利用工作之余为小伙伴写了份作业，关于进程间通信的。题目如下：
“父进程从键盘上接受1000个数据，对其求和sum1，子进程对这1000个数平方和sum2，结果传给父进程，父进程将sum1+sum2后，打印结果。”
要求：用大小为10的共享区传递1000个数据；子进程用消息机制将sum2传给父进程。
主要利用共享内存实现进程间通信，使用管道实现进程间竞争关系，freebsd下测试通过。代码如下：时间有限，有可能有些不足，希望高手给予指点。
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#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/shm.h>
#include <signal.h>
const int key = 0x12345678;
static int pfd1[2], pfd2[2];
#define shm_len (10*1024)
#define val_num 5
int init_shm() {
int shmid = -1;
shmid = shmget((key_t)key, shm_len, 0666 | ipc_creat);
if (shmid < 0) {
printf("shmget failed!\n");
exit(-1);
}
return shmid;
}
void cancel_shm(int shmid) {
if (shmctl(shmid, ipc_rmid, 0) == -1) {
printf("shmctl failed!\n");
exit(-1);
}
printf("cancel_shm successfully!\n");
}
void *shm_get(int shmid) {
void *mem = null;
mem = shmat(shmid, 0, 0);
if (mem == (void *)-1) {
printf("shmat failed!\n");
exit(-1);
}
return mem;
}
int get_val(int *val, int num) {
int i;
for (i = 0; i < num; i++) {
printf("please input a num:");
scanf("%d", val + i);
}
}
void show_val (int *val, int num) {
int i;
for (i = 0; i < num; i++) {
printf("%d\t", *(val + i));
}
printf("\n");
}
int add_val (int *val, int num) {
int result = 0;
int i;
for (i = 0; i < num; i++) {
result += *(val + i);
}
return result;
}
int square_val (int *val, int num) {
int result = 0;
int i, tmp;
for (i = 0; i < num; i++) {
tmp = *(val + i);
result += (tmp * tmp);
}
return result;
}
void tell_wait (void) {
if (pipe(pfd1) < 0 || pipe(pfd2) < 0) {
printf("pipe error!\n");
exit(-1);
}
}
void tell_parent (pid_t pid) {
if (write(pfd2[1], "c", 1) != 1) {
printf("write error!\n");
exit(-1);
}
}
void wait_parent (void) {
char c;
if (read(pfd1[0], &c, 1) != 1) {
printf("read error!\n");
exit(-1);
}
}
void tell_child (pid_t pid) {
if (write(pfd1[1], "p", 1) != 1) {
printf("write error!\n");
exit(-1);
}
}
void wait_child (void) {
char c;
if (read(pfd2[0], &c, 1) != 1) {
printf("read error!\n");
exit(-1);
}
}
int main(int argc, char *argv[]) {
void *mem = null;
int shmid = -1;
pid_t pid = -1;
int val[val_num];
int result = 0;
shmid = init_shm();
tell_wait();
if ((pid = fork()) < 0) { //error
printf("fork error!\n");
exit(-1);
} else if (pid == 0) { //child
int result = 0;
wait_parent();
mem = shm_get(shmid); //get share memery
memcpy(val, mem, sizeof(int) * val_num);
result = square_val(val, val_num);
*(int *)((void *)mem + shm_len - 4) = result;
tell_parent(pid);
exit(1);
} else { //parent
int child_result = 0;
mem = shm_get(shmid); //get share memery
get_val(val, val_num); //get user input
memcpy(mem, val, sizeof(int) * val_num); //copy user input to share memery
tell_child(pid);
result = add_val(val, val_num);
wait_child();
child_result = *(int *)((void *)mem + shm_len - 4);
printf("result:%d, child_result:%d, all:%d\n", result, child_result, result + child_result);
}
cancel_shm(shmid);
return 0;
}
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HDU 4417 Super Mario（莫队 + 树状数组 + 离散化）
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