sync image的问题

Glen · 发表于 2023-10-4 20:42:46

[Fortran] 纯文本查看 复制代码

program test2
    implicit none
    integer :: me
    me = this_image() ! 需要先声明me再赋值,如果直接在声明初始化，会保存到常量区，变成大家共有的变量
    if(me > 1) sync images(me-1)!编号高的image需要等待前一个image彻底运行完到程序结束？
    write(*,*) "Image from ", this_image(), "out of", num_images()
    if(me < num_images()) sync images(me+1)!image1 这里不需要等待image2？
    write(*,*) "Image from ", this_image(), "out of", num_images()
end program test2

最诡异的是当image1运行到第二个if时，不需要与其他image同步，于是最终输出结果为

Image from  1 out of 4
Image from  1 out of 4
Image from  2 out of 4
Image from  2 out of 4
Image from  3 out of 4
Image from  3 out of 4
Image from  4 out of 4
Image from  4 out of 4

而我预想的结果是

Image from  1 out of 4
Image from  2 out of 4
Image from  3 out of 4
Image from  4 out of 4
Image from  4 out of 4
Image from  3 out of 4
Image from  2 out of 4
Image from  1 out of 4

风平老涡 · 发表于 2023-10-4 22:10:25

因为对于 image1 和 image2 俩个if语句产生竟速，结果不可预测。

Glen · 发表于 2023-10-6 10:48:33

本帖最后由 Glen 于 2023-10-6 14:01 编辑

风平老涡发表于 2023-10-4 22:10
因为对于 image1 和 image2 俩个if语句产生竟速，结果不可预测。

解释在这里SYNC IMAGES (intel.com)

Description

When SYNC IMAGES statements are executed on images C and G, they correspond if the number of times image C has executed a SYNC IMAGES statement in the current team with G in its image set is the same as the number of times image G has executed a SYNC IMAGES statement with C in its image set in this team.

就是说images之间互相等待的次数要扯平，然后images才会听你的sync images命令，否则就不执行。所以image1对于第7行直接无视，因为第7行的时候image1发现image2等了我一次，但是我没有等待过image2，因此没有扯平，既然这样image1不会听从第7行的指令，虽然第7行起到了计数扯平的作用，但是image1不会去correspond(respond?)，而是会直接执行第8行。想要实现我预计的结果，需要把sync命令成对出现，即多写2行sync命令（新增第8、10行），如下：

[Fortran] 纯文本查看 复制代码

program test2
    implicit none
    integer :: me
    me = this_image() ! 需要先声明me再赋值,如果直接在声明初始化，会保存到常量区，变成大家共有的变量
    if(me > 1) sync images(me-1)
    write(*,*) "Image from ", this_image(), "out of", num_images()
    if(me < num_images()) sync images(me+1)   
    if(me < num_images()) sync images(me+1)
    write(*,*) "Image from ", this_image(), "out of", num_images()
    if(me > 1) sync images(me-1)
end program test2

Image from 1 out of 4

Image from 2 out of 4

Image from 3 out of 4

Image from 4 out of 4

Image from 3 out of 4

Image from 2 out of 4

Image from 1 out of 4

[并行] sync image的问题

规矩勋章