Thread: convert int to char and concatenate

Your answer is completely implementation-defined, and your program is wrong.

First, you assume that the machine is using a character encoding that's a superset of ASCII. There's no guarantee that # and $ have the codes 35 and 36, respectively. If you run your program on a different architecture, say, and IBM mainframe, it won't work.

But let's assume the computer uses ASCII.

This line:
var1 = static_cast<char>(var1);
does nothing. Since var1 is within the range of a char, it really does nothing. If var1 were outside, it would truncate the value. (char is just a small integer with a special interpretation in some contexts.)

To get char types, you need variables of type char. Types are static in C++: a variable can't change its type.
Therefore:
char cvar1 = static_cast<char>(cvar1);

Finally, the cout.
var1 + var2 simply adds the two integers.
If they were chars, it would still simply add the two integers.
That's because they're integers. Doing anything but adding makes no sense. Note that char can only represent a single character anyway.

If you want to concatenate with +, you need C++ strings.
std::string s1 = "#";
std::string s2 = "$";
std::string s3 = s1 + s2;

But that's not necessary here. Since you only want to output them, you can do that.
cout << cvar1 << cvar2;

or if you don't want the extra variables, you can cast directly.

cout << static_cast<char>(var1) << static_cast<char>(var2);