sync image的问题
program test2implicit 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 from1 out of 4
Image from1 out of 4
Image from2 out of 4
Image from2 out of 4
Image from3 out of 4
Image from3 out of 4
Image from4 out of 4
Image from4 out of 4
而我预想的结果是
Image from1 out of 4
Image from2 out of 4
Image from3 out of 4
Image from4 out of 4
Image from4 out of 4
Image from3 out of 4
Image from2 out of 4
Image from1 out of 4
因为对于 image1 和 image2 俩个if语句产生竟速，结果不可预测。 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)
DescriptionWhen 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：
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 from1 out of 4 Image from2 out of 4 Image from3 out of 4 Image from4 out of 4 Image from4 out of 4 Image from3 out of 4 Image from2 out of 4 Image from1 out of 4
页:
[1]