[英]Fibonacci Sequence
#define MAX_SEQUENCE 10 // Max values to store in shared memory
#define MIN_SEQUENCE 2 // Min value the user can enter
//shared memory:
// 1) holds an array of numbers
// 2) holds how many numbers are in the array
typedef struct {
int fib_seq[MAX_SEQUENCE];
int sequence_size;
} shared_data;
//MAIN function
int main(int argc, char *argv[]) {
pid_t pid; //process ID
int segment_id; //Shared Memory ID
shared_data *mem; //Shared Memory Pointer
//check to validate atleast two arguments
if(argc != 2) {
printf("USAGE ERROR: [0-9]\n");
exit(0);
}
//validate the input is not larger then the MAX
if(atoi(argv[1]) > MAX_SEQUENCE) {
printf("Max Input Size: %d\n", MAX_SEQUENCE);
exit(0);
}
//validate the input is not smaller then the MIN
if(atoi(argv[1]) < MIN_SEQUENCE) {
printf("Min Input Size: %d\n", MIN_SEQUENCE);
exit(0);
}
// 1) create a new shared memory location 'IPC_PRIVATE'
// 2) the size of our shared memory structure 'sizeof(shared_data)'
// 3) Set Modes S_IRUSR and S_IWUSR so the owner can read and write to the shared memory 'S_IRUSR|S_IWUSR'
segment_id = shmget(IPC_PRIVATE, sizeof(shared_data), S_IRUSR|S_IWUSR);
//attach the shared memory and get the pointer to the beginning location in memory
mem = (shared_data *) shmat(segment_id,NULL,0);
//set the size of the sequence to the argument that was passed in via command line
mem->sequence_size = atoi(argv[1]);
// fork a child process
pid = fork();
if(pid < 0) { /* error occured */
fprintf(stderr, "Fork Failed\n");
return 1;
}
else if(pid == 0) { /* child process */
int counter = 0;
printf("Child Fibonacci Sequence: ");
while(counter < mem->sequence_size) {
if(counter == 0){
//FIB of zero is always zero
mem->fib_seq[counter] = 0;
}
else if(counter == 1){
//FIB of one is always one
mem->fib_seq[counter] = 1;
}
else {
//The Fibonacci Sequence formula 'R = fib(n-1) + fib(n-2)'
//The first two numbers in the sequence are always 0 and 1.
//To get a value in the sequence you will want to take the previous
//two numbers and add them together. For example:
// b + a = c
// [fib(d-1) = c] + [fib(d-2) = b] = R
// fib(0) = 0
// fib(1) = 1
// fib(2): 1 + 0 = 1
// fib(3): 1 + 1 = 2
// fib(4): 2 + 1 = 3
// fib(5): 3 + 2 = 5
// The next Fibonacci number in the sequence will be '8'
mem->fib_seq[counter] = mem->fib_seq[counter - 1] + mem->fib_seq[counter - 2];
}
printf("%d ", mem->fib_seq[(counter)]);
counter++;
}
}
else { /* parent process */
/* parent will wait for the child process to complete */
wait(NULL);
//Print out shared memory
int count = 0;
printf("\nParent Fibonacci Sequence: ");
while(count < mem->sequence_size){
printf("%d ", mem->fib_seq[count]);
count++;
}
//detach shared memory
shmdt(mem);
//remove shared memory segment
shmctl(segment_id,IPC_RMID,NULL);
printf("\nComplete\n");
}
return 0;
}
好的,我有一個已經使用了一段時間的程序,問題是數字序列偏離1,而我似乎找不到它的位置。 它沒有為fib(0)打印0。 因此,當我執行Fib(2)時,它給了我0 1而不是0 1 1,有人對我有什么建議嗎?
經典的Obi Wan錯誤(一對一)。 您需要做:
mem->sequence_size = atoi(argv[1]) + 1;
(已編輯,先前的發布導致超出范圍的數組訪問)
您的代碼似乎可以正常工作...
如果您希望Fib(2)打印出三個數字,則可能需要看一下這一行:
while(counter < mem->sequence_size) {
但是,如果這樣做,則需要注意,您將需要11個內存來計算Fib(10)。 您目前只能給自己10個內存。
如果這沒有意義,則將MIN_SEQUENCE
設置為0,並問自己在計算Fib(0)
時代碼的作用。
根據您的代碼,序列大小設置為您要傳遞的參數
mem->sequence_size = atoi(argv[1]);
在while循環中,您正確檢查為:
while(counter < mem->sequence_size)
。
因此, Fib 2
應該只打印斐波那契序列中的2個元素0 1
。 如果您需要0 1 1
的輸出,那么您是否會以fib 3
身份運行程序?
您正在打印counter
編號的項目。 即,當counter == 2時,您將打印兩個數字,而當counter == 0時,您將打印零。
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