Literals

1. The problem here is that the value on the right-hand side is too large to fit in an int. According to the rules about integer literals, its type is long. For that reason it doesn't fit the type of the variable on the left-hand side. There are at least two solutions.

   One solution is to leave the type of the variable to the compiler for example by the auto keyword:

       auto amount = 10_000_000_000;
   

   The type of amount would be deduced to be long from its initial value from the right-hand side.

   Another solution is to make the type of the variable long as well:

       long amount = 10_000_000_000;
   

2. We can take advantage of the special '\r' character that takes the printing to the beginning of the line.

   import std.stdio;
   
   void main() {
       for (int number = 0; ; ++number) {
           write("\rNumber: ", number);
       }
   }
   

   The output of that program may be erratic due to its interactions with the output buffer. The following program flushes the output buffer and waits for 10 millisecond after each write:

   import std.stdio;
   import core.thread;
   
   void main() {
       for (int number = 0; ; ++number) {
           write("\rNumber: ", number);
           stdout.flush();
           Thread.sleep(10.msecs);
       }
   }
   

   Flushing the output is normally not necessary as it is flushed automatically before getting to the next line e.g. by writeln, or before reading from stdin.