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C、Open MPI : segmentation fault from call to MPI_Finalize(). 段错误并不总是发生,尤其是在进程数量较少的情况下

转载 作者:行者123 更新时间:2023-12-01 12:54:48 37 4
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我正在编写一个简单的代码来学习如何定义 MPI_Datatype 并将其与 MPI_Gatherv 结合使用。我想确保我可以在一个进程中组合可变长度、动态分配的结构化数据数组,这似乎工作正常,直到我调用 MPI_Finalize()。我已经通过使用打印语句和 Eclipse PTP 调试器(后端是 gdb-mi)确认这是问题开始显现的地方。我的主要问题是,如何摆脱段错误?

每次运行代码时都不会发生段错误。例如,它没有发生在 2 或 3 个进程中,但是当我运行大约 4 个或更多进程时往往会定期发生。

此外,当我使用 valgrind 运行此代码时,不会发生段错误。但是,我确实收到了来自 valgrind 的错误消息,尽管我在使用 MPI 函数时很难理解输出,即使有大量有针对性的抑制也是如此。我还担心如果我使用更多的抑制,我会使有用的错误消息静音。

我使用这些标志编译普通代码,所以我在两种情况下都使用 C99 标准:
-ansi -pedantic -Wall -O2 -march=barcelona -fomit-frame-pointer -std=c99
和调试代码:
-ansi -pedantic -std=c99 -Wall -g

两者都使用 gcc 4.4 mpicc 编译器,并在使用 Red Hat Linux 和 Open MPI v1.4.5 的集群上运行。如果我遗漏了其他重要的信息,请告诉我。这是代码,提前致谢:

//#include <unistd.h>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
//#include <limits.h>

#include "mpi.h"

#define FULL_PROGRAM 1

struct CD{

int int_ID;
double dbl_ID;
};

int main(int argc, char *argv[]) {

int numprocs, myid, ERRORCODE;

#if FULL_PROGRAM
struct CD *myData=NULL; //Each process contributes an array of data, comprised of 'struct CD' elements
struct CD *allData=NULL; //root will dynamically allocate this array to store all the data from rest of the processes
int *p_lens=NULL, *p_disp=NULL; //p_lens stores the number of elements in each process' array, p_disp stores the displacements in bytes
int MPI_CD_size; //stores the size of the MPI_Datatype that is defined to allow communication operations using 'struct CD' elements

int mylen, total_len=0; //mylen should be the length of each process' array
//MAXlen is the maximum allowable array length
//total_len will be the sum of mylen across all processes

// ============ variables related to defining new MPI_Datatype at runtime ====================================================
struct CD sampleCD = {.int_ID=0, .dbl_ID=0.0};
int blocklengths[2]; //this describes how many blocks of identical data types will be in the new MPI_Datatype
MPI_Aint offsets[2]; //this stores the offsets, in bytes(bits?), of the blocks from the 'start' of the datatype
MPI_Datatype block_types[2]; //this stores which built-in data types the blocks are comprised of
MPI_Datatype myMPI_CD; //just the name of the new datatype
MPI_Aint myStruct_address, int_ID_address, dbl_ID_address, int_offset, dbl_offset; //useful place holders for filling the arrays above
// ===========================================================================================================================
#endif
// =================== Initializing MPI functionality ============================
MPI_Init(&argc, &argv);
MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
MPI_Comm_rank(MPI_COMM_WORLD, &myid);
// ===============================================================================
#if FULL_PROGRAM
// ================== This part actually formally defines the MPI datatype ===============================================
MPI_Get_address(&sampleCD, &myStruct_address); //starting point of struct CD
MPI_Get_address(&sampleCD.int_ID, &int_ID_address); //starting point of first entry in CD
MPI_Get_address(&sampleCD.dbl_ID, &dbl_ID_address); //starting point of second entry in CD
int_offset = int_ID_address - myStruct_address; //offset from start of first to start of CD
dbl_offset = dbl_ID_address - myStruct_address; //offset from start of second to start of CD

blocklengths[0]=1; blocklengths[1]=1; //array telling it how many blocks of identical data types there are, and the number of entries in each block
//This says there are two blocks of identical data-types, and both blocks have only one variable in them

offsets[0]=int_offset; offsets[1]=dbl_offset; //the first block starts at int_offset, the second block starts at dbl_offset (from 'myData_address'

block_types[0]=MPI_INT; block_types[1]=MPI_DOUBLE; //the first block contains MPI_INT, the second contains MPI_DOUBLE

MPI_Type_create_struct(2, blocklengths, offsets, block_types, &myMPI_CD); //this uses the above arrays to define the MPI_Datatype...an MPI-2 function

MPI_Type_commit(&myMPI_CD); //this is the final step to defining/reserving the data type
// ========================================================================================================================

mylen = myid*2; //each process is told how long its array should be...I used to define that randomly but that just makes things messier

p_lens = (int*) calloc((size_t)numprocs, sizeof(int)); //allocate memory for the number of elements (p_lens) and offsets from the start of the recv buffer(d_disp)
p_disp = (int*) calloc((size_t)numprocs, sizeof(int));

myData = (struct CD*) calloc((size_t)mylen, sizeof(struct CD)); //allocate memory for each process' array
//if mylen==0, 'a unique pointer to the heap is returned'

if(!p_lens) { MPI_Abort(MPI_COMM_WORLD, 1); exit(EXIT_FAILURE); }
if(!p_disp) { MPI_Abort(MPI_COMM_WORLD, 1); exit(EXIT_FAILURE); }
if(!myData) { MPI_Abort(MPI_COMM_WORLD, 1); exit(EXIT_FAILURE); }


for(double temp=0.0;temp<1e6;++temp) temp += exp(-10.0);
MPI_Barrier(MPI_COMM_WORLD); //purely for keeping the output organized by give a delay in time

for (int k=0; k<numprocs; ++k) {

if(myid==k) {

//printf("\t ID %d has %d entries: { ", myid, mylen);

for(int i=0; i<mylen; ++i) {

myData[i]= (struct CD) {.int_ID=myid*(i+1), .dbl_ID=myid*(i+1)}; //fills data elements with simple pattern
//printf("%d: (%d,%lg) ", i, myData[i].int_ID, myData[i].dbl_ID);
}
//printf("}\n");
}
}

for(double temp=0.0;temp<1e6;++temp) temp += exp(-10.0);
MPI_Barrier(MPI_COMM_WORLD); //purely for keeping the output organized by give a delay in time

MPI_Gather(&mylen, 1, MPI_INT, p_lens, 1, MPI_INT, 0, MPI_COMM_WORLD); //Each process sends root the length of the vector they'll be sending

#if 1
MPI_Type_size(myMPI_CD, &MPI_CD_size); //gets the size of the MPI_Datatype for p_disp
#else
MPI_CD_size = sizeof(struct CD); //using this doesn't change things too much...
#endif

for(int j=0;j<numprocs;++j) {

total_len += p_lens[j];

if (j==0) { p_disp[j] = 0; }
else { p_disp[j] = p_disp[j-1] + p_lens[j]*MPI_CD_size; }
}

if (myid==0) {

allData = (struct CD*) calloc((size_t)total_len, sizeof(struct CD)); //allocate array
if(!allData) { MPI_Abort(MPI_COMM_WORLD, 1); exit(EXIT_FAILURE); }
}

MPI_Gatherv(myData, mylen, myMPI_CD, allData, p_lens, p_disp, myMPI_CD, 0, MPI_COMM_WORLD); //each array sends root process their array, which is stored in 'allData'

// ============================== OUTPUT CONFIRMING THAT COMMUNICATIONS WERE SUCCESSFUL=========================================
if(myid==0) {

for(int i=0;i<numprocs;++i) {
printf("\n\tElements from %d on MASTER are: { ",i);
for(int k=0;k<p_lens[i];++k) { printf("%d: (%d,%lg) ", k, (allData+p_disp[i]+k)->int_ID, (allData+p_disp[i]+k)->dbl_ID); }

if(p_lens[i]==0) printf("NOTHING ");
printf("}\n");
}
printf("\n"); //each data element should appear as two identical numbers, counting upward by the process ID
}
// ==========================================================================================================

if (p_lens) { free(p_lens); p_lens=NULL; } //adding this in didn't get rid of the MPI_Finalize seg-fault
if (p_disp) { free(p_disp); p_disp=NULL; }
if (myData) { free(myData); myData=NULL; }
if (allData){ free(allData); allData=NULL; } //the if statement ensures that processes not allocating memory for this pointer don't free anything

for(double temp=0.0;temp<1e6;++temp) temp += exp(-10.0);
MPI_Barrier(MPI_COMM_WORLD); //purely for keeping the output organized by give a delay in time
printf("ID %d: I have reached the end...before MPI_Type_free!\n", myid);

// ====================== CLEAN UP ================================================================================
ERRORCODE = MPI_Type_free(&myMPI_CD); //this frees the data type...not always necessary, but a good practice

for(double temp=0.0;temp<1e6;++temp) temp += exp(-10.0);
MPI_Barrier(MPI_COMM_WORLD); //purely for keeping the output organized by give a delay in time

if(ERRORCODE!=MPI_SUCCESS) { printf("ID %d...MPI_Type_free was not successful\n", myid); MPI_Abort(MPI_COMM_WORLD, 911); exit(EXIT_FAILURE); }
else { printf("ID %d...MPI_Type_free was successful, entering MPI_Finalize...\n", myid); }
#endif
ERRORCODE=MPI_Finalize();

for(double temp=0.0;temp<1e7;++temp) temp += exp(-10.0); //NO MPI_Barrier AFTER MPI_Finalize!

if(ERRORCODE!=MPI_SUCCESS) { printf("ID %d...MPI_Finalize was not successful\n", myid); MPI_Abort(MPI_COMM_WORLD, 911); exit(EXIT_FAILURE); }
else { printf("ID %d...MPI_Finalize was successful\n", myid); }

return EXIT_SUCCESS;
}

最佳答案

k 上的外循环是假的,但在技术上并没有错——它只是没用。

真正的问题是您对 MPI_GATHERV 的位移是错误的。如果您运行 valgrind,您将看到如下内容:

==28749== Invalid write of size 2
==28749== at 0x4A086F4: memcpy (mc_replace_strmem.c:838)
==28749== by 0x4C69614: unpack_predefined_data (datatype_unpack.h:41)
==28749== by 0x4C6B336: ompi_generic_simple_unpack (datatype_unpack.c:418)
==28749== by 0x4C7288F: ompi_convertor_unpack (convertor.c:314)
==28749== by 0x8B295C7: mca_pml_ob1_recv_frag_callback_match (pml_ob1_recvfrag.c:216)
==28749== by 0x935723C: mca_btl_sm_component_progress (btl_sm_component.c:426)
==28749== by 0x51D4F79: opal_progress (opal_progress.c:207)
==28749== by 0x8B225CA: opal_condition_wait (condition.h:99)
==28749== by 0x8B22718: ompi_request_wait_completion (request.h:375)
==28749== by 0x8B231E1: mca_pml_ob1_recv (pml_ob1_irecv.c:104)
==28749== by 0x955E7A7: mca_coll_basic_gatherv_intra (coll_basic_gatherv.c:85)
==28749== by 0x9F7CBFA: mca_coll_sync_gatherv (coll_sync_gatherv.c:46)
==28749== Address 0x7b1d630 is not stack'd, malloc'd or (recently) free'd

表明 MPI_GATHERV 以某种方式获得了错误信息。

(还有来自 Open MPI 中的 libltdl 的其他 valgrind 警告,不幸的是这是不可避免的——它是 libltdl 中的一个错误,另一个来自 PLPA,不幸的是,这也是不可避免的,因为它是故意这样做的[原因不值得讨论这里])

看看你的位移计算,我明白了
    total_len += p_lens[j];                                                              

if (j == 0) {
p_disp[j] = 0;
} else {
p_disp[j] = p_disp[j - 1] + p_lens[j] * MPI_CD_size;
}

但是 MPI 收集位移以数据类型为单位,而不是字节。所以它真的应该是:
p_disp[j] = total_len;
total_len += p_lens[j];

进行此更改使 MPI_GATHERV valgrind 警告对我来说消失了。

关于C、Open MPI : segmentation fault from call to MPI_Finalize(). 段错误并不总是发生,尤其是在进程数量较少的情况下,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10406438/

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