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我正在尝试为 iOS 制作一个 Ada 库。但是，需要手动执行 Ada 细化。

我知道编译器可以生成一个初始化符号，以后可以导入和使用。但是，对于以下 GPR 定义，它不会生成（该nm命令未列出它）。命名应该<libname>init与<libname>GPR 指令中定义的值相对应Library_Name

GPR 是按以下方式定义的（这是 windows/style -see DLL references-，但在 Mac 上为 iOS 生产时也存在问题）：

project adalib is
    for Languages use ("Ada");
    for Source_Dirs use (project'Project_Dir & "./src");
    for Library_Kind use "static"; --"static" on iOS will produce a .a file
    for Library_Name use project'Name; -- will produce "libadalib.a"
    for Library_Dir use project'Project_Dir & "./lib";
    for Library_Src_Dir use project'Project_Dir & "./includes";
    -- define your favorite compiler, builder, binder, linker options
end adalib;

我想念它：如何产生那个符号？

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1 回答 1

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我找到了解决方案。我的 GPR 缺少这个简单的指令：

    for Library_Interface use ("mypackage"); -- put whatever packages you want to expose, without .adb/.ads since we're talking about packages

使用上面的指令，我可以通过命令找到adalibinit符号。nm当我在我的 ada 代码中导入它时，我也可以使用它，请参阅：

package body mypackage is
    procedure Init_My_Lib
    is
       -- I want to call elaboration;
       pragma import (C, ada_elaboration, "adalibinit");
    begin
       ada_elaboration;
       -- further code
    end Init_My_Lib;
-- rest of package

所以，完整的 GPR 应该是：

project adalib is
    for Languages use ("Ada");
    for Source_Dirs use (project'Project_Dir & "./src");
    for Library_Kind use "static"; -- will produce a .a file
    for Library_Name use project'Name; -- will produce "libadalib.a"

    for Library_Interface use ("mypackage"); -- <=== THIS IS HERE

    for Library_Dir use project'Project_Dir & "./lib";
    for Library_Src_Dir use project'Project_Dir & "./includes";
    -- define your favorite compiler, builder, binder, linker options
end adalib;
于 2017-06-28T16:57:21.240 回答