Pointing to a some characters in a string using pointer - c++

Question

#include<iostream>
using namespace std;
int main()
{
char s1[80]={"This is a developed country."};
char *s2[8];
s2[0]=&s1[10];
cout<<*s2; //Predicted OUTPUT: developed
// Actual OUTPUT: developed country.
return 0;
}
I want that the cout<<*s2; should print only the letters {"developed"} in it, so I gave *s2[8] length as 8 characters. What can I do so that the variable cout<<*s2 will only print upto the length of 8 characters. I'm using dmc, lcc and OpenWatcom compilers. This is only a small part of other bigger program where I'm using string data type, so what can I do now, well extremely thanks for answering my question :)

Answer 1

s2 is a length 8 array of pointers to char. You are making its first element point to s1 starting at position 10. That is all. You are not using the remaining elements of that array. Therefore the length of s2 is irrelevant.
You could have done this instead:
char* s2 = &s1[10];
If you want to create a string out of part of s1, you can use std::string:
std::string s3(s1+10, s1+19);
std::cout << s3 << endl;
Note that this allocates its own memory buffer and holds a copy or the original character sequence. If you only want a view of part of another string, you can easily implement a class holding a begin and one-past the end pointer to the original. Here's a rough sketch:
struct string_view
{
typedef const char* const_iterator;
template <typename Iter>
string_view(Iter begin, Iter end) : begin(begin), end(end) {}
const_iterator begin;
const_iterator end;
};
std::ostream& operator<<(std::ostream& o, const string_view& s)
{
for (string_view::const_iterator i = s.begin; i != s.end; ++i)
o << *i;
return o;
}
then
int main()
{
char s1[] = "This is a developed country.";
string_view s2(s1+10, s1+19);
cout << s2 << endl;
}

Answer 2

s2 is an array of pointers to char*. You are only ever using the zeroth element in this array.
&s1[10] points to the 11th character in the string s1. That address is assigned to the zeroth element of s2.
In the cout statement, *s2 is equivalent to s2[0];. So cout << *s2; outputs the zeroth element of s2, which has been assigned to the 11th character of s1. cout will trundle along the memory until the null-terminator of your string is reached.

Answer 3

Strings must be NULL terminated aka \0. The start of s2 is fine but cout will continue reading until the end. You have to actually copy the data rather than simply setting the pointer if you want to be able to output.

Answer 4

Your mistake is thinking that
char *s2[8];
declares a pointer to an array of 8 characters (or, not equivalently, a pointer to a string-with-exactly-8-characters). It doesn't do either of those. Instead of declaring a pointer-to-an-array, it declares an array-of-pointers.
If you want s2 to be a pointer to an array-of-8-characters, you need:
char (*s2)[8];
But, that's still messed up. You ask:
What can I do so that the variable *s2 will store only up to its length?
Do you think its length is 8? Before trying to answer that, return to your definition of s1:
char s1[80]={"This is a developed country."};
Is the length 80, or 28? The answer is either, depending on how you define 'length' - the length of the array or the length up to the null terminator?
All of these misconceptions about size are unhelpful. As #n.m. has pointed out in a comment, the solution to all pointer problems in C++ is to stop using pointers. (Apologies if I've mis-paraphrased n.m.!)
#include<iostream>
using namespace std;
int main()
{
string s1="This is a developed country.";
string s2;
s2 = s1.substr(10, 9);
cout << s2;
return 0;
}

Answer 5

If you want to do it ghetto style and skip the std::string for some reason you can always use strncpy, memcpy or strstr etc.
int main()
{
char s1[80]="This is a developed country.";
char s2[10];
strncpy(s2,s1+10,9);
s2[9] = '\0';
std::cout << s2 << std::endl;
std::cin.get();
return 0;
}

Answer 6

s2 is an char type arry, the element of the arry is char *,so you can't use it to store a string. if you want to get the "developed" in the strings,you can write code like it:
#include<iostream>
using namespace std;
int main()
{
char *s1[]={"This", "is", "a", "developed", "country."};
char *s2[8];
s2[0]= s1 + 3;
cout<<s2[0]; //Predicted OUTPUT: developed
// Actual OUTPUT: developed country.
return 0;
}