外部函数使用的设备内存中的变量如何？ [英] How is variable in device memory used by external function?

问题描述

在此代码中：

#include <iostream>

void intfun(int * variable, int value){
    #pragma acc parallel present(variable[:1]) num_gangs(1) num_workers(1)
    {
        *variable = value;
    }
}

int main(){
    int var, value = 29;

    #pragma acc enter data create(var) copyin(value)
        intfun(&var,value);
    #pragma acc exit data copyout(var) delete(value)

    std::cout << var << std::endl;
}

如何 int value 在 intfun 中识别为在设备内存中？如果我在 present（variable [：1]，value）中替换 present（variable [：1]） intfun pragma，我得到以下运行时错误：

How is int value recognized to be on device memory in intfun? If I replace present(variable[:1]) by present(variable[:1],value) in the intfun pragma, I get the following runtime error:

FATAL ERROR: data in PRESENT clause was not found on device 1: name=_43144_33_value
 file:/opt/pgi/linux86-64/14.9/include/CC/iostream intfun__FPii line:5
Present table dump for device[1]: NVIDIA Tesla GPU 1, compute capability 3.5
host:0x7fffc11faa28 device:0x2303f20200 size:4 presentcount:1 line:14 name:_43152_14_value
host:0x7fffc11faa34 device:0x2303f20000 size:4 presentcount:2 line:14 name:_43152_9_var

我不明白为什么要指定 value 是 present 导致上面的失败。我使用NVVP检查了 value 只在输入数据指令中复制一次，即它不会在并行指令 intfun 。 OpenACC是如何工作的？

I don't understand why specifying that value is present leads to the failure above. I checked with NVVP that value is only copied once in the enter data directive, i.e. it is not copied again in the parallel directive in intfun. How does OpenACC work its magic?

推荐答案

你会被你自己的语法和你指出的已经。

You're getting confused again by your own syntax and over what was pointed out to you already.

在 intfun 中的值 code> main 中的 copyin（value）函数调用通过值传递 value ，这意味着它会对其进行复制。因此，将它添加到 present（）语句没有意义，因为它在设备上不存在。编译器必须复制它。（当你根本没有提到它时，编译器会自动识别它是需要的，并为你复制它。）

value in intfun is not the same as value that you did copyin(value) in main. The function call passes value by value, meaning it makes a copy of it. Therefore adding it to a present() clause makes no sense because it is not present on the device. The compiler must copy it in. (And when you don't mention it at all, the compiler automatically recognizes that it is needed and copies it in for you.)

设备上存在的项目是 main -scope变量 value 是 intfun 不使用。给所有的变量同名可能不会帮助你理解这一点。

The item that is present on the device is the main-scope variable value, which is not used by intfun. Giving all your variables the same name is probably not helping you understand this.

为了演示这个，让我们通过 value 而非 ：

To demonstrate this, let's pass value by reference instead of by value:

$ cat main8.cpp
#include <iostream>

void intfun(int * variable, int &value){
    #pragma acc parallel present(variable[:1],value) num_gangs(1) num_workers(1)
    {
        *variable = value;
    }
}

int main(){
    int var, value = 29;

    #pragma acc enter data create(var) copyin(value)
        intfun(&var,value);
    #pragma acc exit data copyout(var) delete(value)

    std::cout << var << std::endl;
}
[user2@dc12 misc]$ pgcpp -acc -Minfo main8.cpp
intfun(int *, int &):
      5, Generating present(variable[:1])
         Generating present(value[:])
         Accelerator kernel generated
         Generating Tesla code
main:
     14, Generating enter data copyin(value)
         Generating enter data create(var)
     17, Generating exit data delete(value)
         Generating exit data copyout(var)
$ ./a.out
29
$

现在主范围变量 value 在设备上，编译器知道它，它将直接使用 intfun ，并将它添加到 present 条款是合法和有效的。

Now the main-scope variable value is on the device, the compiler knows it, it will be used by intfun directly, and adding it to the present clause is legal and functional.

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