在这里,我尝试使用管道创建一个简单的客户端和服务器。我 fork 了一个进程,使子进程充当客户端,父进程充当服务器。下面是代码:
#include<unistd.h>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
void errorMsg(char* msg)
{
printf("\n%s\n", msg);
// exit(0);
}
int main()
{
int servfd[2];
int clntfd[2];
int chldpid;
if(pipe(servfd) < 0)
errorMsg("Unable to create server pipe.!!");
if(pipe(clntfd) < 0)
errorMsg("Unable to create client pipe.!!!");
if((chldpid = fork()) == 0)
{
char* txt = (char*) calloc(256, sizeof(char));
close(servfd[1]);
close(clntfd[0]);
printf("@Client: Enter a string: ");
//scanf("%s", txt); //or fgets
printf("Entered.!!");
int n;
txt = "Anything that you want will not be considered no matter what you do!!";
char txt1[256];
write(clntfd[1], txt, 256);
//if(txt1[strlen(txt1) - 1] = '\n')
//{ printf("asdasdas");
//txt[strlen(txt) - 1] = '\0';}
//int i = 0;
//for(i = 0; i < 256; i++)
//printf("%c", txt1[i]);
while((n = read(servfd[0], txt1, 256)) > 0)
printf("\nAt client: %d bytes read\n\tString: %s\n", n, txt1);
}
else
{
//printf("Parent: \n\n");
close(servfd[0]);
close(clntfd[1]);
char* txt = NULL;
int n, n1;
n = read(clntfd[0], &txt, 256);
printf("Server read: %d", n);
n1 = write(servfd[1], &txt, 256);
printf("Server write: %d", n1);
wait(chldpid);
}
exit(0);
}
问题 1:
这就是正在发生的事情。当我运行该程序时,它只打印 Anything that yo
(正好 16 个字符)而没有其他内容。当我尝试使用注释中显示的for
循环查看txt1
的完整内容时,我发现yo之后有空值(上帝知道从哪里来)
在 txt1
中。之后就是正常的内容了。知道为什么会发生这种情况吗?
编辑:
当我尝试在适当的位置打印它们时,读取和写入的字节数都是正确的。它打印 256 bytes read
。但是,strlen 的 txt1
的大小为“16”。此外,程序在打印部分字符串后挂起。
问题 2:
当我尝试使用注释中显示的 scanf
或 fgets
从用户那里获取字符串时,只要我按 Enter 键,程序就会终止。也不知道为什么会发生这种情况。
对行为的任何见解都会有所帮助。抱歉有多个问题。谢谢你的时间。!我正在使用 ubuntu 12.04,如果这有任何帮助的话。
最佳答案
我已向您的代码添加了各种注释和更正。现在它按预期工作了。 正如 codeaddict 所指出的,您的主要问题是您没有分配缓冲区。我很惊讶你没有因 SIGSEGV 崩溃。
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void errorMsg(char* msg)
{
printf("\n%s\n", msg);
// exit(0);
}
// move this into global space and make it const (non modifiable, easyer to debug)
const char text_to_send[] = "Anything that you want will not be considered no matter what you do!!";
int main()
{
int servfd[2];
int clntfd[2];
int chldpid;
if(pipe(servfd) < 0)
errorMsg("Unable to create server pipe.!!");
if(pipe(clntfd) < 0)
errorMsg("Unable to create client pipe.!!!");
if((chldpid = fork()) == 0)
{
char txt[256]; // You have to actually allocate a buffer (aka enough memory to hold your string. You have allocated a pointer to a buffer, but no actual buffer)
close(servfd[1]);
close(clntfd[0]);
printf("@Client: Enter a string: ");
scanf("%s", txt); // since you now actually have a buffer to put the input into this no longer fails
printf("Entered.!!\n");
int n;
char txt1[256];
write(clntfd[1], text_to_send, sizeof(text_to_send)); // write only as much as you actually have to say, not the whole size of your buffer
while((n = read(servfd[0], txt1, 256)) > 0)
printf("\nAt client: %d bytes read\n\tString: %s\n", n, txt1);
// this is not nessesary at this point, but it is good style to clean up after yourself
close(servfd[0]);
close(clntfd[1]);
}
else
{
//printf("Parent: \n\n");
close(servfd[0]);
close(clntfd[1]);
char txt[256]; // same here, you need to actually allocate a buffer.
int n, n1;
n = read(clntfd[0], txt, 256); // read into txt, not &txt. you want to read into your buffer pointed to by txt, not into the part of memory that contains the pointer
printf("Server read: %d\n", n);
n1 = write(servfd[1], txt, n); // do not send the whole buffer, just as much as you have actually useful information in it
printf("Server write: %d\n", n1);
// close the loose file descriptors, else your child will read on them forever
close(servfd[1]);
close(clntfd[0]);
int status;
wait(&status); // this is called like this. if you want to use the pid you call waitpid(chldpid, &status, 0);
}
exit(0);
}
关于c - 使用 C 中的管道创建客户端和服务器时的奇怪行为,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12788070/