如何在 C++ 中移动字符数组的某些位置

Question

我有一个字符数组，我想从固定位置开始移动它的某些位置，并为移动的位置数量放置空格。

例如，如果我有数组：

(read A var := 5)

如果我想将它从第一个位置移动一个位置，我想得到：

( read A var := 5)

但是我得到了：

((ead A var := 5)

你可以在下面找到我的源代码：

#include <iostream>
#include <cctype>
#include <cstdio>
#include <cstring>
using namespace std;

char output[100] = "(read A var := 5)";

void move(int spos, int places)
{
    int i = places, j = spos;

    for (int k = 0; k < places; k++) {
        output[j+i] = output[j];
        i++;
        j++;
    }
}

int main(void)
{
    for (i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    move(0, 1);
    output[18] = '\0';

    for (i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    return 0;
}

Answer 1

这是我的移动功能版本，试试看：

#include <iostream>
#include <cctype>
#include <cstdio>
#include <cstring>
using namespace std;

char output[100] = "(read A var := 5)";

/* my version of move ;) */
void move2(int spos, int places)
{
    for(int x = strlen(output); x >=spos; x--)
    {
        output[x+places] = output[x]; /* move character from position x to position x+places */
    }

    for(int x = 0; x < places; x++)
        output[x+spos] = ' ' ; /* now we are adding white spaces to the begining */

}

int main(void)
{
    int i = 0;
    for(i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    move2(1, 1);    
    for (i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    return 0;
}

Answer 2

试试这个：

#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <cstdio>
#include <cstring>
using namespace std;

char output[100] = "(read A var := 5)";

void move(int spos, int shift_Len)
{
    int  j = spos;
    int  i = shift_Len;
    for (int k = strlen(output); k >j; k--) {
        output[k+i]= output[k-1];
    }
    for (int len=0 ; len <=i ; len++ ) {
        output[j+len]=' '; //fill shift position with space char
    }
}

int main(void)
{
    int i;
    for (i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    move(5,10);        //first argument is shift position and second argumnet is number of shift
    output[strlen(output)] = '\0';

    for (i = 0; i < strlen(output); i++)
        cout << output[i];
    cout << endl;

    return 0;
}

Answer 3

我实际上已经将这个写到了现在已删除的问题中，这非常相似：

• https://stackoverflow.com/questions/23715073/how-to-swap-words-inside-char-array-in-c

希望对您有所帮助：Live On Coliru

#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iostream>

using It         = std::string::iterator;
using tokref     = boost::iterator_range<It>;
using swappables = std::pair<tokref, tokref>;

namespace qi = boost::spirit::qi;

int main()
{
    std::string input = "read A var := 5";

    It first(input.begin()), last(input.end());

    using boost::phoenix::construct;
    using boost::phoenix::push_back;
    using namespace qi;

    qi::rule<It, tokref()> token_;
    qi::rule<It, swappables(), qi::space_type> swap_;

    token_ = qi::raw[+qi::graph];
    swap_  = (token_ >> token_) [ _val = construct<swappables>(_1, _2) ];

    std::vector<swappables> to_swap;

    std::cout << "Before: " << input << "\n";
    if (qi::phrase_parse(first, last, 
                *(
                     (&as_string [token_] [ _pass = ("var" == _1) ] >> swap_)
                   | omit[ token_ ]
                 ), space, to_swap))
    {
        // swap all swappables!
        for (auto& pair : to_swap)
        {
            std::cout << "Swap '" << pair.first << "' and '" << pair.second << "'\n";
            auto& a = pair.first;
            auto& b = pair.second;

            input.replace(a.begin(), b.end(), 
                    std::string(b.begin(), b.end()) +
                    std::string(a.end(), b.begin()) +
                    std::string(a.begin(), a.end()));
        }
    }
    std::cout << "After:  " << input << "\n";
}

输出：

read A var := 5
Swap 'var' and ':='
read A := var 5