string - Delphi 字符串写入时复制

Question

我有这样的代码:

function Test: string;
var
   r, s: string;
begin
   r := 'Hello';
   Writeln(NativeInt(PChar(@r[1])));
   s := r;
   Writeln(NativeInt(PChar(@s[1])));
   Result := r;
   Writeln(NativeInt(PChar(@Result[1])));
end;

人们说delphi对字符串使用写时复制。但上面的函数为变量、r、s 和结果打印 3 个不同的地址。所以这很令人困惑。内存中是否只有“Hello”字符串的副本？

Answer 1

每当您获取字符串元素的地址时，在编译器看来，这都算作一次写入。就其而言，您现在有了一个指向字符串内部的原始指针，谁知道您打算用它做什么。所以，从它的角度来看，它是安全的。它决定制作该字符串的唯一副本，以便您可以自由地对原始指针执行任何您计划执行的卑鄙行为。

您的代码编译为:

Project2.dpr.13: r := 'Hello';00419EF8 8D45FC           lea eax,[ebp-$04]00419EFB BAA89F4100       mov edx,$00419fa800419F00 E827D2FEFF       call @UStrLAsgProject2.dpr.14: Writeln(NativeInt(PChar(@r[1])));00419F05 8D45FC           lea eax,[ebp-$04]00419F08 E883D3FEFF       call @UniqueStringU00419F0D 8BD0             mov edx,eax00419F0F A18CE64100       mov eax,[$0041e68c]00419F14 E853B2FEFF       call @Write0Long00419F19 E82EB5FEFF       call @WriteLn00419F1E E845A1FEFF       call @_IOTestProject2.dpr.15: s := r;00419F23 8D45F8           lea eax,[ebp-$08]00419F26 8B55FC           mov edx,[ebp-$04]00419F29 E8FED1FEFF       call @UStrLAsgProject2.dpr.16: Writeln(NativeInt(PChar(@s[1])));00419F2E 8D45F8           lea eax,[ebp-$08]00419F31 E85AD3FEFF       call @UniqueStringU00419F36 8BD0             mov edx,eax00419F38 A18CE64100       mov eax,[$0041e68c]00419F3D E82AB2FEFF       call @Write0Long00419F42 E805B5FEFF       call @WriteLn00419F47 E81CA1FEFF       call @_IOTestProject2.dpr.17: Result := r;00419F4C 8BC3             mov eax,ebx00419F4E 8B55FC           mov edx,[ebp-$04]00419F51 E88ED1FEFF       call @UStrAsgProject2.dpr.18: Writeln(NativeInt(PChar(@Result[1])));00419F56 8BC3             mov eax,ebx00419F58 E833D3FEFF       call @UniqueStringU00419F5D 8BD0             mov edx,eax00419F5F A18CE64100       mov eax,[$0041e68c]00419F64 E803B2FEFF       call @Write0Long00419F69 E8DEB4FEFF       call @WriteLn00419F6E E8F5A0FEFF       call @_IOTest

对 UniqueStringU 的调用正在执行写时复制中的复制。