數組元素都指向同一個對象無故

-1

我解決一類項目，我遇到這個問題這裏（當我追加新對象好像所有元素的數組是在同一個對象指向）是我的代碼數組元素都指向同一個對象無故

import os 
from Process import Process 

processes = [] 

# Read the file 
file = open("file.txt","r") 
# Read the lines 
for line in file.readlines() : 
    # split the line components 
    process = line.split(";") 
    global processes 
    # append a new element to the array 
    pr = Process(process[0], process[1], "Ready") 
    # find how many elements in the line to loop through 
    process_len = len(process) 
    # loop through the line elements 
    for element in range(2,process_len) : 
     # if it's a CPU 
     if process[element].find("CPU") == 0: 
      s = process[element].split(":") 
      pr.add_cpu(s[1]) 
     # if it's an IO 
     elif process[element].find("IO") == 0: 
      s = process[element].split(":") 
      pr.add_io(s[1]) 
    #  End of elif 
    processes.append(pr) 
file.close() 
print "-------------------------- \n"; 
for p in processes : 
    print p.getcpu()

這裏，我從讀文件：

1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35;IO:5;CPU:1 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11; 
1;500;CPU:20;IO:5;CPU:10;IO:15;CPU:30;IO:11;CPU:35 
1;500;CPU:20;IO:5;CPU:10;IO:15;

，這裏是輸出：

-------------------------- 

['20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10'] 
['20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10'] 
['20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10'] 
['20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10'] 
['20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10', '20', '10', '30', '35', '1\n', '20', '10', '30', '20', '10', '30', '35\n', '20', '10'] 
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Process finished with exit code 0

我認爲這個問題是我要創建同一個對象的新引用和我一直把它們添加到數組作爲新的對象，但我不知道如何確保，如果是這樣的話還是我錯了如果是這種情況，我該如何解決這個問題？

編輯：下面是該過程類：

class Process : 

    pid = 0; 
    pat = 0; 
    wt = 0; 
    global ion; 
    ion = 0; 
    global cpun; 
    cpun = 0; 
    global pState; 
    pState = "Ready"; 
    global pNature; 
    pNature = "I\O bound"; 
    global turnAroundTime; 
    turnAroundTime = 0; 
    global cpu; 
    cpu = [] 
    global io; 
    io = [] 

    # constructor 
    def __init__(self,id,at,state) : 
     pid = id; 
     pat = at; 
     pState = state; 

    # define the process nature 
    @staticmethod 
    def nature() : 
     global pNature; 
     global ion; 
     global cpun; 
     if ion > cpun : 
      pNature = "I\O bound"; 
     else : 
      pNature = "CPU bound"; 

    # add new cpu 
    @staticmethod 
    def add_cpu(cp) : 
     cpu.append(cp); 
     global cpun; 

     cpun = cpun + int(cp); 

    # add new io 
    @staticmethod 
    def add_io(i) : 
     io.append(i); 
     global ion; 
     ion = ion + int(i); 

    # get the cpu array size 
    @staticmethod 
    def get_cpu_size() : 
     return len(cpu); 

     # get the IO array size 
    @staticmethod 
    def get_io_size() : 
     return len(io) 

    # returns the cpu array 
    @staticmethod 
    def getcpu(): 
     return cpu

來源

2016-12-30 keloa

@凱丹格我得到相同的輸出 – keloa

我保證它不是沒有原因的。但是你需要顯示Process類的代碼。 –

BTW，做'在全球範圍內的全球processes'是毫無意義的。請參閱[文檔]（https://docs.python.org/3/reference/simple_stmts.html#grammar-token-global_stmt） –

類過程的變量是共同的PR的各instansiation。所以你每次都在壓制價值。

import os 

processes = [] 

class Process(): 
    # constructor 
    def __init__(self, id, at, state): 
     self.pid = id 
     self.pat = at 
     self.pState = state 
     self.wt = 0 
     self.ion = 0 
     self.cpun = 0 
     self.pNature = "I\O bound" 
     self.turnAroundTime = 0 
     self.cpu = [] 
     self.io = [] 

    # define the process nature 
    def nature(self): 
     if self.ion > self.cpun: 
      self.pNature = "I\O bound" 
     else: 
      self.pNature = "CPU bound" 

    # add new cpu 
    def add_cpu(self, cp): 
     self.cpu.append(cp) 
     self.cpun = self.cpun + int(cp) 

    # add new io 
    def add_io(self, i): 
     self.io.append(i) 
     self.ion = self.ion + int(i) 

    # get the cpu array size 
    def get_cpu_size(self): 
     return len(self.cpu) 

    # get the IO array size 
    def get_io_size(self): 
     return len(self.io) 

    # returns the cpu array 
    def getcpu(self): 
     return self.cpu 


# Read the file 
file = open("file.txt", "r") 
# Read the lines 
for line in file.readlines(): 
    # split the line components 
    process = line.split(";") 
    # append a new element to the array 
    pr = Process(process[0], process[1], "Ready") 
    # find how many elements in the line to loop through 
    process_len = len(process) 
    # loop through the line elements 
    for element in range(2, process_len): 
     # if it's a CPU 
     if process[element].find("CPU") == 0: 
      s = process[element].split(":") 
      pr.add_cpu(s[1]) 
     # if it's an IO 
     elif process[element].find("IO") == 0: 
      s = process[element].split(":") 
      pr.add_io(s[1]) 
    #  End of elif 
    processes.append(pr) 
file.close() 
print("-------------------------- \n") 
for p in processes: 
    print(p.getcpu())

PS：不要使用;在Python。並嘗試使用Pep8。這是一個Python風格指南。

來源

2016-12-30 10:51:47 Wilfried

職能是分開的每個對象，但我會後你期望的輸出是什麼類 – keloa

？ – Wilfried

這工作！謝謝 – keloa

數組元素都指向同一個對象無故

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