在过去的几个小时里,我一直被困在两个程序之间工作,我被卡住了,不知道我做错了什么。我的程序的想法是我将使用 interface.c 打开管道,然后执行 db.c 。我想使用两个管道在两个不同的程序之间进行通信。现在,将interface.c作为父母'并且db.c是孩子',我不确定我是否通过execl命令正确地将参数传递给我的管道。所有内容都正确编译,但是当我尝试运行界面程序时,我收到一条错误消息:'文件编号错误。'我可能没有正确使用管道吗?目前,我只是试图让我的程序通过管道向db.c发送一个整数值。任何帮助将不胜感激。
interface.c的代码
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <errno.h>
#include <sys/wait.h>
//PIPES:
//
//Parent: reads from P1_READ, writes on P1_WRITE
//Child: reads from P2_READ, writes on P2_WRITE
#define P1_READ 0
#define P2_WRITE 1
#define P2_READ 2
#define P1_WRITE 3
// the total number of pipe *pairs* we need
#define NUM_PIPES 2
int main(int argc, char *argv[])
{
//Create Pipe Array
int fd[2*NUM_PIPES];
//For Parameter Passing:
char param0[20]; //P1_Read
char param1[20]; //P2_Write
char param2[20]; //P2_Read
char param3[20]; //P1_Write
snprintf(param0, sizeof(param0), "%d" , fd[0]);
snprintf(param1, sizeof(param1), "%d" , fd[1]);
snprintf(param2, sizeof(param2), "%d" , fd[2]);
snprintf(param3, sizeof(param3), "%d" , fd[3]);
//Variables
pid_t pid;
int val = 42;
//Allocate the PIPES
for (int i=0; i<NUM_PIPES; ++i)
{
if(pipe(fd+(i*2)) < 0)
{
perror("Failed to allocate the pipes");
exit(EXIT_FAILURE);
}
}
//If the fork of the program does not work:
if ((pid = fork()) < 0)
{
perror("Failed to fork process");
return EXIT_FAILURE;
}
if(pid == 0)
{ //Child Process
execl("./db", "db", param0, param1, param2, param3, (char *)NULL);
}
else
{ //Parent Process
//SENDING VALUES HERE
close(fd[P2_READ]);
close(fd[P2_WRITE]);
printf("Interface is sending|%d| to DB\n", val);
if(write(fd[P1_WRITE],&val, sizeof(val)) != sizeof(val))
{
perror("Interfae failed to send value to DB");
exit(EXIT_FAILURE);
}
}
return 0;
}
这是针对db.c
的#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>
#include <sys/wait.h>
#include <sys/types.h>
//Typedef-Class-
typedef struct Information
{
int accountId;
int checkNumber;
int date;
float amount;
} Information;
int main(int argc, char *argv[])
{
//For Input
//Account Data
Information acctData[25];
int dataStorageLooper = 0; //How many db entries
//For File Input
int aVal;
int bVal;
int cVal;
float dVal;
//Prepare for file input:
FILE * fp;
fp = fopen ("accountData.txt", "r");
//Reads Input
while(1)
{
if (fscanf(fp, "%d %d %d %f", &aVal, &bVal, &cVal, &dVal)!=4)
{
break;
}
//Puts data into appropriate arrays
acctData[dataStorageLooper].accountId= aVal;
acctData[dataStorageLooper].checkNumber= bVal;
acctData[dataStorageLooper].date= cVal;
acctData[dataStorageLooper].amount= dVal;
dataStorageLooper++;
}
//Decrement index to point to last item
dataStorageLooper--;
//Displays all values
printf("\nDisplaying AccountData.txt\n");
for( int i = 0; i < dataStorageLooper; i++)
{
printf("Line|%d|: Account|%d|: Check|%d|: Date|%d|: Amount|%.2f|\n",i,acctData[i].accountId,acctData[i].checkNumber,acctData[i].date,acctData[i].amount);
}
//Closes File
fclose(fp);
//End Input
//Parameter Receiving:
int pipes[4]; //Pipe Array
int value = 7;
int test;
//Build the pipes
pipes[0] = atoi(argv[1]); //P1_Read
pipes[1] = atoi(argv[2]); //P2_Write
pipes[2] = atoi(argv[3]); //P2_Read
pipes[3] = atoi(argv[4]); //P1_Write
//Troubleshooting
printf("The number of parameters: %d\n",argc);
printf("Parameter 1: %s\n", argv[0]);
printf("I stared correctly\n");
//Testing
close(pipes[0]);
close(pipes[3]);
//SHOULD RECEIVE VALUE HERE
test = read(pipes[2], &value, sizeof(value));
if (test < 0)
{
perror("DB: Failed to read data from parent");
exit(EXIT_FAILURE);
}
else if (test == 0)
{
//Unexpected
fprintf(stderr, "DB: Read End-Of-File from pipe");
}
else
{
//What did the child receive?
printf("DB: Received Value:(%d)\n", value);
}
close(pipes[2]);
close(pipes[1]);
return 0;
}
答案 0 :(得分:0)
在您为其分配任何价值之前,snprintf
fd
中LIBBSON_DIR
中各种元素的价值LIBMONGOC_DIR
C:\mongo-c-driver
。这种未定义的行为,以及您作为参数传递的值完全没有意义(充其量)。
但这让我觉得这是一种非常奇怪的做事方式。通常你只需要dup2 fds 0和1,这样孩子的stdin和stdout就会被重定向到适当的管道fds。