英文:
When creating a single pointer array for representing a 2D array, what is the size of the pointer array? Is it row or column of the 2D array?
问题
这段代码是我的老师给的，它运行正常，但我对为什么是 int (*x1)[3]; 而不是 int (*x1)[2];Feeling confused，因为 y 只有2行。我认为这个数组 x1 的每个元素将存储到 y 的行的地址，所以只需要两个这样的元素，而不是3个。请帮我解释一下。谢谢
int main(){
int (*x1)[3];
int y[2][3]={{1,2,3},{4,5,6}};
x1 = y;
for (int i = 0; i<2; i++)
for (int j = 0; j<3; j++)
printf("\n The X1 is %d and Y is %d",*(*(x1+i)+j), y[i][j]);
return 0;
}
我尝试运行了这段代码，它运行正常。但我不明白它是如何工作的。内部发生了什么？内存是如何分配的？
英文:
So, this code was given by my teacher and it works fine, but I am confused as to why it is int (*x1)[3]; and not int (*x1)[2]; as y has 2 rows only. I think that each element of this array x1 will store the address to the row of y so only two such elements are required and not 3. Kindly help me. Thank you
int main(){
int (*x1)[3];
int y[2][3]={{1,2,3},{4,5,6}};
x1 = y;
for (int i = 0; i<2; i++)
for (int j = 0; j<3; j++)
printf("\n The X1 is %d and Y is %d",*(*(x1+i)+j), y[i][j]);
return 0;
}
I tried running this code and it is working fine. I don't understand how it is working though. Like what is going on internally? How is the memory being allocated?
答案1
得分: 1
int y[2][3]={{1,2,3},{4,5,6}};
每个`y`中的元素都是`int[3]`，`y`有2个这样的元素。这声明了一个指向这种元素的指针：
int (*x1)[3];
而在这里，`y`会衰减为指向第一个元素的指针，这就是为什么赋值有效的原因：
x1 = y;
建议稍微修改以使其更容易阅读：
x1 = y;
for (int i = 0; i < 2; ++i, ++x1) // 在这里步进 x1
for (int j = 0; j < 3; ++j)
printf("\n X1 是 %d 而 Y 是 %d", (*x1)[j], y[i][j]);
// 并且在这里对其进行解引用变得更加清晰 ^^^^^^^^
英文:
int y[2][3]={{1,2,3},{4,5,6}};
Every element in y is an int[3] and y has 2 such elements. This declares a pointer to one such element:
int (*x1)[3];
And in this, y decays into a pointer to the first element which is why the assignment works:
x1 = y;
Suggestion for making it a little easier to read:
x1 = y;
for (int i = 0; i < 2; ++i, ++x1) // step x1 here
for (int j = 0; j < 3; ++j)
printf("\n The X1 is %d and Y is %d", (*x1)[j], y[i][j]);
// and dereferencing it becomes nicer here ^^^^^^^^
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