性病::复制二维数组

问题描述

您好，我尝试使用的std ::复制（）函数来复制一个二维数组。我在想，如果有可能做到这一点，像这样！我不断收到一个段错误，但数组被正确地复制。我试过减了一些，并加入了一些到结束的情况下的复制功能，但没有成功。

  const int的列= 3;
    const int的列= 3;
    INT敏[行] [列] = {{1,2,3}，{4,5,6}，{7,8,9}};
    INT favint [行] [列];
    性病::复制（敏，敏+行*列，favint）;
 

很明显，敏行+ *列是不正确的，事实证明，这个值对应于整个行，使得敏行+ *列= 1意味着它将复制整个第一行。如果敏行+ *列= 2会复制前两行等有人能解释一下作业本给我吗？

解决方案

 的std ::复制（敏，敏+行*列，favint）;
 

应该是：

 的std ::复制（安培;敏[0] [0]，＆安培;敏[0] [0] +行*列，和放大器; favint [0] [0] ）;
 

原型的std ::复制：

 模板＆LT;类InputIt，类OutputIt＆GT;
OutputIt副本（InputIt第一，InputIt最后OutputIt d_first）;
 

请注意，指针数组元素可以作为包装器的迭代器。

Hello I'm trying to use the std::copy() function to copy a two dimensional array. I was wondering if it's possible to do it like so! I keep getting a "Segmentation Fault" but the array is copied correctly. I've tried subtracting a few and adding a few to the end case for the copy function, but with no success.

    const int rows = 3;
    const int columns = 3;
    int myint[rows][columns]={{1,2,3},{4,5,6},{7,8,9}};
    int favint[rows][columns];
    std::copy(myint, myint+rows*columns,favint);

It's obvious that "myint+rows*columns" is incorrect, and it turns out that this value corresponds to entire rows such that "myint+rows*columns=1" means it will copy the entire first row. if "myint+rows*columns=2" it copies the first two rows etc. Can someone explain the operation of this for me?

解决方案

std::copy(myint, myint+rows*columns,favint);

should be:

std::copy(&myint[0][0], &myint[0][0]+rows*columns,&favint[0][0]);

prototype of std::copy:

template< class InputIt, class OutputIt >
OutputIt copy( InputIt first, InputIt last, OutputIt d_first );

Notice that pointer to array element could be wrapper as an iterator.

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