Question

The set functions’ idea:

First argument is a reference, allocates space to hold copy of testing, sets str member of beany to point to the new block, copies testing to new block, and sets ct member of beany.

Problem:

1) Line that contains:

for (int i = 0; i < temp.length(); i++)

Error:expression must have a class
type

2) Line that contains:

temp[i] = cstr[i];

Error: expression must have
pointer-to-object type

3) overload of function show() for stringy type can’t find matching function signature due to presence of const

Very new to these concepts, could someone explain the reason for the errors?

#include "stdafx.h"
#include <iostream>
using namespace std;
#include <cstring>
#include <cctype>
struct stringy {
    char * str; //points to a string
    int ct; //length of string(not counting '\0')
};

void set(stringy & obj,  char cstr);
void show(const stringy & obj, int times=1);
void show(const char * cstr, int times = 1);

int _tmain(int argc, _TCHAR* argv[])
{
    string beany;
    char testing[] = "Reality isn't what it used to be.";

    set(beany, testing);
    show(beany);
    show(beany, 2);
    testing[0] = 'D';
    testing[1] = 'u';
    show(testing);
    show(testing, 3);
    show("Done");
    return 0;
}

void set(stringy & obj, char cstr)
{
    char * temp = new char[cstr];
    obj.str = temp;
    for (int i = 0; i < temp.length(); i++)
    temp[i] = cstr[i];
}

void show(const stringy & obj, int times)
{
    for (int i = 0; i < times; i++)
        cout << obj.str;
}

void show(const char * cstr, int times)
{
    for (int i = 0; i < times; i++)
        cout << cstr;
}

Answer 1

temp is a const char*. That type does not provide any kind of length facilities- it is not an object and does not have a length() member method. Use a std::string– that is what it is for.