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<DIV dir=ltr align=left><SPAN class=713233622-29012007><FONT face=Arial
color=#0000ff size=2>It means that the user must call the function from all
processes. The implementation may implement it by having each process read a
distinct portion of the data and send it to other processes.</FONT></SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=713233622-29012007><FONT face=Arial
color=#0000ff size=2></FONT></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=713233622-29012007><FONT face=Arial
color=#0000ff size=2>Rajeev</FONT></SPAN></DIV><BR>
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<FONT face=Tahoma size=2><B>From:</B> owner-mpich-discuss@mcs.anl.gov
[mailto:owner-mpich-discuss@mcs.anl.gov] <B>On Behalf Of </B>Luiz
Mendes<BR><B>Sent:</B> Monday, January 29, 2007 7:41 AM<BR><B>To:</B>
mpich-discuss<BR><B>Subject:</B> [MPICH] MPI_File_write_all or read_all
doubts<BR></FONT><BR></DIV>
<DIV></DIV>Hi all,<BR><BR>I read about MPI_File_read_all and write
correpondant, and the text says this operation perform reads using all
processes defined in a certain group.<BR><BR>Well i have saw this operation is
slower than file_read non collective version. Is it right? <BR><BR>If i define
3 processes during mpirun, if i put this operation MPI_File_read_all in
algorithm, this one will be executed 3 times, 3 times all processes will read
all file?<BR><BR>Or it will read once a time? I recorded the time, and it
shows three diferent times, then i think it have executed 3 times. <BR>The
part of code is below<BR><BR> MPI_File_open(MPI_COMM_WORLD,
"Wtupi.dat", MPI_MODE_RDONLY, MPI_INFO_NULL, &fh2);<BR>
MPI_File_set_view(fh2, 0, MPI_INT, MPI_INT, "native",MPI_INFO_NULL);
<BR> start=MPI_Wtime();<BR>
MPI_File_read_all(fh2, buf2, 900000, MPI_INT,
MPI_STATUS_IGNORE);<BR>
tempo=MPI_Wtime()-start;<BR> printf("\nTime spent for read
operation %f",tempo);<BR> }
<BR><BR>Thanks<BR>Luiz<BR><BR><BR></BLOCKQUOTE></BODY></HTML>