我一直在从事这项学校作业，我们正在C中创建Collatz Conjecture。我已经完成了所有代码，剩下的唯一事情就是传递数字并同时打印数字我从逻辑中得到回报。我尝试在每次更改输入之前和之后使用printf()，但这对于每次迭代仅打印两次。以下是其工作方式的示例。我还将代码放在下面。
For example：
如果我输入数字4。我应该回来4,2,1
Input Passed : ./hw1 4
Expected Output: 4 , 2, 1
Child ID is : 0
Child ID 1 is : 17488
Parent ID is 17487
Parent PID 1 is : 17488
Actual Output: , 2, 1
Child ID is : 0
Child ID 1 is : 17488
Parent ID is 17487
Parent PID 1 is : 17488
#include <stdio.h>
#include <unistd.h>
#include <sys/types.h>
int main(int argc, char *argv[]) {
// this is the number the user passes in
int n;
int pid;
int my_status;
pid_t my_pid, my_secondpid;
my_pid = fork();
if (argc == 1) {
fprintf(stderr, "Usage: ./hw1 <starting value>\n");
return -1;
}
n = atoi(argv[1]); // n is the input starting value
// Error checking
if (n < 0 || n == 0) {
printf("Number cannot be less than 0");
return 1;
}
//pid_t my_pid = fork();
if (my_pid < 0) {
printf("Unsuccesfull in creating the child process");
return 1;
} else if (my_pid == 0) {
my_secondpid = getpid();
//pid_t my_child = getpid();
while (n != 1) {
//pid_t my_child = getpid();
// if the number is even
if (n % 2 == 0) {
//printf("%d " , n);
n = n / 2;
printf(" , %d", n);
// printf("\n child id is : %d", my_child);
// printf(" \n parent : %d" , my_pid);
//printf("\n parent id is : %d" , my_pid);
}
// if the number is odd
else if (n % 2 != 0) {
//printf("%d" , n);
n = 3 * n + 1;
printf(" , %d", n);
//printf("\n child id is : %d", my_child);
// printf(" \n parent : %d" , my_pid);
//printf("\n parent id is : %d" , my_pid);
}
//printf("%d \n", n);
} // end of while
printf("\n");
printf("\nChild ID is : %d", my_pid);
printf("\nChild ID 1 is : %d", my_secondpid);
//printf("\nProcess ID : %d", pid);
} // end of else if
else {
wait(NULL);
my_secondpid = getpid();
//printf("\n The starting pid is : %d \n" , my_pid);
printf("\nParent ID is %d", my_secondpid);
printf("\nParent PID 1 is : %d", my_pid);
printf("\n");
}
return 0;
}