please wait
pointer to a string object
Nov 2, 2011 at 4:13am
Hi everybody, I have an exercise that says: Write a program with a pointer to a string object. Use the pointer to the pointer to call the size() member function of the string object.
I did this but it doesn't work.
#include <iostream>
#include <string>
using namespace std;
int main()
{
cout << "Assigning &str to pStr\n";
string str = "score";
string* pStr = &str; //pointer to string object
cout << "str is: " << str << "\n";
cout << "*pStr is: " << *pStr << "\n";
cout << "(*pStr).size() is: " << (*pStr).size() << "\n";
return 0;
}
I need help, thank you in advance
I did this but it doesn't work.
#include <iostream>
#include <string>
using namespace std;
int main()
{
cout << "Assigning &str to pStr\n";
string str = "score";
string* pStr = &str; //pointer to string object
cout << "str is: " << str << "\n";
cout << "*pStr is: " << *pStr << "\n";
cout << "(*pStr).size() is: " << (*pStr).size() << "\n";
return 0;
}
I need help, thank you in advance
Last edited on Nov 2, 2011 at 4:14am
Nov 2, 2011 at 6:38am
How does it not work?
Though if the assignment says to "use the pointer to the pointer", you probably need another variable
Though if the assignment says to "use the pointer to the pointer", you probably need another variable
string** ppStr = &pStr;
Nov 2, 2011 at 10:42pm
What about now?.. the thing is in the chapter it never teaches about using **. There is another way?
#include <iostream>
#include <string>
using namespace std;
int main()
{
cout << "Assigning &str to pStr\n";
string str = "score";
string* pStr = &str; //pointer to string object
string** ppStr = &pStr;
cout << "str is: " << str << "\n";
cout << "*pStr is: " << *pStr << "\n";
cout << "(*pStr).size() is: " << (**ppStr).size() << "\n";
return 0;
}
#include <iostream>
#include <string>
using namespace std;
int main()
{
cout << "Assigning &str to pStr\n";
string str = "score";
string* pStr = &str; //pointer to string object
string** ppStr = &pStr;
cout << "str is: " << str << "\n";
cout << "*pStr is: " << *pStr << "\n";
cout << "(*pStr).size() is: " << (**ppStr).size() << "\n";
return 0;
}
Nov 2, 2011 at 10:47pm
If I had to guess, it meant to say "use the pointer to the string" rather than "use the pointer to the pointer".
Nov 2, 2011 at 10:55pm
it's what i understood! so what it's wrong, because, i submit the first exercise and I got 0 points.
I need hepl!
I need hepl!
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