在,您的代码说:
printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3);
(i) x+3 (一) x+3
Here x is your base address, which is 3000(assumed), and +3 means the 3rd row.这里 x 是你的基地址,它是 3000(假设),+3 表示第三行。 Since there is nothing after that,ie, x+3, we will find out the address of the first element of the 3rd row.
由于之后没有任何内容,即x+3,我们将找出第3行第一个元素的地址。
=3000+36=3036 =3000+36=3036
(refer the 2D array visualisation that WE see, the first one) (参考我们看到的二维数组可视化,第一个)
(ii) *(x+3) (ii) *(x+3)
This is the same as x+3, ie, we have to find the address of the first element of the third row.这与 x+3 相同,即我们必须找到第三行的第一个元素的地址。 I say the first element because if something was mentioned after *(x+3)+'something', then we would be concerned about finding it.
我说第一个元素是因为如果在 *(x+3)+'something' 之后提到了什么,那么我们会担心找到它。
=3000+36=3036. =3000+36=3036。
(refer the 2D array visualisation that WE see, the first one) (参考我们看到的二维数组可视化,第一个)
(iii) *(x+2)+3 (iii) *(x+2)+3
Just for understanding , let us break this into two parts.只是为了理解,让我们把它分成两部分。
First, *(x+2): as we have seen in the above two parts, *(x+2) says that "I have started from the base address, now give me the address of the first element of the 2nd row. So, we reach element with value 7 and address 3024 . (refer the 2D array visualisation that WE see, the first one)首先,*(x+2):正如我们在上面两部分中看到的,*(x+2)表示“我已经从基地址开始,现在给我第2行第一个元素的地址。因此,我们到达值为 7 且地址为 3024 的元素。 (参考我们看到的二维数组可视化,第一个)
Second, +3 as in *(x+2)+3 means that since we have reached element with value 7 and address 3024, now see/move to the 3rd element after the element with value 7 and address 3024 (where we are right now).其次,*(x+2)+3 中的 +3 表示由于我们已经到达值为 7 且地址为 3024 的元素,现在查看/移动到值为 7 且地址为 3024 的元素之后的第三个元素(我们是对的)现在)。 So, we move past/see-through elements with values 8 and 9, with addresses 3028 and 3032 respectively, and reach/land on the element with value 10 and address
因此,我们移动过去/透视元素的值为 8 和 9,地址分别为 3028 和 3032,并到达/降落在值为 10 和地址的元素上
=3000+36=3036 =3000+36=3036
(refer the 2D array visualisation that WE see, the first one) (参考我们看到的二维数组可视化,第一个)
For your second code:对于您的第二个代码:
printf("\n%u, %u, %u", x,&x,*x);
I go with @ n.我和@n一起去。 'pronouns' m.
'代词' m. answer.
回答。
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