# HD5dump **URL:** \ **Category:** Hdf-forum archives\ **Created:** [February 27, 2013, 12:32pm UTC](https://forum.hdfgroup.org/t/hd5dump/2830 "2013-02-27T12:32:38Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![Pradeep\_Jha1](https://forum.hdfgroup.org/letter_avatar_proxy/v4/letter/p/ad7895/32.png) [@Pradeep\_Jha1](https://forum.hdfgroup.org/u/Pradeep_Jha1)\ **Post date:** [February 27, 2013, 12:32pm UTC](https://forum.hdfgroup.org/t/hd5dump/2830/1 "2013-02-27T12:32:38Z")
Hello, I am trying to convert a unformatted data file created by fortran into the \*h5 format. To do this I am using this program\<[http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5\_rdwt.f90](http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5_rdwt.f90)\>provided on the website as the base. I am dealing with a data of size Nx by Ny by Nz. So I changed the "dims" declaration in the code to something like > **···** > > ------------------------------------------- > > \*\*\*\* code \*\*\*\* > > INTEGER :: Nx, Ny, Nz > > INTEGER(HSIZE\_T), DIMENSION(3) :: dims > > \*\*\*\* code \*\*\*\* > >      dims(1) = Nx > >      dims(2) = Ny > >      dims(3) = Nz > > \*\*\*\* code \*\*\*\* > > -------------------------------------------- > > This code is working perfectly fine. But when I am doing a "h5dump -H" on > the output file, the output is: > > HDF5 "output\_file" { > GROUP "/" { >    DATASET "variable" { >       DATATYPE H5T\_IEEE\_F32LE >       DATASPACE SIMPLE { ( Nz,Ny,Nx ) / ( Nz,Ny,Nx) } >    } > } > } > > Can you please explain is this how the order of Nz,Ny and Nx should be? > Does this represent a cube of size Nx by Ny by Nz? Or does dims(1) actually > represent the z dimension and I should assign it the value Nz? > > Thank you, > Pradeep ---
**Author:** ![brtnfld](https://forum.hdfgroup.org/user_avatar/forum.hdfgroup.org/brtnfld/32/1444_2.png) [@brtnfld](https://forum.hdfgroup.org/u/brtnfld)\ **Post date:** [February 27, 2013, 3:25pm UTC](https://forum.hdfgroup.org/t/hd5dump/2830/2 "2013-02-27T15:25:39Z")
The HDF5 file uses C storage conventions, which is why the matrix is transposed. See [http://www.hdfgroup.org/HDF5/doc/UG/UG\_frame12Dataspaces.html](http://www.hdfgroup.org/HDF5/doc/UG/UG_frame12Dataspaces.html) section: 7.3.2.5. C versus Fortran Dataspaces > **···** > > On 2013-02-27 06:32, Pradeep Jha wrote: > > > Hello, > > > > I am trying to convert a unformatted data file created by fortran > > into the \*h5 format. To do this I am using this program [1] provided > > on the website as the base. > > > > I am dealing with a data of size Nx by Ny by Nz. So I changed the > > "dims" declaration in the code to something like > > > > ------------------------------------------- > > > > \*\*\*\* code \*\*\*\* > > > > INTEGER :: Nx, Ny, Nz > > > > INTEGER(HSIZE\_T), DIMENSION(3) :: dims > > > > \*\*\*\* code \*\*\*\* > > > > ``` > > dims\(1\) = Nx > > > > dims\(2\) = Ny > > > > dims\(3\) = Nz > > > > ``` > > > > \*\*\*\* code \*\*\*\* > > > > -------------------------------------------- > > > > This code is working perfectly fine. But when I am doing a "h5dump > > -H" on the output file, the output is: > > > > HDF5 "output\_file" { > > GROUP "/" { > > DATASET "variable" { > > DATATYPE H5T\_IEEE\_F32LE > > DATASPACE SIMPLE { ( Nz,Ny,Nx ) / ( Nz,Ny,Nx) } > > } > > } > > > > Can you please explain is this how the order of Nz,Ny and Nx should > > be? Does this represent a cube of size Nx by Ny by Nz? Or does dims(1) > > actually represent the z dimension and I should assign it the value > > Nz? > > > > Thank you, > > Pradeep > > > > Links: > > ------ > > [1] [http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5\_rdwt.f90](http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5_rdwt.f90) > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > Hdf-forum is for HDF software users discussion. > > Hdf-forum@hdfgroup.org > > [http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org](http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org) ---
**Author:** ![Pradeep\_Jha1](https://forum.hdfgroup.org/letter_avatar_proxy/v4/letter/p/ad7895/32.png) [@Pradeep\_Jha1](https://forum.hdfgroup.org/u/Pradeep_Jha1)\ **Post date:** [February 28, 2013, 3:01am UTC](https://forum.hdfgroup.org/t/hd5dump/2830/3 "2013-02-28T03:01:45Z")
Thanks for the response. So from what I understand, the HD5 fortran wrapper automatically transposes the matrix to store it in the C storage conventions. So puttings "dims(1) = Nx" and "dims(3) = Nz" is correct. HDF5 fortran wrapper is just transposing the data inside the program before storing it in the final h5 format. But this is confusing me about something. I convert my original unformatted data written by fortran to h5 format so that I can visualize the original data using a software (Paraview). Does that mean that the data I will visualize using Paraview and the h5 data file will be a transposed data of what I originally intended to visualize? And if that is true, are there any simple ways to make sure that the HDF5 stores untransposed data? Will I need to pass a transposed data to HDF5 fortran wrapper to ensure this? Thank you again for the response, Pradeep > **···** > > 2013/2/28 \ > > > The HDF5 file uses C storage conventions, which is why the matrix is > > transposed. > > > > See [http://www.hdfgroup.org/HDF5/\*\*doc/UG/UG\_frame12Dataspaces.\*\*html<http://www.hdfgroup.org/HDF5/doc/UG/UG\_frame12Dataspaces.html>](http://www.hdfgroup.org/HDF5/ **doc/UG/UG_frame12Dataspaces.** html<http://www.hdfgroup.org/HDF5/doc/UG/UG_frame12Dataspaces.html>); > > > > section: 7.3.2.5. C versus Fortran Dataspaces > > > > On 2013-02-27 06:32, Pradeep Jha wrote: > > > > > Hello, > > > > > > I am trying to convert a unformatted data file created by fortran > > > into the \*h5 format. To do this I am using this program [1] provided > > > > > > on the website as the base. > > > > > > I am dealing with a data of size Nx by Ny by Nz. So I changed the > > > "dims" declaration in the code to something like > > > > > > ------------------------------\*\*------------- > > > > > > \*\*\*\* code \*\*\*\* > > > > > > INTEGER :: Nx, Ny, Nz > > > > > > INTEGER(HSIZE\_T), DIMENSION(3) :: dims > > > > > > \*\*\*\* code \*\*\*\* > > > > > >      dims(1) = Nx > > > > > >      dims(2) = Ny > > > > > >      dims(3) = Nz > > > > > > \*\*\*\* code \*\*\*\* > > > > > > ------------------------------\*\*-------------- > > > > > > This code is working perfectly fine. But when I am doing a "h5dump > > > -H" on the output file, the output is: > > > > > > HDF5 "output\_file" { > > > GROUP "/" { > > >    DATASET "variable" { > > >       DATATYPE H5T\_IEEE\_F32LE > > >       DATASPACE SIMPLE { ( Nz,Ny,Nx ) / ( Nz,Ny,Nx) } > > >    } > > > } > > > } > > > > > > Can you please explain is this how the order of Nz,Ny and Nx should > > > be? Does this represent a cube of size Nx by Ny by Nz? Or does dims(1) > > > actually represent the z dimension and I should assign it the value > > > Nz? > > > > > > Thank you, > > > Pradeep > > > > > > Links: > > > ------ > > > [1] [http://www.hdfgroup.org/ftp/\*\*HDF5/examples/introductory/](http://www.hdfgroup.org/ftp/**HDF5/examples/introductory/)\*\* > > > F90/h5\_rdwt.f90\<[http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5\_rdwt.f90>](http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5_rdwt.f90>); > > > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\*\*\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > > Hdf-forum is for HDF software users discussion. > > > Hdf-forum@hdfgroup.org > > > [http://mail.hdfgroup.org/\*\*mailman/listinfo/hdf-forum\_\*\*hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org>](http://mail.hdfgroup.org/ **mailman/listinfo/hdf-forum_** hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org>); > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\*\*\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > Hdf-forum is for HDF software users discussion. > > Hdf-forum@hdfgroup.org > > [http://mail.hdfgroup.org/\*\*mailman/listinfo/hdf-forum\_\*\*hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org>](http://mail.hdfgroup.org/ **mailman/listinfo/hdf-forum_** hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org>); ---
**Author:** ![Pradeep\_Jha1](https://forum.hdfgroup.org/letter_avatar_proxy/v4/letter/p/ad7895/32.png) [@Pradeep\_Jha1](https://forum.hdfgroup.org/u/Pradeep_Jha1)\ **Post date:** [March 4, 2013, 2:42am UTC](https://forum.hdfgroup.org/t/hd5dump/2830/4 "2013-03-04T02:42:56Z")
Hello, can someone please clarify my doubt please? I would really appreciate it. Thanks, Pradeep > **···** > > 2013/2/28 Pradeep Jha \ > > > Thanks for the response. So from what I understand, the HD5 fortran > > wrapper automatically transposes the matrix to store it in the C storage > > conventions. So puttings "dims(1) = Nx" and "dims(3) = Nz" is correct. HDF5 > > fortran wrapper is just transposing the data inside the program before > > storing it in the final h5 format. > > > > But this is confusing me about something. > > > > I convert my original unformatted data written by fortran to h5 format so > > that I can visualize the original data using a software (Paraview). Does > > that mean that the data I will visualize using Paraview and the h5 data > > file will be a transposed data of what I originally intended to visualize? > > > > And if that is true, are there any simple ways to make sure that the HDF5 > > stores untransposed data? Will I need to pass a transposed data to HDF5 > > fortran wrapper to ensure this? > > > > Thank you again for the response, > > Pradeep > > > > 2013/2/28 \ > > > > The HDF5 file uses C storage conventions, which is why the matrix is > > > > > transposed. > > > > > > See [http://www.hdfgroup.org/HDF5/\*\*doc/UG/UG\_frame12Dataspaces.\*\*html<http://www.hdfgroup.org/HDF5/doc/UG/UG\_frame12Dataspaces.html>](http://www.hdfgroup.org/HDF5/ **doc/UG/UG_frame12Dataspaces.** html<http://www.hdfgroup.org/HDF5/doc/UG/UG_frame12Dataspaces.html>); > > > > > > section: 7.3.2.5. C versus Fortran Dataspaces > > > > > > On 2013-02-27 06:32, Pradeep Jha wrote: > > > > > > > Hello, > > > > > > > > I am trying to convert a unformatted data file created by fortran > > > > into the \*h5 format. To do this I am using this program [1] provided > > > > > > > > on the website as the base. > > > > > > > > I am dealing with a data of size Nx by Ny by Nz. So I changed the > > > > "dims" declaration in the code to something like > > > > > > > > ------------------------------\*\*------------- > > > > > > > > \*\*\*\* code \*\*\*\* > > > > > > > > INTEGER :: Nx, Ny, Nz > > > > > > > > INTEGER(HSIZE\_T), DIMENSION(3) :: dims > > > > > > > > \*\*\*\* code \*\*\*\* > > > > > > > >      dims(1) = Nx > > > > > > > >      dims(2) = Ny > > > > > > > >      dims(3) = Nz > > > > > > > > \*\*\*\* code \*\*\*\* > > > > > > > > ------------------------------\*\*-------------- > > > > > > > > This code is working perfectly fine. But when I am doing a "h5dump > > > > -H" on the output file, the output is: > > > > > > > > HDF5 "output\_file" { > > > > GROUP "/" { > > > >    DATASET "variable" { > > > >       DATATYPE H5T\_IEEE\_F32LE > > > >       DATASPACE SIMPLE { ( Nz,Ny,Nx ) / ( Nz,Ny,Nx) } > > > >    } > > > > } > > > > } > > > > > > > > Can you please explain is this how the order of Nz,Ny and Nx should > > > > be? Does this represent a cube of size Nx by Ny by Nz? Or does dims(1) > > > > actually represent the z dimension and I should assign it the value > > > > Nz? > > > > > > > > Thank you, > > > > Pradeep > > > > > > > > Links: > > > > ------ > > > > [1] [http://www.hdfgroup.org/ftp/\*\*HDF5/examples/introductory/](http://www.hdfgroup.org/ftp/**HDF5/examples/introductory/)\*\* > > > > F90/h5\_rdwt.f90\<[http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5\_rdwt.f90>](http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5_rdwt.f90>); > > > > > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\*\*\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > > > Hdf-forum is for HDF software users discussion. > > > > Hdf-forum@hdfgroup.org > > > > [http://mail.hdfgroup.org/\*\*mailman/listinfo/hdf-forum\_\*\*hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org>](http://mail.hdfgroup.org/ **mailman/listinfo/hdf-forum_** hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org>); > > > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\*\*\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > > Hdf-forum is for HDF software users discussion. > > > Hdf-forum@hdfgroup.org > > > [http://mail.hdfgroup.org/\*\*mailman/listinfo/hdf-forum\_\*\*hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org>](http://mail.hdfgroup.org/ **mailman/listinfo/hdf-forum_** hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org>); ---
**Author:** ![brtnfld](https://forum.hdfgroup.org/user_avatar/forum.hdfgroup.org/brtnfld/32/1444_2.png) [@brtnfld](https://forum.hdfgroup.org/u/brtnfld)\ **Post date:** [March 5, 2013, 5:37am UTC](https://forum.hdfgroup.org/t/hd5dump/2830/5 "2013-03-05T05:37:47Z")
> Thanks for the response. So from what I understand, the HD5 fortran > wrapper automatically transposes the matrix to store it in the C > storage conventions. So puttings "dims(1) = Nx" and "dims(3) = Nz" is > correct. HDF5 fortran wrapper is just transposing the data inside the > program before storing it in the final h5 format. > > But this is confusing me about something. > > I convert my original unformatted data written by fortran to h5 > format so that I can visualize the original data using a software > (Paraview). Does that mean that the data I will visualize using > Paraview and the h5 data file will be a transposed data of what I > originally intended to visualize? Yes, if you are writing a multidimensional array using the fortran APIs and you want to read the data back using the same dimensional array then you need to use the fortran APIs. If you were to use a C program to read the hdf5 data back then you need to account for the transposing in the C program, or handle it when writing from fortran. Here is an explanation from the archives: > **···** > > On 2013-02-27 21:01, Pradeep Jha wrote: > > > HDF5 is a "self-describing" format, which means that HDF5 metadata > > stored in a dataset object header allows the HDF5 C library and any > > other non-C applications built on top of it, to retrieve a raw data > > (i.e. elements of a multidimensional array) in the correct order. > > > > (Let's for a second forget about HDF5, C and Fortran, Python and > > Matlab 🙂 ) > > > > If we have a matrix A(N,M,K), we usually count dimensions from left > > to right saying that the first dimension has size N, the second > > dimension has size M, the third dimension has size K, and so on. > > > > (Now let's talk about HDF5 but without referring to any language.) > > > > When we describe a matrix using HDF5 datatspace object, we use the > > same convention (i.e. specifying dimensions from left to right): the > > first dimension has size N, the second dimension has size M, the > > third dimension has size K. (Aside: Please notice that this > > description is valid for both C and Fortran HDF5 applications, i.e. C > > and Fortran dims array needed by H5Screate\_simple > > (h5screate\_simple\_f) will have the values dims = {N,M,K}). > > > > The question is: how does HDF5 know how to interpret a blob of {N x > > M x K x by sizeof(datatype)} bytes of dataset raw data stored in the > > file? Was A(N,M,K) stored? Or was it A(K,N,M) stored? Or any other > > permutation of (K,N,M)? > > > > HDF5 file has no clue about matrices and their dimensions, and the > > languages they were written from. This is application's > > responsibility to interpret data correctly and pass the correct > > interpretation to the HDF5 C library to store in a file. > > > > As it was mentioned above, dimensions of the matrix are described > > using HDF5 dataspace object and are stored in the file. d integers > > P1, ..., Pd, where d is a rank of a matrix, are stored in a dataspace > > object header according to the following convention: the last value > > - Pd is the size of the FASTEST changing dimension of the matrix, > > i.e. HDF5 file spec and HDF5 C library follow C storage convention > > (no wonder, it is a C library :-). Therefore there is no ambiguity in > > interpreting {N x M x K x sizeof(datatype)} bytes, and HDF5 file has > > enough information to interpret data correctly by any "row-major" or > > "column-major" application (including bypassing HDF5 C library and > > reading directly from the HDF5 file!) > > > > Here is what is happening when HDF5 Fortran library is used: > > > > Suppose we want to write A(N,M,K) matrix to the HDF5 file. HDF5 > > Fortran API describes dataspace with the first dimension being N, the > > second dimension being M, the third dimension being K (as we would do > > it in C and any other language). But HDF5 Fortran API also knows > > that the fastest changing dimension has size N (i.e. we have > > column-major order). Therefore HDF5 Fortran library instructs C > > library to store K,M,N values in the dataspace object header instead > > of N,M,K, since N is the size of the fastest changing dimension. > > > > So, if we read matrix A(N,M,K) ((i.e. N x M x K x sizeof(datatype) > > blob) written from Fortran by a C application, we will read it to > > the matrix B(K,M,N) ( C API that requests sizes of the first, second > > and third dimensions will return values K,M,N stored in the dataspace > > header) > > > > If we read matrix A(N,M,K) written from Fortran by Fortran > > application, we will read it once again into B(N,M,K) ( Fortran API > > that requests sizes of the first, second and third dimension will > > flip an array K,M,N stored in the file and return N,M,K) > > > > In other words: HDF5 library stores information about how to > > interpret data. Interpretation follows C storage convention: the last > > dimension specified for the dataspace object is the fastest changing > > one. It is the responsibility of the application (in this case > > FORTRAN HDF5 library) to interpret correctly the order of dimensions > > and pass to/ from the HDF5 C library. > > > > Please notice that there is no need to transpose data itself: one > > only has to pass a correct interpretation of the data to the HDF5 C > > Library and to make sure it is done according to the HDF5 C library > > convention - the first value stored in the dataspace header > > corresponds to the slowest changing dimension, ...., the last value > > stored in the dataspace header corresponds to the fastest changing > > dimension). ---
**Author:** ![Pradeep\_Jha1](https://forum.hdfgroup.org/letter_avatar_proxy/v4/letter/p/ad7895/32.png) [@Pradeep\_Jha1](https://forum.hdfgroup.org/u/Pradeep_Jha1)\ **Post date:** [March 5, 2013, 11:19am UTC](https://forum.hdfgroup.org/t/hd5dump/2830/6 "2013-03-05T11:19:17Z")
Thank you for the response. > **···** > > 2013/3/5 \ > > > On 2013-02-27 21:01, Pradeep Jha wrote: > > > > > Thanks for the response. So from what I understand, the HD5 fortran > > > wrapper automatically transposes the matrix to store it in the C > > > storage conventions. So puttings "dims(1) = Nx" and "dims(3) = Nz" is > > > correct. HDF5 fortran wrapper is just transposing the data inside the > > > program before storing it in the final h5 format. > > > > > > But this is confusing me about something. > > > > > > I convert my original unformatted data written by fortran to h5 > > > format so that I can visualize the original data using a software > > > (Paraview). Does that mean that the data I will visualize using > > > Paraview and the h5 data file will be a transposed data of what I > > > originally intended to visualize? > > > > Yes, if you are writing a multidimensional array using the fortran APIs > > and you want to read the data back using the same dimensional array then > > you need to use the fortran APIs. If you were to use a C program to read > > the hdf5 data back then you need to account for the transposing in the C > > program, or handle it when writing from fortran. > > > > Here is an explanation from the archives: > > > > HDF5 is a "self-describing" format, which means that HDF5 metadata > > > > > stored in a dataset object header allows the HDF5 C library and any > > > other non-C applications built on top of it, to retrieve a raw data > > > (i.e. elements of a multidimensional array) in the correct order. > > > > > > (Let's for a second forget about HDF5, C and Fortran, Python and > > > Matlab 🙂 ) > > > > > > If we have a matrix A(N,M,K), we usually count dimensions from left > > > to right saying that the first dimension has size N, the second > > > dimension has size M, the third dimension has size K, and so on. > > > > > > (Now let's talk about HDF5 but without referring to any language.) > > > > > > When we describe a matrix using HDF5 datatspace object, we use the > > > same convention (i.e. specifying dimensions from left to right): the > > > first dimension has size N, the second dimension has size M, the > > > third dimension has size K. (Aside: Please notice that this > > > description is valid for both C and Fortran HDF5 applications, i.e. C > > > and Fortran dims array needed by H5Screate\_simple > > > (h5screate\_simple\_f) will have the values dims = {N,M,K}). > > > > > > The question is: how does HDF5 know how to interpret a blob of {N x > > > M x K x by sizeof(datatype)} bytes of dataset raw data stored in the > > > file? Was A(N,M,K) stored? Or was it A(K,N,M) stored? Or any other > > > permutation of (K,N,M)? > > > > > > HDF5 file has no clue about matrices and their dimensions, and the > > > languages they were written from. This is application's > > > responsibility to interpret data correctly and pass the correct > > > interpretation to the HDF5 C library to store in a file. > > > > > > As it was mentioned above, dimensions of the matrix are described > > > using HDF5 dataspace object and are stored in the file. d integers > > > P1, ..., Pd, where d is a rank of a matrix, are stored in a dataspace > > > object header according to the following convention: the last value > > > - Pd is the size of the FASTEST changing dimension of the matrix, > > > i.e. HDF5 file spec and HDF5 C library follow C storage convention > > > (no wonder, it is a C library :-). Therefore there is no ambiguity in > > > interpreting {N x M x K x sizeof(datatype)} bytes, and HDF5 file has > > > enough information to interpret data correctly by any "row-major" or > > > "column-major" application (including bypassing HDF5 C library and > > > reading directly from the HDF5 file!) > > > > > > Here is what is happening when HDF5 Fortran library is used: > > > > > > Suppose we want to write A(N,M,K) matrix to the HDF5 file. HDF5 > > > Fortran API describes dataspace with the first dimension being N, the > > > second dimension being M, the third dimension being K (as we would do > > > it in C and any other language). But HDF5 Fortran API also knows > > > that the fastest changing dimension has size N (i.e. we have > > > column-major order). Therefore HDF5 Fortran library instructs C > > > library to store K,M,N values in the dataspace object header instead > > > of N,M,K, since N is the size of the fastest changing dimension. > > > > > > So, if we read matrix A(N,M,K) ((i.e. N x M x K x sizeof(datatype) > > > blob) written from Fortran by a C application, we will read it to > > > the matrix B(K,M,N) ( C API that requests sizes of the first, second > > > and third dimensions will return values K,M,N stored in the dataspace > > > header) > > > > > > If we read matrix A(N,M,K) written from Fortran by Fortran > > > application, we will read it once again into B(N,M,K) ( Fortran API > > > that requests sizes of the first, second and third dimension will > > > flip an array K,M,N stored in the file and return N,M,K) > > > > > > In other words: HDF5 library stores information about how to > > > interpret data. Interpretation follows C storage convention: the last > > > dimension specified for the dataspace object is the fastest changing > > > one. It is the responsibility of the application (in this case > > > FORTRAN HDF5 library) to interpret correctly the order of dimensions > > > and pass to/ from the HDF5 C library. > > > > > > Please notice that there is no need to transpose data itself: one > > > only has to pass a correct interpretation of the data to the HDF5 C > > > Library and to make sure it is done according to the HDF5 C library > > > convention - the first value stored in the dataspace header > > > corresponds to the slowest changing dimension, ...., the last value > > > stored in the dataspace header corresponds to the fastest changing > > > dimension). > > > > \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\*\*\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ > > Hdf-forum is for HDF software users discussion. > > Hdf-forum@hdfgroup.org > > [http://mail.hdfgroup.org/\*\*mailman/listinfo/hdf-forum\_\*\*hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum\_hdfgroup.org>](http://mail.hdfgroup.org/ **mailman/listinfo/hdf-forum_** hdfgroup.org<http://mail.hdfgroup.org/mailman/listinfo/hdf-forum_hdfgroup.org>);