如何将数组（或c串）的元素向左移动给定数字的索引 [英] how to shift elements of array (or c-string) left by a given number indexes

问题描述

我正在写一个函数，将我的c字符串的字符向左移动给定数量的字符。目前的功能将移动字符左，但我失去了一个。我知道这是一个索引问题与我的for循环，但我不能固定下来。

I am writing a function to shift the characters of my c-string left by a given number of characters. Currently the function will shift the characters left but I am losing one. I know it is some sort of indexing issue with my for loop, but I can't pin it down.

编辑：按shift left我的意思是：

By shift left I mean:

给定a，b，c，d
的起始c字符串如果向左移一个索引，这个数组将等于b，c，d，a
如果向左移两个索引，这个相同的c字符串将等于c，d，a，b

Given a starting c-string of a, b, c, d if shifted left one index this same array would equal b, c, d, a if shifted left two indexes this same c-string would equal c, d, a, b

这里是我的代码到目前为止：

Here is my code so far:

#include <iostream>
using namespace std;

void shiftleft (char myarray[], int size, int shiftBy)
{

char temp;

for (int i=size-1; i > 0; i--)
{
    temp = myarray[size+shiftBy];
    myarray[size+shiftBy] = myarray[i];
    myarray[i] = temp;
}


}

int main() {
    char myarray[20] = "test";
    int size = 4;
    shiftleft (myarray, size, 1);

    for(int i = 0; i < size+1; i++){
    cout << myarray[i];
    }


    return 0;
}

这里是我的工作功能，将每个元素向右移动，要做的是逆向这个循环，并移动元素离开，如下所示：<----

Here is my working function that shifts each element to the right, all I need to do is reverse this loop, and move the elements left, as in this way: <----

//function bloack
void shiftright (char myarray[], int size, int shiftBy)
{
    if(shiftBy > size){
        shiftBy = shiftBy - size;
    }
    if(size == 1){
        //do nothing
    }
    else{
        char temp;
        //for loop to print the array with indexes moved up (to the right) --> by 2
        for (int i=0; i < size; i++)
        {
            temp = myarray[size-shiftBy];
            myarray[size-shiftBy] = myarray[i];
            myarray[i] = temp;
        }

    }
}

推荐答案

有了一些风扇，我能够让它工作。这是我的功能函数：）

With a little fanangling, I was able to get it to work. Here is my functioning function :)

问题是，我需要分配元素i到i + shiftBy，并且只重复循环，而i < size-shiftBy。

The issue was that I needed to assign element i to i+shiftBy, and only repeat the loop while i < size-shiftBy.

//function bloack
void shiftLeft (char myarray[], int size, int shiftBy)
{
    if(shiftBy > size){
        shiftBy = shiftBy - size;
    }

    if(size == 1){
        //do nothing
    }
    else{
        char temp;
        //for loop to print the array with indexes moved up (to the left) <-- by 2
        for (int i=0; i < size-shiftBy; i++)
        {//EXAMPLE shift by 3  for a c-string of 5
            temp = myarray[i];//temp = myarray[0]
            myarray[i] = myarray[i + shiftBy];//myarray[0] == myarray[2]
            myarray[i + shiftBy] = temp;//myarray[2] = temp(value previously at index i)
        }

    }
}

这篇关于如何将数组（或c串）的元素向左移动给定数字的索引的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！