Linux fork与printf缓冲问题

printf输出条件:

(1) 调用fflush;

(2) 缓冲区满了;

(3) 遇到\n \r这些字符

(4) 遇到scanf这些要取缓冲区的;

(5) 线程或者进程退出;

 

fork之后会拷贝父进程的缓冲区;

 

代码:

 1 #include <unistd.h>
 2 #include <stdio.h>
 3 
 4 
 5 int glob = 6;
 6 char buf[ ] = "a write to stdout\n";
 7 
 8 int main()
 9 {
10     int var;
11     pid_t pid;
12 
13     var = 88;
14 
15     if (write(STDOUT_FILENO, buf, sizeof(buf) - 1) != sizeof(buf) - 1){
16         perror("write error\n");
17         return -1;
18     }
19 
20     printf("before fork\n");
21 
22     if ((pid = fork()) < 0){
23         perror("fork error\n");
24         return -1;
25     } else if (pid == 0){
26         glob++;
27         var++;
28     } else {
29         sleep (2);
30     }
31 
32     printf("pid = %d, glob = %d, var = %d\n", getpid(), glob, var);
33 
34     return 0;
35 }

 

结果:

[root@AY1404041716143649d3Z forkPrintf]# gcc forkPrintf.c -o forkPrintf
[root@AY1404041716143649d3Z forkPrintf]# ./forkPrintf
a write to stdout
before fork
pid = 17098, glob = 7, var = 89
pid = 17097, glob = 6, var = 88
[root@AY1404041716143649d3Z forkPrintf]# ./forkPrintf > tmp.out
[root@AY1404041716143649d3Z forkPrintf]# cat tmp.out
a write to stdout
before fork
pid = 17106, glob = 7, var = 89
before fork
pid = 17105, glob = 6, var = 88

 

(1) 第一次输入到控制台,write无缓冲直接输出,printf("before fork\n")存在\n行缓冲标识直接输出,而子进程拷贝父进程代码段,随后也会输出最后的printf;

(2) 第二次输入到文件,write无缓冲直接输出,printf("before fork\n")成了全缓冲的并没有输出,子进程拷贝了父进程的缓冲区和代码段,所以当父子进程退出时,刷新缓冲区,均有两条printf输出;

 

posted @ 2016-04-01 18:57  AlexAlex  阅读(1023)  评论(0编辑  收藏  举报