c编程的shmat（）权限被拒绝 [英] c programming shmat ( ) permission denied

问题描述

我有一个问题，当我运行我的code。我的shmat失败，并打印权限被拒绝。我搜索在谷歌如何解决这个问题，但我不能。我的code是以下内容：

I have a problem when I run my code. My shmat fails and prints permission denied. I searched on google how to solve it but I can't. My code is the following:

#include <stdio.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <stdlib.h>
#include <unistd.h>
#define ERROR -1

int main ( int argc, char *argv[] ) {
    int shmid,key=50;
    int *val;
    int *x;
    int rw = -1;

    // 0 for write and 1 for read 

    shmid = shmget ( key, sizeof( int ), IPC_CREAT );

    if ( shmid == -1 ) {
        perror ( "Error in shmget\n" );
        return ( ERROR );
    }

    val = ( int * ) shmat ( shmid, NULL, 0 );

    if ( val == -1 ) {
        perror ( "Error in shmat\n" );
        return ( ERROR );
    }

    scanf ( "%d", &rw);

    while ( rw >= 0 ) {
        if ( rw == 0 ) {
            //write in the shared memory
            x = ( int * ) malloc ( sizeof ( int ) );

            if ( x == NULL ) {
                perror ( "Error in malloc" );
                return ( ERROR );
            }

            scanf ( "%d", x );

            val = x;

        }
        else {
            // read from the shared memory
            if ( rw == 1 ) {
                printf ( "%d\n", *val );
            }
        }

        scanf ( "%d", &rw );
    }

    return ( 0 );

}

在此code我要测试的共享内存。我写在共享内存中的一个整数，当我给RW = 1否则我读共享内存的值，然后我打印此值。我找不到哪里出了问题......

In this code I want to test the shared memory. I write an integer in the shared memory when I give rw = 1 else I read the value of the shared memory and then I print this value. I can't find where is the problem....

推荐答案

您创建的共享内存段设置权限 0000 ：

You created the shared memory segment with permissions set to 0000:

shmid = shmget ( key, sizeof( int ), IPC_CREAT );

应

shmid = shmget ( key, sizeof( int ), IPC_CREAT | 0660 );

或相似的。

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