数组转移到下一个元素

How can I move elements in an array to the next element

如何将数组中的元素移动到下一个元素

eg: x[5] = { 5, 4, 3, 2, 1 }; // initial values
    x[0] = 6; // new values to be shifted
    x[5] = { 6, 5, 4, 3, 2 }; // shifted array, it need to be shifted, 
                              // not just increment the values.

This what I've done so far. It's wrong, that's why I need help here. Thanks in advance.

这就是我到目前为止所做的。这是错的，这就是我需要帮助的原因。提前致谢。

#include <iostream>

using namespace std;

int main() 
{
  int x[5] = { 5, 4, 3, 2, 1 };

  int array_size = sizeof(x) / sizeof(x[0]);

  x[0] = 6;

  int m = 1;

  for(int j = 0; j < array_size; j++) {
    x[m+j] = x[j];
    cout << x[j] << endl;
  }

  return 0;
}

7 个解决方案

#1

#include <iostream>

int main () {

  int x[5] = { 5, 4, 3, 2, 1 };

  int array_size = sizeof (x) / sizeof (x[0]);

  for (int j = array_size - 1; j > 0; j--) {

      x[j] = x[j - 1];
  }

  x[0] = 6;

  for (int j = 0; j < array_size; j++) {

      std::cout << x[j];
  }

  return 0;
}

#2

#include<algorithm>

// ...
std::rotate(x, x+4, x+5);
x[0] = 6;

#3

To "move rightwards" you have to iterate from the end of array:

要“向右移动”，你必须从数组的末尾迭代：

for(int j = array_size - 2; j >= 0; j--) {
   x[m+j] = x[j];
   cout << x[j] << endl;
}

otherwise you just overwrite all the elements with the 0th element.

否则你只需用第0个元素覆盖所有元素。

Please note array_size - 2 - otherwise you have "off by one" trying to access the element beyond the array end and that's undefined behavior.

请注意array_size - 2 - 否则你有“off by one”试图访问数组末尾之外的元素，这是未定义的行为。

#4

First of all, you should shift the old values in the array before you write the new value. But instead of a loop, you are better of using memmove(). Or even better with std::vector instead of an array - it handles all these low-level issues for you, including automatically resizing the array when needed.

首先，在写入新值之前，应该在数组中移动旧值。但不是循环，你最好使用memmove（）。或者甚至更好地使用std :: vector而不是数组 - 它会为您处理所有这些低级问题，包括在需要时自动调整数组大小。

#5

In the general case where you need to shift m elements (where 0 <= m <n): Start from the end of the array. If you start at the begining (index 0) then you overwrite and then move that overridden value.

在一般情况下，你需要移动m个元素（其中0 <= m ）：从数组的末尾开始。如果从开头（索引0）开始，则覆盖然后移动该重写值。

Studying the source code of std::memmove may be instructive as well.

研究std :: memmove的源代码也可能具有指导意义。

#6

You can start from the end of the array. You copy the

您可以从数组的末尾开始。你复制了

element in 2nd last position to the last position,
第二个位置到最后一个位置的元素，
element in 3rd last position to the 2nd last position,
元素在第3个最后位置到第2个最后位置，
....
....
element in first position(index 0) to the 2nd position and finally
元素在第一个位置（索引0）到第二个位置，最后
copy the new number in the first position. .
将新号码复制到第一个位置。。

。

for(j = array_size-1; j >0; j--) {
 x[j] = x[j-1];
}
x[0] = 6;

#7

    #include <iostream>

    using namespace std;

    int main() 
    {
       int x[5] = { 5, 4, 3, 2, 1 };

        int array_size = sizeof(x) / sizeof(x[0]);

        int m = 1;

        for(int j = array_size-1; j > 0; j--) {
           x[j] = x[j-m];
           cout << x[j] << endl;
        }

       x[0] = 6;
       return 0;
    }

#1

#include <iostream>

int main () {

  int x[5] = { 5, 4, 3, 2, 1 };

  int array_size = sizeof (x) / sizeof (x[0]);

  for (int j = array_size - 1; j > 0; j--) {

      x[j] = x[j - 1];
  }

  x[0] = 6;

  for (int j = 0; j < array_size; j++) {

      std::cout << x[j];
  }

  return 0;
}

#2

#include<algorithm>

// ...
std::rotate(x, x+4, x+5);
x[0] = 6;

#3