C ++获取临时地址-在将引用分配给指针时出错 [英] C++ Taking address of temporary - error while assigning reference to pointer

问题描述

该程序是用C ++编写的.我正在尝试使用void函数来扩展Line结构，该结构由整数长度和指向下一条连接的线的指针组成.有一个void Expand函数可以为结构中的行指针分配行引用.新行将是当前行大小的两倍.使用我正在使用的代码，我得到一个g ++错误获取临时[-fpermissive]的地址".谁能建议该函数将行引用的有效实例添加到行指针nextLine的方法吗?

This program is written in C++. I am trying to use a void function to expand a Line structure which consists of an integer length and a pointer to the next connected line. There is a void Expand function made to assign a line reference to the line pointer in the struct. The new line is to be double the size of the current line. With the code I am using, I get a g++ error of 'Taking address of temporary [-fpermissive]'. Could anyone suggest a way in which the function adds a valid instance of a line reference to the Line pointer nextLine?

struct Line
{
    int length;
    Line* nextLine;
};

Line NewLine(Line& lineRef)
{
    Line newLine;
    newLine.length = lineRef.length * 2;
    return newLine;
}

void Expand(Line& lineRef)
{
    //Error here states: Taking address of temporary [-fpermissive]
    lineRef.nextLine = &NewLine(lineRef);
}

int main() {

    Line line;

    Expand(line);

    cout << line.length << endl;
    cout << line.nextLine->length << endl;

    return 0;
}

推荐答案

此方法有效

    struct Line
    {
            int length;
            Line* nextLine;
            ~Line(){delete nextLine;}
             //Make copy constructor and assignment operator private
    };

    void Expand(Line* lineRef)
    {
            lineRef->nextLine = new Line;
            lineRef->nextLine->length = 2*(lineRef->length) ;
    }

int main() 
{

        Line* line = new Line;
        line->length = 5;

        Expand(line);

        cout << line->length << endl;
        cout << line->nextLine->length << endl;

        delete line;
        return 0;
}

这篇关于C ++获取临时地址-在将引用分配给指针时出错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！