blocks|key|1432950|text|使用malloc(3)分配第一个数组并将malloc(3)创建的指针放入其中应该适用于array[r][c]，因为它应该等同于*(*(array+%2B+r)+%2B+c)，它是在C标准中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1432951|entityMap^0|2|9|K|9|17|B|1R|J|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

Using <code>malloc(3)</code> for allocate the first array and putting in there pointers created by <code>malloc(3)</code> should work with <code>array[r][c]</code> because it should be equivalent to <code>*(*(array + r) + c)</code>, it is in the C standard.

blocks|key|1647881|text|您的语法mline是正确的。但是为了在C中使用二维数组，你必须为它分配内存。例如，此代码将为给定行和列的表分配内存。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1647882|int**+AllocateArray(int+line,+int+column)+{
++int**+pArray+=+(int**)malloc(line*sizeof(int*));
++for+(+int+i+=+0;+i+<+line;+i%2B%2B+)+{
++++pArray[i]+=+(int*)malloc(column*sizeof(int));
++}
++return+pArray;
}|code-block|syntax|javascript|1647883|注意，为了简洁起见，我省略了malloc的错误检查。一个真正的解决方案应该包括它们。|1647884|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|$]]

Your syntax m[line][colummn] is correct. But in order to use a 2D array in C, you must allocate memory for it. For instance this code will allocated memory for a table of given line and column.

<pre><code>int** AllocateArray(int line, int column) {
 int** pArray = (int**)malloc(line*sizeof(int*));
 for ( int i = 0; i &lt; line; i++ ) {
 pArray[i] = (int*)malloc(column*sizeof(int));
 }
 return pArray;
}
</code></pre>

Note, I left out the error checks for malloc for brevity. A real solution should include them.

blocks|key|1647935|text|它不是二维数组--它是一个数组的数组--因此它需要多次分配。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1647936|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|$]]

It's not a 2d array - it's an array of arrays - thus it needs the multiple allocations.

blocks|key|1433174|text|尽管我同意其他答案，但在大多数情况下，一次分配整个数组更好，因为malloc相当慢。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1433175|int+**
array_new(size_t+rows,+size_t+cols)
{
++++int+**array2d,+**end,+**cur;
++++int+*array;

++++cur+=+array2d+=+malloc(rows+*+sizeof(int+*));
++++if+(!array2d)
++++++++return+NULL;

++++array+=+malloc(rows+*+cols+*+sizeof(int));
++++if+(!array)
++++{
++++++++free(array2d);
++++++++return+NULL;
++++}

++++end+=+array2d+%2B+rows;
++++while+(cur+!=+end)
++++{
++++++++*cur+=+array;
++++++++array+%2B=+cols;
++++++++cur%2B%2B;
++++}

++++return+array2d;
}|code-block|syntax|javascript|1433176|要释放数组，只需执行以下操作：free(*array);+free(array);|offset|length|style|CODE|1433177|注意:只有当您不想更改行的顺序时，此解决方案才有效，因为您可能会丢失第一个元素的地址，您需要它稍后释放数组。|1433178|entityMap^0|0|0|F|Q|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Although I agree with the other answers, it is in most cases better to allocate the whole array at once, because malloc is pretty slow.

<pre><code>
int **
array_new(size_t rows, size_t cols)
{
 int **array2d, **end, **cur;
 int *array;

 cur = array2d = malloc(rows * sizeof(int *));
 if (!array2d)
 return NULL;

 array = malloc(rows * cols * sizeof(int));
 if (!array)
 {
 free(array2d);
 return NULL;
 }

 end = array2d + rows;
 while (cur != end)
 {
 *cur = array;
 array += cols;
 cur++;
 }

 return array2d;
}
</code></pre>

To free the array simply do:
<code> free(*array); free(array); </code>

Note: this solution only works if you don't want to change the order of the rows, because you could then lose the address of the first element, which you need to free the array later.

blocks|key|1433247|text|嗯。老式的烟雾和镜子作为一种选择怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1433248|#define+ROWS++5
#define+COLS+13
#define+X(R,+C)+*(p+%2B+((R)+*+ROWS)+%2B+(C))

int+main(void)
{
++++int+*p+=+(int+*)+malloc+(ROWS+*+COLS+*+sizeof(int));
++++if+(p+!=+NULL)
++++{
++++++++size_t+r;
++++++++size_t+c;
++++++++for+(r+=+0;+r+<+ROWS;+r%2B%2B)
++++++++{
++++++++++++for+(c+=+0;+c+<+COLS;+c%2B%2B)
++++++++++++{
+++++++++++++++++X(r,c)+=+r+*+c;++/*+put+some+silly+value+in+that+position+*/+
++++++++++++}
++++++++}

++++++++/*+Then+show+the+contents+of+the+array+*/+
++++++++for+(r+=+0;+r+<+ROWS;+r%2B%2B)
++++++++{
++++++++++++printf("%25d+",+r);+++/*+Show+the+row+number+*/+

++++++++++++for+(c+=+0;+c+<+COLS;+c%2B%2B)
++++++++++++{
+++++++++++++++++printf("%25d",+X(r,c));
++++++++++++}

++++++++++++printf("\n");
++++++++}

++++++++free(p);
++++}
++++else
++++{
++++++++/*+issue+some+silly+error+message+*/+
++++}

++++return+0;
}|code-block|syntax|javascript|1433249|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Humm. How about old fashion smoke and mirrors as an option?

<pre><code>#define ROWS 5
#define COLS 13
#define X(R, C) *(p + ((R) * ROWS) + (C))

int main(void)
{
 int *p = (int *) malloc (ROWS * COLS * sizeof(int));
 if (p != NULL)
 {
 size_t r;
 size_t c;
 for (r = 0; r &lt; ROWS; r++)
 {
 for (c = 0; c &lt; COLS; c++)
 {
 X(r,c) = r * c; /* put some silly value in that position */ 
 }
 }

 /* Then show the contents of the array */ 
 for (r = 0; r &lt; ROWS; r++)
 {
 printf("%d ", r); /* Show the row number */ 

 for (c = 0; c &lt; COLS; c++)
 {
 printf("%d", X(r,c));
 }

 printf("\n");
 }

 free(p);
 }
 else
 {
 /* issue some silly error message */ 
 }

 return 0;
}
</code></pre>

I don't truly understand some basic things in C like dynamically allocating array of arrays.
I know you can do:

<pre><code>int **m;
</code></pre>

in order to declare a 2 dimensional array (which subsequently would be allocated using some *alloc function). Also it can be "easily" accessed by doing <code>*(*(m + line) + column)</code>. But how should I assign a value to an element from that array? Using gcc the following statement <code>m[line][column] = 12;</code> fails with a segmentation fault.

Any article/docs will be appreciated. :-)

dynamic allocating array of arrays in C

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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我并不真正理解C中的一些基本东西，比如动态分配数组的数组。我知道你能做到：int **m;以便声明一个二维数组(随后将使用某个*alloc函数进行分配)。此外，可以通过执行*(*(m + line) + column)“轻松”访问它。但是我应该如何给该数组中的元素赋值呢？使用gcc，下面的语句m[line][colum...

问在C++中动态分配数组数组
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