学习了mmap以后，实现一个简单的小程序，进行多个进程对一个文件进行拷贝。
Linux mmap共享内存学习可以参考我的另一篇博客：传送门
实现思想
我们可以将原来的文件利用mmap分成多个段分别进行传输。
实现代码
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<unistd.h>
#include<sys/types.h>
#include<sys/wait.h>
#include<sys/mman.h>
#include<fcntl.h>int main(int argc,char *argv[])
{char *input,*output;if(argc<2){printf("You should input file name\n");exit(1);}else if(argc>3){printf("too many arguments\n");exit(1);}else if(3 == argc){input=argv[1];output=argv[2];}else if(2 == argc){input=argv[1];output=strcat(input,".out");}int fd1 = open(input,O_RDONLY);if(-1 == fd1){perror("input file open error:");exit(1);}int fd2 = open(output,O_RDWR | O_CREAT | O_TRUNC,0644);if(-1 == fd2){perror("output file open error:");exit(1);}int file_size = lseek(fd1,0,SEEK_END);int ret = ftruncate(fd2,file_size);if(-1 == ret){perror("ftruncate ouput file error:");exit(1);}input = mmap(NULL, file_size, PROT_READ, MAP_SHARED, fd1, 0);if(MAP_FAILED == input){perror("input mmap error:");exit(1);}close(fd1);output = mmap(NULL, file_size, PROT_READ | PROT_WRITE, MAP_SHARED, fd2, 0);if(MAP_FAILED == output){perror("output mmap error:");exit(1);}close(fd2);int psize = file_size / 4;pid_t pid;for(int son=0; son<4; ++son){pid = fork();if(-1 == pid){perror("fork error:");exit(1);}else if(0 == pid){memcpy(output+son*psize, input+son*psize, psize);exit(0);}}if(file_size > psize * 4){memcpy(output+4*psize,input+4*psize,file_size-psize*4);}ret = munmap(input, file_size);if(-1 == ret){perror("close input mmap error:");exit(1);}ret = munmap(output, file_size);if(-1 == ret){perror("close output mmap error:");exit(1);}return 0;
}和折半查找)
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