Julia ccall 接口和符号的问题 [英] Issue with Julia ccall interface and symbols

问题描述

我正在尝试使用 Julia 的 ccall 函数与 C 库进行交互.所有类型和指针都是正确的，并且下面的函数调用成功地返回了正确的答案(为简洁起见，此处未显示变量定义和设置).

I'm attemping to use Julia's ccall function to interface with a C library. All the types and pointers are correct, and the below function call successfully returns the correct answer (variable defintion and setup not shown here for brevity).

    ccall((:vDSP_convD, libacc),  Void,
          (Ptr{T}, Int64,  Ptr{T},  Int64,  Ptr{T},  Int64, UInt64, UInt64),
          x_padded, 1, pointer(K, Ksize), -1, result, 1,  Rsize, Ksize)

但是，如果我希望将函数名生成为符号，然后将其用作 ccall 的参数，则会失败.

However, if I wish to generate the function name as a symbol, and then use it as an argument to ccall, it fails.

fname = symbol(string("vDSP_conv", "D"))
ccall((fname , libacc),  Void,
          (Ptr{T}, Int64,  Ptr{T},  Int64,  Ptr{T},  Int64, UInt64, UInt64),
          x_padded, 1, pointer(K, Ksize), -1, result, 1,  Rsize, Ksize)

错误是:

ERROR: LoadError: TypeError: conv: in ccall: first argument not a 
pointer or valid constant expression, expected Ptr{T}, 
got Tuple{Symbol,ASCIIString}

如果我打印这两个命名版本的类型，我会得到

If I print the type of each of these two naming versions, I get

julia> println(typeof(:vDSP_convD))
       Symbol
julia> println(typeof(fname))
       Symbol

有没有办法让它工作?我猜我必须将它包装在宏或 @eval 中才能完成这项工作，但我很好奇为什么上述功能不能如图所示工作?

Is there a way to get this to work? I'm guessing that I will have to wrap this in a macro or @eval to get this work, but I am curious as to why the above functionality doesn't work as shown?

任何帮助将不胜感激！

编辑

我最终将其包装在 @eval 块中以使其工作；但是，我仍然对后端逻辑感到好奇，为什么上述语法不起作用(为什么它有时将符号解释为指针，而其他时候则不)

I ended up wrapping this in an @eval block to get it to work; however, I'm still curious as to the backend logic as to why the above syntax doesn't work (why it interprets a symbol as a pointer sometimes, and then not other times)

推荐答案

ccall 并不是真正的函数——它是一种使用 C ABI 转换为 C 函数调用的语法形式.要发出对 C 函数的调用，您需要能够静态解析函数的地址——这就是这个要求的来源.请注意，在 C 和 Julia 中，您还可以使用变量函数指针调用函数.在 Julia 中，有几种方法可以获得这样的指针，通常使用 dlopen 和 dlsym.ccall 不会做的是通过非常量名称解析函数:这在 C 中是不可能的(无需自己构建查找表)；在 Julia 中，您可以通过使用 eval 来做到这一点，正如您所知道的那样——但是这样做会产生编译器开销.这就是 ccall 不会自动执行此操作的原因:例如，您不想冒在循环中意外引入编译器开销的风险.

ccall is not really a function – it's a syntactic form that is translated to a C function call using the C ABI. To emit a call to a C function, you need to be able to statically resolve the address of the function – that's where this requirement comes from. Note that in both C and Julia you can also call a function using a variable function pointer. In Julia, there are a few ways to get such a pointer, typically by using dlopen and dlsym. What ccall won't do is resolve a function by non-constant name: this is impossible in C (without a building yourself a lookup table); in Julia you can do this, as you've figured out, by using eval – but there is compiler overhead to doing this. That is why ccall won't do this automatically: you don't want to run the risk of accidentally introducing compiler overhead in a loop, for example.

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