我有3个文件-SwimMill.c，Fish.c和Pellets.c-每个文件都被编译成可执行文件。运行SwimMill时，它使用fork()和exec()运行Fish和Pellets。但是，由于某种原因，当我使用终端时，使用make编译程序并运行SwimMill，文件Fish首先运行。有人可以帮我吗？
Makefile
all: SwimMill Fish Pellets
SwimMill: SwimMill.c
gcc -o SwimMill SwimMill.c
Fish: Fish.c
gcc -o Fish Fish.c -lm
Pellets: Pellets.c
gcc -o Pellets Pellets.c
SwimMill.c
// Uses both fish and pellets, 30 seconds, then print it out
// Create pellets at random intervals, from 0x80
// Eating --> Get rid of most significant bit
// Use shared memory for fish and pellet position only
#include <stdio.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <sys/types.h>
#include <signal.h>
#include <stdlib.h>
#include <time.h>
#include <unistd.h>
#define SHM_SIZE 1000
/*
TODO: SwimMIll FIRST, draw and get everything working
*/
/* 1. Create share memory using shmget
2. Attach to shared memory using shmat
3. Do operations
4. Detach using shmdt
*/
void printGrid(int*);
void handler(int);
int main(int argc, char* argv[]) {
printf("Hello");
signal(SIGINT, handler);
key_t key;
int shmid;
int *shm;
int timer = 0;
int fish = 0;
int pellet[20];
key = ftok("SwimMill.c", 'b'); //generate random ke
shmid = shmget(key, SHM_SIZE, IPC_CREAT|0666);
shm = shmat(shmid, NULL, 0); // Attach
// Initializing the shared memory to prevent segmentation fault
for (int i = 0; i < SHM_SIZE; i++){
shm[i] = -1;
}
int index = 0;
while(timer <= 30){
sleep(1); // Slow process down
fish = fork();
execv("Fish", argv);
pellet[index] = fork();
execv("Pellets", argv);
printGrid(shm);
printf("\n");
timer++;
index++;
}
shmdt(shm);
shmctl(shmid, IPC_RMID, NULL);
printf("Program finished! \n");
getchar(); // Pause consol
return 0;
}
void printGrid(int* shm) {
int row = 10;
int column = 10;
char stream[row][column]; //2D Dimensional array, fish can only move last row of 2d
//Initializing grid first
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
stream[i][j] = '~';
}
}
//Printing out grid with fish and pellet
for (int i = 0; i < row; i++) {
for (int j = 0; j < column; j++) {
stream[i][j] = '~'; // water
for (int k = 0; k < 20; k++) {
stream[shm[k]/10][shm[k]%10] = 'O'; // pellets
stream[shm[0]/10][shm[0]%10] = 'Y'; // Fish
}
printf("%c ", stream[i][j] );
}
printf("\n");
}
}
void handler(int num) {
perror(" Interrupt signal is pressed!! \n");
exit(1);
}
Fish.c
// 1 fish
// Scan the entire array, and focus on one pellet
// Arrange itself
#include <stdio.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <math.h>
#include <unistd.h>
int findClosestPellet(int*);
void moveLeft(int, int*);
void moveRight(int, int*);
int main() {
printf("printing from fish");
key_t key = ftok("SwimMill.c", 'b');
int shmid = shmget(key, 1024, IPC_CREAT|0666);
int *shm = (int*) shmat(shmid, NULL, 0);
int fish = 94; // Middle position
shm[0] = fish; // Store first shared memor space to fish
int columnToMoveTo = 0;
while(1) {
int closestPellet = shm[findClosestPellet(shm)];
if ((closestPellet % 10) > (fish % 10) ) {
moveRight(fish, shm);
}
else if ((closestPellet % 10) < (fish % 10)) {
moveLeft(fish, shm);
}
sleep(1);
}
shmdt(shm);
return 0;
}
int findClosestPellet(int* shm) {
// Using distance formula to find closest pellet
// (x2 - x1)^2 + (y2 - y1)^2
int closestPellet = 0;
int distance[20] = {0}; // Distance of all 20 pellets
int minimumDistance = 0;
// shm[1] = 11;
// shm[2] = 14;
// shm[3] = 10;
// shm[4] = 55;
int x2 = shm[0] % 10;
int y2 = shm[0] / 10;
for (int i = 1; i < 20; i++) {
int x1 = shm[i] % 10;
int y1 = shm[i] / 10;
distance[i] = pow(x2-x1,2) + pow(y2-y1,2); // Storing them
}
minimumDistance = distance[1];
//Finding smallest distance
for (int i = 2; i < 20; i++) {
if (distance[i] <= minimumDistance) {
closestPellet = i;
}
}
printf("Closest pellet %d \n", closestPellet);
return shm[closestPellet];
}
void moveLeft(int fish, int* shm) {
if (shm[0] <= 90) {
}
else{
fish--;
shm[0]--;
}
}
void moveRight(int fish, int* shm) {
if (shm[0] >= 99){
}
else{
fish++;
shm[0]++;
}
}
Pellets.c
// Multiple pellets
//Process ID, position, eaten/misse
#include <stdio.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <unistd.h>
#include <stdlib.h>
void eatPellet();
void missPellet();
int main() {
key_t key = ftok("SwimMill.c", 'b');
int shmid = shmget(key, 1024, IPC_CREAT|0666);
int *shm = (int*) shmat(shmid, NULL, 0);
int i = 1; // 1 - 19 are pellets
for (; i < 20; i++) {
int pelletPosition = rand() % 9 + 0; // random number from 0 - 9
shm[i] = pelletPosition;
break;
}
while(1) {
if (shm[i] < 90) {
shm[i] += 10;
}
else if (shm[i] == shm[0]) {
eatPellet();
printf("Position: %d\n", shm[i] );
break;
// EATEN and KILL
}
else {
// KIll process, terminate
missPellet();
printf("Position: %d\n", shm[i] );
break;
}
// printf("%d\n",shm[i] );
i++;
sleep(1);
}
shmdt(shm);
return 0;
}
void eatPellet() {
printf("Pellet eaten!");
printf("PID: %d \n", getpid());
}
void missPellet() {
printf("Pellet missed");
printf("PID: %d \n", getpid());
}
对于makefile，我运行“ make”。然后，我运行./SwimMill。但是，由于某种原因，它运行Fish。
答案 0 :(得分：9)
您没有正确使用fork / exec：
while(timer <= 30){
sleep(1); // Slow process down
fish = fork();
execv("Fish", argv);
pellet[index] = fork();
execv("Pellets", argv);
printGrid(shm);
printf("\n");
timer++;
index++;
}
回想一下fork函数返回了两次：一次返回到父进程，在该子进程中它返回子进程的pid，一次返回到子进程，在该子进程中它返回0。
您正在分叉一个新进程，但未检查返回值。因此，孩子和都调用execv来启动“ Fish”程序，因此您有两个“ Fish”运行副本，而没有“ SwimMill”运行副本。
您需要检查fork的返回值以查看该进程是父进程还是子进程，并采取相应措施。
while(timer <= 30){
sleep(1); // Slow process down
fish = fork();
if (fish == -1) {
perror("fork failed");
exit(1);
} else if (fish == 0) {
execv("Fish", argv);
perror("exec failed");
exit(1);
}
pellet[index] = fork();
if (pellet[index]== -1) {
perror("fork failed");
exit(1);
} else if (pellet[index] == 0) {
execv("Pellets", argv);
perror("exec failed");
exit(1);
}
printGrid(shm);
printf("\n");
timer++;
index++;
}
答案 1 :(得分：2)
是什么让您认为Fish在SwimMill之前奔跑？
如果您依赖于输出到stdout的顺序，则会被"hello"的缓冲所迷惑，直到写'\n'，fflush(stdout)之前，缓冲printf("Hello");
才立即输出调用，否则过程终止。
也就是说SwimMill首先运行，而Fish首先显示输出。
更改：
printf("Hello\n");
到
printf("printing from fish");
类似地：
printf("printing from fish\n");
到
system.windows.control.images