How to stop checking when a string has empty index - c++

Question

/* Write a program that would mix-and-merge two given strings (s1 and s2) into string s3 as follows:
first character of s1, first character of s2, second character of s1, second character of s2, etc. */
But the problem with my code is: that if i type for s1 "John" and then for s2 "Stevens"
Result will be = JSothenv e n s.
How do i fix the spaces that are left after one of the strings ends?
The way i thought i would fix it is i would check with the ifs i have below in the for loop To see if the index is null or '\0' but that doesn't work as the string holds random values after the string ends.
#include <iostream>
#include <string>
using namespace std;
int main()
{
string s1, s2;
string s3;
int i; // For index
int j = 0; // For second index and loop checking
cout << "Type first string: ";
getline(cin, s1);
cout << "Type second string: ";
getline(cin, s2);
s3.resize(s1.size() + s2.size() + 100); // The + 100 is used so we have space for all the characters. The + 100 is not needed if i fix my problem.
for(i = 0; j <= s1.size(); i += 2)
{
if(s1[j] == null) // With what do i check it?
{
break;
}
else
{
s3[i] = s1[j];
++j;
}
}
j = 0;
for(i = 1; j <= s2.size(); i += 2)
{
if(s2[j] == null)
{
break;
}
else
{
s3[i] = s2[j];
}
++j;
}
for( i = 0; i <= s3.size(); ++i)
{
cout << s3[i];
}
return 0;

Answer 1

Try this:
#include <iostream>
#include <string>
using namespace std;
int main(int argc, char const *argv[])
{
string s1, s2, s3;
int i,j,k;
cout << "Type first string: ";
getline(cin, s1);
cout << "Type second string: ";
getline(cin, s2);
s3.resize(s1.size()+s2.size());
for(i = 0, j = 0, k = 0; j < s1.size() && k < s2.size(); i++) {
if(i & 1) {
s3[i] = s2[k++];
} else {
s3[i] = s1[j++];
}
}
if(j == s1.size()) {
while(k < s2.size()) {
s3[i++] = s2[k++];
}
} else {
while(j < s1.size()) {
s3[i++] = s1[j++];
}
}
cout << s3 << endl;
return 0;
}

Answer 2

You can replace the code after getline with this:
string::const_iterator i1 = s1.begin(), i2 = s2.begin();
bool useFirst = true;
while (i1 != s1.end() || i2 != s2.end())
{
if (useFirst && i1 != s1.end())
{
s3.push_back(*i1++);
}
else if (i2 != s2.end())
{
s3.push_back(*i2++);
}
useFirst = !useFirst; // switch over for next iteration
}
cout << s3 << endl;
It's a bit simpler, and basically iterates alternately over the two strings until they're both exhausted.

Answer 3

It's quite simple to write a generic interleave algorithm that iterates over both ranges once:
template<typename InputIterator1, typename InputIterator2, typename OutputIterator>
OutputIterator interleave(InputIterator1 first1, InputIterator1 end1,
InputIterator2 first2, InputIterator2 end2,
OutputIterator out)
{
while(first1 != end1 && first2 != end2)
{
out = *first1;
out = *first2;
++first1;
++first2;
}
// copy remaining elements from both ranges
while(first1 != end1)
{
out = *first1;
++first1;
}
while(first2 != end2)
{
out = *first2;
++first2;
}
return out;
}
Live demo here. Usage is like std::set_merge:
interleave(begin(s1), end(s1),
begin(s2), end(s2),
std::back_inserter(result));
It iterates over both ranges until one runs out, then copies the remaining elements in both ranges. Iterates once over both containers and has one double end check I couldn't get rid of without assuming one range to be larger than the other. Can handle two different containers now