共享内存段内的指针 [英] Pointers inside shared memory segment

问题描述

我一直想这几个小时，而谷歌所有的事情我想那种，但我要疯了。

我有一个结构：

  typedef结构{
  诠释行;
  INT collumns;
  INT *垫;
  字符* IDs_row;
} MEM;
 

我不知道为int *（矩阵）和char *的大小，直到后来。

当我这样做，我创建共享内存是这样的：

 存储* CTRL;
INT大小=（2 +（（I-1）* num_cons））*的sizeof（int）的+ I * 26 *的sizeof（char）的; //我现在有真正的大小
shmemid = shmget的（KEY，大小，IPC_CREAT | 0666）;
如果（shmemid℃，）{
    PERROR（哈fallado拉creacion德拉MEMORIA compartida。）;
    出口（1）;
}
CTRL =（MEM *）的shmat（shmemid，0，0）;
如果（按Ctrl＆LT; =（MEM *）（0））{
    PERROR（哈fallado埃尔acceso一个MEMORIA compartida）;
    出口（2）;
}
 

这里没有问题。然后，我给一个值CTRL->行和collumns，并分配0至所有矩阵。

但在那之后，我写了char *与BAM，一些分段错误。

调试程序，我看到两个指针，垫和IDs_row了空。我怎么给他们共享内存段??

里面的正确的价值观

我试图消除的char *指针，只给它一个尝试，然后分段错误是在连接到所述共享存储器的其他程序，只是检查了矩阵（内部的价值观检查 - >行 - > collumns是succesfull）

解决方案

  CTRL =（MEM *）的shmat（shmemid，0，0）;
 

这仅分配有效的内存到 CTRL 指针，而不是 CTRL-＆GT;垫或 CTRL-＆GT; IDs_row

。

您可能希望：

 存储* CTRL;
shmemid = shmget的（KEY，sizeof的（CTRL），IPC_CREAT | 0666）;
//为结构分配内存
CTRL =（MEM *）的shmat（shmemid，0，0）;//为INT分配内存*
shmemid = shmget的（KEY，（（I-1）* num_cons））*的sizeof（int）的，IPC_CREAT | 0666）;
CTRL-＆GT;垫子=为（int *）的shmat（shmemid，0，0）;//为字符分配内存*
shmemid = shmget的（KEY，我* 26 * sizeof的（炭），IPC_CREAT | 0666）;
CTRL-＆GT; IDs_row =（字符*）的shmat（shmemid，0,0）;
 

I've been trying this for hours, and google all the things I kind think of, but I'm going crazy.

I have a struct:

typedef struct {
  int rows;
  int collumns;
  int* mat;
  char* IDs_row;
} mem;

I don't know the sizes of the int* (a Matrix) and char* untill later.

When I do, I create the shared memory like this:

mem *ctrl;
int size = (2 + ((i-1)*num_cons))*sizeof(int) + i*26*sizeof(char); //I have the real size now
shmemid = shmget(KEY, size, IPC_CREAT | 0666);
if (shmemid < 0) {
    perror("Ha fallado la creacion de la memoria compartida.");
    exit(1);
}
ctrl = (mem *)shmat(shmemid, 0, 0);
if (ctrl <= (mem *)(0)) {
    perror("Ha fallado el acceso a memoria compartida");
    exit(2);
}

No problem here. Then I give a value to ctrl->rows and collumns, and assign 0 to all the matrix.

But after that, I write something in the char* and bam, segmentation fault.

Debugging the program I saw that both pointers, mat and IDs_row where null. How do I give them the correct values inside the shared memory segment??

I tried removing the char* pointer, just to give it a try, and then the segmentation fault error was in the other program that connected to said shared memory and just checked the values inside the matrix (checking ->rows and ->collumns was succesfull)

解决方案

ctrl = (mem *)shmat(shmemid, 0, 0); 

This only assigns valid memory to the ctrl pointer, not to ctrl->mat or ctrl->IDs_row.

You probably want:

mem *ctrl;
shmemid = shmget(KEY, sizeof(ctrl), IPC_CREAT | 0666);
//allocate memory for the structure
ctrl = (mem *)shmat(shmemid, 0, 0);

//allocate memory for the int*
shmemid = shmget(KEY,((i-1)*num_cons))*sizeof(int), IPC_CREAT | 0666);
ctrl->mat = (int*)shmat(shmemid, 0, 0);

//allocate memory for the char*
shmemid = shmget(KEY,i*26*sizeof(char), IPC_CREAT | 0666);
ctrl->IDs_row = (char*)shmat(shmemid,0,0);

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