我有一项作业,要求我编写一个多处理程序,该程序可处理包含字符串的内存映射文件。父进程将文件映射到内存后,会生成2个子进程来修改文件。子进程 1 输出文件的内容,将文件的内容转换为大写形式,然后输出文件的新内容。子进程 2 等待 1 秒让子进程 1 完成,输出文件的内容,删除所有连字符“-”,然后输出文件的新内容。我对两个子进程的问题是,在第一次显示文件的内容后,进程尝试修改文件的内容,但两个子进程都不会输出文件的新内容。我在运行或编译时没有收到错误,所以我无法找出问题所在。当然,我是内存映射的新手,所以请随时让我知道我做错了什么。这是我的源代码:
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/mman.h>
#include <fcntl.h>
#include <sys/stat.h>
#include <signal.h>
#include <string.h>
int main (int argc, char *argv[]) {
struct stat buf;
int fd, length, status, i, j, k;
char *mm_file;
char *string = "this is a lowercase-sentence.";
length = strlen(string);
fd = open(argv[1], O_CREAT | O_RDWR, 0666); //Creates file with name given at command line
write(fd, string, strlen(string)); //Writes the string to be modified to the file
fstat(fd, &buf); //used to determine the size of the file
//Establishes the mapping
if ((mm_file = mmap(0, (size_t) buf.st_size, PROT_READ|PROT_WRITE, MAP_SHARED, fd, 0)) == (caddr_t) - 1) {
fprintf(stderr, "mmap call fails\n");
}
//Initializes child processes
pid_t MC0;
pid_t MC1;
//Creates series of child processes which share the same parent ID
if((MC0 = fork()) == 0) {
printf("Child 1 %d reads: \n %s\n", getpid(), mm_file);
//{convert file content to uppercase string};
for (i = 0; i < length; i++) {
string[i] = toupper(string[i]);
}
//sync the new contents to the file
msync(0, (size_t) buf.st_size, MS_SYNC);
printf("Child 1 %d reads again: \n %s\n", getpid(), mm_file);
exit(EXIT_SUCCESS); //Exits process
} else if ((MC1 = fork()) == 0) {
sleep(1); //so that child 2 will perform its task after child 1 finishes
("Child 2 %d reads: \n %s\n", getpid(), mm_file);
//{remove hyphens}
for (j = 0; j < length; i++) {
if (string[i] == '-') {
string[i] = ' ';
}
}
//sync the new contents to the file
msync(0, (size_t) buf.st_size, MS_SYNC);
printf("Child 2 %d reads again: \n %s\n", getpid(), mm_file);
exit(EXIT_SUCCESS); //Exits process
}
// Waits for all child processes to finish before continuing.
waitpid(MC0, &status, 0);
waitpid(MC1, &status, 0);
return 0;
}
那么我的输出如下:
**virtual-machine:~$** ./testt file
Child 1 3404 reads:
this is a lowercase-sentence.
Child 2 3405 reads:
this is a lowercase-sentence.
All child processes have finished. Now exiting program.
**virtual-machine:~$**
但我想要的结果是:
**virtual-machine:~$** ./testt file
Child 1 3404 reads:
this is a lowercase-sentence.
Child 1 3404 reads again:
THIS IS A LOWERCASE-SENTENCE.
Child 2 3405 reads:
THIS IS A LOWERCASE-SENTENCE.
Child 2 3405 reads:
THIS IS A LOWERCASE SENTENCE.
All child processes have finished. Now exiting program.
**virtual-machine:~$**
非常感谢任何帮助。
最佳答案
这里有一些错误。首先,写入文件,然后将其映射到内存中。映射是正确的,但书写不正确。如果字符串有 n 个字符,则必须编写 n+1 个字符,因为 C 中的字符串以 null 结尾。现在你只有 n,所以所有 C 字符串函数都会尝试访问至少一个字节,这是不好的。如果那个额外的字节不为空(零),则函数将走得更远。在更多调试中,它们可能会被清零,但在优化代码中通常不会。所以你必须使用
write(fd, string, strlen(string)+1); //Writes the string to be modified to the file
然后你这样做:
for (i = 0; i < length; i++) {
string[i] = toupper(string[i]);
}
这只改变指针string
引用的数据,与内存映射文件无关。你应该有:
for (i = 0; i < length; i++) {
mm_file[i] = toupper(mm_file[i]);
}
第二个子进程也是如此。
此外,您的 msync()
调用也有点可疑。您将内存地址指定为 0,该地址不在内存映射文件内,因此它不会同步内容。您需要调用msync(mm_file, (size_t) buf.st_size, MS_SYNC);
此外,许多编译器会将常量字符串放入只读内存中,因此您甚至可能不允许更改 string
引用的数据。在这种情况下,您似乎被允许。
还请记住,文件的长度比字符串的长度大一个字节,因此请正确使用变量。目前您可以这样做,因为您将文件与文件长度同步并使用字符串长度处理字符串。
关于C - 多进程的内存映射,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29910878/