如何移动数组中的第一个元素并将其移动到c编程中的最后一个元素？

If I have an array, let's say [1,2,3,4,5,6], how can I make it to be [2,3,4,5,6,1]? Which loop should I use for this? I tried using a for loop, however since I am fairly new to C programming I was unable to produce the output.

如果我有一个数组,让我们说[1,2,3,4,5,6],我怎么能让它成为[2,3,4,5,6,1]?我应该使用哪个循环?我尝试使用for循环,但由于我对C编程很新,我无法生成输出。

3 个解决方案

#1

int array[6] = { 1, 2, 3, 4, 5, 6 };
int array_size = 6;

// array is now { 1, 2, 3, 4, 5, 6 }

int array = array[0];                        // #1

for(int i = 1; i < array_size; i++) {        // #2
   array[i - 1] = array[i];
}

array[array_size - 1] = tmp;                 // #3

// array is now { 2, 3, 4, 5, 6, 1 }

It works like this:

它的工作原理如下:

Take the first value and store it in a special variable, tmp.

取第一个值并将其存储在一个特殊变量tmp中。

Take each of the remainding values and move one step to the front of the array.

获取每个剩余值并将一步移动到数组的前面。

Take the value stored in tmp and move it to the back of the array.

获取存储在tmp中的值并将其移动到数组的后面。

#2

Whoops. I misread the problem the first time

哎呦。我第一次误解了这个问题

int len = 5;
char *pOriginal = {0,1,2,3,4}
char ch0 = pOrigina[0];
memmove(pOriginal, &pOrigina[1], len-2) // Thanks guys!
pOriginal[len-1] = ch0;

#3

-1

If you know the size of the array then you can use swapping, Just save the first variable in a temporary variable and then do a swapping of i element with the i+1 element and then at the last position do the swapping between last element and temporary variable. thank you...

如果您知道数组的大小,那么您可以使用交换,只需将第一个变量保存在临时变量中,然后用i + 1元素交换i元素,然后在最后一个位置进行最后一个元素之间的交换。临时变量。谢谢...

#1

int array[6] = { 1, 2, 3, 4, 5, 6 };
int array_size = 6;

// array is now { 1, 2, 3, 4, 5, 6 }

int array = array[0];                        // #1

for(int i = 1; i < array_size; i++) {        // #2
   array[i - 1] = array[i];
}

array[array_size - 1] = tmp;                 // #3

// array is now { 2, 3, 4, 5, 6, 1 }