How to concatenate additional character to a string in c++ - c++

Question

Suppose there is a array..and the contents of the array="ironman" Now, i need to add some additional character to this string like "i*r%o#n#m^a!n"
out[i]="ironman"
Outputs:
out[]=i
out[]=*
out[]=r
out[]=%
out[]=o
out[]=#
out[]=n
out[]=#
out[]=m
out[]=^
out[]=a
out[]=!
out[]=n
I have written a code which concatenate at the end of the string, but i want to concatenate between the string.
char in[20] = "ironman";
const unsigned int symbol_size = 5;
std::string symbols[symbol_size];
std::string out(in);
out = out + "#" + "*" + "#";

Answer 1

You can use string.insert(pos, newString). Example below:
std::string mystr
mystr.insert(6,str2);
If you know the index, specify it directly as 'pos'. Otherwise, you may want to do somekind of str.find() and pass in the result.

Answer 2

If I have understood correctly what you need then you can use the following straightforward approach
#include <iostream>
#include <string>
#include <cstring>
int main()
{
char in[] = "ironman";
char symbols[] = "*%##^!";
std::string s;
s.reserve( std::strlen( in ) + std::strlen( symbols ) );
char *p = in;
char *q = symbols;
while ( *p && *q )
{
s.push_back( *p++ );
s.push_back( *q++ );
}
while ( *p ) s.push_back( *p++ );
while ( *q ) s.push_back( *q++ );
std::cout << s << std::endl;
}
The program output is
i*r%o#n#m^a!n
You can write a separate function. For example
#include <iostream>
#include <string>
#include <cstring>
std::string interchange_merge( const char *s1, const char *s2 )
{
std::string result;
result.reserve( std::strlen( s1 ) + std::strlen( s2 ) );
while ( *s1 && *s2 )
{
result.push_back( *s1++ );
result.push_back( *s2++ );
}
while ( *s1 ) result.push_back( *s1++ );
while ( *s2 ) result.push_back( *s2++ );
return result;
}
int main()
{
char in[] = "ironman";
char symbols[] = "*%##^!";
std::string s = interchange_merge( in, symbols );
std::cout << s << std::endl;
}
The output will be the same as above
i*r%o#n#m^a!n