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2011-07-16T17:57:21.103
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One way to check for yourself whether the string is concatenated at compile time is to compile the code and examine the object file. Assuming your file is called <code>test.d</code>: <pre><code>dmd -c test.d objdump test.o | grep -C3 "012345" </code></pre> ...should produce something like: <pre><code>Contents of section .rodata: 0000 2d000000 00000000 00000000 00000000 -............... 0010 01000000 00000000 00000000 00000000 ................ 0020 30313233 34353637 38394142 43444546 0123456789ABCDEF 0030 4748494a 4b4c4d4e 4f505152 53545556 GHIJKLMNOPQRSTUV 0040 5758595a 00000000 00000000 00000000 WXYZ............ 0050 24000000 00000000 20000000 00000000 $....... ....... </code></pre> (This is on Linux; on other platforms, you'll need different tools to inspect the object file.) If you change your <code>enum</code> to <code>const</code> or <code>string</code>, you will (probably) get no output: there will be no concatenated string for <code>grep</code> to find. But the compiler may concatenate strings at compile-time even when <code>enum</code> is not used. Consider this program: <pre><code> import std.stdio; enum a = "Aaaa"; enum b = "Bbbb"; enum c = "Cccc"; void main() { enum x = a ~ b; const y = b ~ a; string z = a ~ c; writeln(x, y, z); } </code></pre> Now, compile it, and examine the object file: <pre><code>% dmd -c test2.d && objdump -s test2.o | egrep "(Aaaa|Bbbb)" 0000 42626262 41616161 00000000 00000000 BbbbAaaa........ 0020 41616161 43636363 00000000 00000000 AaaaCccc........ 0040 41616161 42626262 00000000 00000000 AaaaBbbb........ </code></pre> We see that <code>x</code>, <code>y</code> and <code>z</code> are all static literals. (Mark <code>a</code>, <code>b</code> and <code>c</code> as <code>const</code> instead of <code>enum</code>, and you may see different behaviour.) So, while <code>enum</code> is a guarantee of compile-time evaluation, the absence of <code>enum</code> doesn't prevent compile-time evaluation.
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