String&在传递参数时不反映所做的更改，但String会....？

Question

void gen(string& str, int index)
{
    int i;
    if(str.length() == index)
    {
        cout << str << endl;
        return;
    }
    else
    {
        for(i=index; str[i]!='\0' && str[i]!='?';i++);

        if(str[i] == '?')
        {
            str[i] ='0';
            gen(str,i+1);
            str[i] ='1';
            gen(str,i+1);
        }
    }
    return;
}

int main()
{
    string s ="0??";
    gen(s, 0);
    return 0;
}

谁能给我解释一下....

Answer 1

引用将维护在函数调用中更改的值

after set str[1] ='0';

sub call: gen(str,2); will output combination: 000 001

reset str[1] ='1';

str is still 011

gen(str,i+1); output nothing 

希望这段代码能有所帮助

#include <string>
#include <iostream>
#include <stdio.h>
using namespace std;
void gen(string& str, int index)
{
    int i;
    if(str.length() == index)
    {
        cout << str << endl;
        return;
    }
    else
    {
        for(i=index; str[i]!='\0' && str[i]!='?';i++);

        if(str[i] == '?')
        {
            printf("before set pos %d to 0: %s\n",i,str.c_str());
            str[i] ='0';
            printf("after set pos %d to 0: %s\n",i,str.c_str());
            gen(str,i+1);
            printf("before set pos %d to 1: %s\n",i,str.c_str());
            str[i] ='1';
            printf("after set pos %d to 1: %s\n",i,str.c_str());

            gen(str,i+1);
        }
    }
    return;
}

int main()
{
    string s ="0??";
    gen(s, 0);
    return 0;
}

它输出：