multiplying a string by an integer returns integers?

Question

So I was trying to make an asterisk pyramid using D. First of all I noticed that concatenation seems to be impossible. Writing out something like writeln("foo" + "bar") will give you a syntax error. So instead I tried multiplying the strings like in python, that didn't work with double quoted strings, but with single quoted strings something weird happens.

If you type in this

import std.stdio;
void main()
{
    foreach (i; 0 .. 10)
    {
        writeln(i*'0');
    }
}

it will return a bunch of integers. Could anyone explain why this happens? And letting me know how to concatenate strings will also be really helpful.

thanks!

Answer 1

The '0' is not a string, it is a character, which uses ASCII encoding. The number is being multiplied with the encoding's integer id. For example, the encoding for ASCII's 'A' is 65.

import std.stdio;
int main()
{
        writeln( cast(int)'A' );
        writeln( 10 * 'A' );
        return 0;
}

This program will print 65 and 650 because the character is being converted to an integer in both cases.

To solve the original concatenation problem you can use the '~' operator for concatenating two arrays, or use "array1 ~= array2" to append array2 onto array1 in one statement.

Answer 2

First solution that comes to mind:

char[5] arr3 = 's';
writeln(arr3);

Two alternatives are std.array.replicate and std.range.repeat:

import std.array;
import std.stdio;

void main() {
    auto arr = replicate(['s'], 5); // lazy version: http://dlang.org/phobos/std_range.html#repeat
    // or
    auto arr2 = ['s'].replicate(5);
    writeln(arr);
    writeln(arr2);
}