Fix to_string to call cpp2::to_string

Calling std::to_string makes it impossible to use to_string
on optional that contains std::string.

The solution is to call cpp2::to_string instead of std::to_string.

Added examples of use in regression tests with results of execution.

Author: filipsajdak

include/cpp2util.h

@@ -741,7 +741,7 @@ auto to_string(std::string const& s) -> std::string const&
 template<typename T>
 auto to_string(std::optional<T> const& o) -> std::string {
     if (o.has_value()) {
-        return std::to_string(o.value());
+        return cpp2::to_string(o.value());
     }
     return "(empty)";
 }

regression-tests/mixed-string-interpolation.cpp2

@@ -7,4 +7,7 @@ main: () -> int = {
 
     b = 42;
     std::cout << "a^2 + b = (a * a + b.value())$\n";
+
+    os: std::optional<std::string> = "testing";
+    std::cout << "os = (os)$\n";
 }

regression-tests/test-results/mixed-string-interpolation.cpp

@@ -17,4 +17,7 @@
 
     b = 42;
     std::cout << "a^2 + b = " + cpp2::to_string(a * a + b.value()) + "\n";
+
+    std::optional<std::string> os { "testing" }; 
+    std::cout << "os = " + cpp2::to_string(os) + "\n";
 }

regression-tests/test-results/mixed-string-interpolation.cpp2

@@ -7,4 +7,7 @@ main: () -> int = {
 
     b = 42;
     std::cout << "a^2 + b = (a * a + b.value())$\n";
+
+    os: std::optional<std::string> = "testing";
+    std::cout << "os = (os)$\n";
 }