小小缓冲区问题

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原文链接：http://www.cnblogs.com/helei/archive/2011/01/11/1933070.html

小小缓冲区问题

代码

#include < sys / wait.h >
#include < unistd.h >
#include < stdio.h >

#define MAXLINE 3

int
main( void )
{
char buf[MAXLINE];
pid_t pid;
int status;

printf( " %% " );
while (fgets(buf, MAXLINE, stdin) != NULL)
{
int i;
for (i = 0 ; i < strlen(buf); i ++ )
printf( " %c " ,buf[i]);
printf( " Done " );
if (buf[strlen(buf) - 1 ] == ' \n ' )
{
buf[strlen(buf) - 1 ] = 0 ;
printf( " Test %c " ,buf[strlen(buf) - 1 ]);
}
if ((pid = fork()) < 0 )
{
printf( " fork error\n " );
}
else if (pid == 0 )
{
printf( " The current buf is %s " , buf);
execlp(buf, buf, ( char * ) 0 );
}

if ((pid = waitpid(pid, & status, 0 )) < 0 )
printf( " waitpid error " );
printf( " %% " );
}

exit( 0 );

}

分析过程：

(gdb) b main
Breakpoint 1 at 0x80485be: file 1-5.c, line 14.
(gdb) r 1-5
Starting program: /home/helei/UnixProgram/1-5 1-5

Breakpoint 1, main () at 1-5.c:14
14    printf("%% ");
(gdb) n
15    while(fgets(buf, MAXLINE, stdin) != NULL)
(gdb) n
% pwd
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
19    printf("%c",buf[i]);
(gdb) watch buf[i]
Hardware watchpoint 2: buf[i]
(gdb) watch i
Hardware watchpoint 3: i
(gdb) n
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
Hardware watchpoint 2: buf[i]

Old value = 112 'p'
New value = 119 'w'
Hardware watchpoint 3: i

Old value = 0
New value = 1
0x080485f3 in main () at 1-5.c:18
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
19    printf("%c",buf[i]);
(gdb) n
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
Hardware watchpoint 2: buf[i]

Old value = 119 'w'
New value = 0 '\000'
Hardware watchpoint 3: i

Old value = 1
New value = 2
0x080485f3 in main () at 1-5.c:18
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
20    printf("Done");
(gdb) n
21    if(buf[strlen(buf) -1] == '\n')
(gdb) n
26    if((pid = fork()) < 0)
(gdb) n
30    else if(pid == 0)
(gdb) n
36    if((pid = waitpid(pid, &status, 0)) < 0)
(gdb) n
38    printf("%% ");
(gdb) n
15    while(fgets(buf, MAXLINE, stdin) != NULL)
(gdb) n
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
Hardware watchpoint 2: buf[i]

Old value = 0 '\000'
New value = 100 'd'
Hardware watchpoint 3: i

Old value = 2
New value = 0
0x080485d8 in main () at 1-5.c:18
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
19    printf("%c",buf[i]);
(gdb) n
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
Hardware watchpoint 2: buf[i]

Old value = 100 'd'
New value = 10 '\n'
Hardware watchpoint 3: i

Old value = 0
New value = 1
0x080485f3 in main () at 1-5.c:18
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
19    printf("%c",buf[i]);
(gdb) n
pwDone% d
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
Hardware watchpoint 2: buf[i]

Old value = 10 '\n'
New value = 0 '\000'
Hardware watchpoint 3: i

Old value = 1
New value = 2
0x080485f3 in main () at 1-5.c:18
18    for(i = 0; i<strlen(buf); i++)
(gdb) n
20    printf("Done");
(gdb) n
21    if(buf[strlen(buf) -1] == '\n')
(gdb) n
23    buf[strlen(buf)-1] = 0;
(gdb) n
24    printf("Test %c",buf[strlen(buf)-1]);
(gdb) n
26    if((pid = fork()) < 0)
(gdb) n
DoneTest dThe current buf is dwaitpid error% 30    else if(pid == 0)

结论：

（１）fget函数在已达到文件结尾或出错时返回ＮＵＬＬ；

（２）若缓冲区不能足够容纳读入的一行，则缓冲区存储的是不完整的行，且以null字符结尾；

（３）输入的行未读完，剩余部分再从头写缓冲区，将覆盖原来的内容；

（4）读完最后一个字符\n后，仍填null字符。

困惑：

（１）在第一次写缓冲区和第二次写的间隔，子进程部分为何不执行（如调试部分加粗标记)？

posted on 2011-01-11 17:12 巢北阅读( ...) 评论( ...) 编辑收藏

转载于:https://www.cnblogs.com/helei/archive/2011/01/11/1933070.html