blocks|key|370495|text|您可以通过将其封装在一个类中来实现您想要的效果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|370496|class+Board+{
public:
++++int&+sec1(int+r,+int+c)+{+return+board[r][c]);+}
++++int&+sec2(int+r,+int+c)+{+return+board[r][c%2B3];+}
++++//+etc.

private:
++++int+board[9][9];
};|code-block|syntax|javascript|370497|但是，我不确定这是代表数独棋盘的最好方式。你可能会发现，一旦你开始研究逻辑，你就会得到一个更好的表示。|370498|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could achieve the effect you want by encapsulating it in a class:

<pre><code>class Board {
public:
 int&amp; sec1(int r, int c) { return board[r][c]); }
 int&amp; sec2(int r, int c) { return board[r][c+3]; }
 // etc.

private:
 int board[9][9];
};
</code></pre>

but, I'm not sure this is the best way to represent a Sudoku board. You may find that once you start working on the logic, you'll come up with a better representation.

blocks|key|3110693|text|这是行不通的，就像写的那样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3110694|联合的行为就像其所有成员都位于距联合起始位置的偏移量0处一样。|3110695|这意味着sec9将与sec1具有相同的偏移量(零)，因此会重叠。|offset|length|style|CODE|3110696|我认为使用联合也不可能做到这一点，因为您需要表示在每个部分结束后需要发生特定的“跳过”，以到达该部分中的下一个单元格。您不能使用C或C%2B%2B数组来做到这一点。|3110697|entityMap^0|0|0|4|4|A|4|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@$F|Q|G|R|H|I]|$F|S|G|T|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

This won't work, as written.

A union behaves as if all its members are at offset 0 from the union's start.

That means that <code>sec9</code> will have the same offset (zero) as <code>sec1</code>, thus overlap.

I don't think it's possible to do it using a union either, since you would need to express that there's a certain "skip" that needs to happen after the end of each section, to get to the next cell in that section. You can't do that using C or C++ arrays.

blocks|key|3110822|text|答案是否定的，内存布局将不会是您所期望的。请注意，无论何时在C/C%2B%2B中定义数组，内存都是连续的，因此对于9x9数组，第四个元素不是第二行的第一个元素，而是第二个3x3块的第一行的第一个元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3110823|您的联合的内存布局将在full对象的9行中的每一行上都有sec块。|offset|length|style|CODE|3110824|entityMap^0|0|B|4|S|3|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

The answer is no, the memory layout will not be what you expect. Note that whenever you define an array in C/C++ the memory is contiguous, so for the 9x9 array the 4th element is not the first element of the second row, but rather the first element of the first row of the second 3x3 block.

The memory layout of your union will have the <code>sec</code> blocks over each one of the 9 lines of the <code>full</code> object.

blocks|key|370581|text|每个联盟成员，因此每个sec1..sec9非常秒将在相同的位置。您可以尝试将所有sects包装在一个struct中，但它们仍然不会对应于3x3正方形，而是原始结构中的9*1行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|370582|union+Board
{
++++int+board[9][9];
++++struct+{
++++++++int+sec1[3][3];
++++++++int+sec2[3][3];
++++++++int+sec3[3][3];
++++++++int+sec4[3][3];
++++++++int+sec5[3][3];
++++++++int+sec6[3][3];
++++++++int+sec7[3][3];
++++++++int+sec8[3][3];
++++++++int+sec9[3][3];+++++
++++}+sects;

}|code-block|syntax|javascript|370583|总而言之，真正的类将是最好的方法。|370584|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Every union member, thus every sec1..sec9 very sec will be in the same location. You could try wrap all sects in a struct, but still they will not correspond to 3x3 squares, but rather 9*1 rows in original structure:

<pre><code>union Board
{
 int board[9][9];
 struct {
 int sec1[3][3];
 int sec2[3][3];
 int sec3[3][3];
 int sec4[3][3];
 int sec5[3][3];
 int sec6[3][3];
 int sec7[3][3];
 int sec8[3][3];
 int sec9[3][3]; 
 } sects;

}
</code></pre>

To sum up, real class would be the best approach.

blocks|key|1702155|text|您不能仅仅使用C%2B%2B语言特性来解决这个问题--您需要考虑解决问题所需的数据结构和该结构上的操作--换句话说，您需要设计一个类(或者更有可能是几个类)。这是编程中最有趣的部分，所以我不打算提出解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1702156|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You are not going to solve this purely using C++ language features - you need to think about the data structure and operations on that structure that are needed to solve your problem - in other words you need to design a class (or more likely several classes). This is the fun part of programming, so I'm not going to suggest a solution.

blocks|key|1118921|text|另一种解决方案(除了Ferruccio提出的方案)可以是定义指向in的3个指针的数组-每个段一个，并在构造函数中相应地初始化这些数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1118922|class+Board+{
public:
++++int+*sec1[3];+//+sec1[0]+=+&(board[0][0]),+sec1[1]+=+&(board[1][0]),sec1[2]+=+&(board[2][0])
++++...

++++int+board[9][9];
};|code-block|syntax|javascript|1118923|但坦率地说，访问方法可能要好得多。|1118924|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another solution (besides the one proposed by Ferruccio) could be to define arrays of 3 pointers to ints - one for each section, and initialise those arrays accordingly in the constructor.

<pre><code>class Board {
public:
 int *sec1[3]; // sec1[0] = &amp;(board[0][0]), sec1[1] = &amp;(board[1][0]),sec1[2] = &amp;(board[2][0])
 ...

 int board[9][9];
};
</code></pre>

But frankly methods for access are probably much better.

blocks|key|3111018|text|这是行不通的，因为9*9电路板中的每个3*3区域无论如何都会占用不连续的内存。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3111019|我所做的是：|3111020|Cell+grid[9][9];
Cell*+cell_ptr[3][81];+//0+=+by+rows;+1+=+by+columns;+2+=+by+box|code-block|syntax|javascript|3111021|其中cell_ptr填充了指向网格的指针，因此cell_ptr[0][0...80]允许迭代行，cell_ptr[1][0...80]将迭代列，cell_ptr[2][0...80]允许迭代3*3个区域。|offset|length|style|CODE|3111022|entityMap^0|0|0|0|2|8|N|J|1C|J|20|J|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]|$K|Y|L|Z|M|N]|$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|P|$]]

That will not work, because each 3 * 3 region in the 9 * 9 board will be occupying non-contiguous memory anyway.

What I have done is:

<pre><code>Cell grid[9][9];
Cell* cell_ptr[3][81]; //0 = by rows; 1 = by columns; 2 = by box
</code></pre>

where <code>cell_ptr</code> is populated with pointers into the grid so that <code>cell_ptr[0][0...80]</code> allows iterating over rows, <code>cell_ptr[1][0...80]</code> will be iterating over columns, and <code>cell_ptr[2][0...80]</code> allows iterating over 3 * 3 regions.

I'm trying to describe a Sudoku Board in C++ with a union statement:

<pre><code>union Board
{
 int board[9][9];
 int sec1[3][3];
 int sec2[3][3];
 int sec3[3][3];
 int sec4[3][3];
 int sec5[3][3];
 int sec6[3][3];
 int sec7[3][3];
 int sec8[3][3];
 int sec9[3][3]; 
}
</code></pre>

Would each section of the board correspond with the correct part of the array? IE,

Would sec4 correspond with board[4-6][0-3]? Is there a better way to do this sort of thing (specifically describing a sudoku board)?

Incremental Union?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试着用一个联盟声明来描述C++中的数独董事会：union Board{    int board[9][9];    int sec1[3][3];    int sec2[3][3];    int sec3[3][3];    int sec4[3][3];    int sec5[3][3];    int sec6[3][3];    int sec7[3][3];    int sec

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