Question

我正在使用Round Robin调度程序模拟器，所以我有4个文件A.txt，B.txt，C.txt，D.txt。每个代表一个过程，其中的每一行代表一条指令。每行都有o，p，q，r行。所以我想从这些文件中创建一个文件RR.txt。即。量子= 8。

8 lines of file A
8 lines of file B
8 lines of file C 
8 lines of file D
next 8 lines of file A
next 8 lines of file B
next 8 lines of file C
next 8 lines of file D
...
and so on until run out of lines on all files.

我已经在阵列上找到了文件名

string names_processes[number_of_processes];

并从我创建的每个函数中获取行数

int countLines(const string& inputFile){
    int counter=0;ifstream file(inputFile);string str;
    while (getline(file, str)) {counter++;}
    return counter;
  }

并像使用它

int o = countLines(names_processes[0]); //file A.txt
int p = countLines(names_processes[1]); //file B.txt
int q = countLines(names_processes[2]); //file C.txt
int r = countLines(names_processes[3]); //file D.txt

Answer 1

假设您需要从文件中获得8行，则可以执行以下操作：

def collide():
    arrow_pos = (arrow.x + arrow.w // 2, arrow.y + arrow.h // 2)
    mpos = (-1, -1)
    if arrow.alive:
        for event in pygame.event.get():
            print(event)
            if event.type == pygame.MOUSEBUTTONUP:
                mpos = pygame.mouse.get_pos()
                print(mpos)
        for square in squares:
            if square.rect.collidepoint(mpos):
                square.act = not square.act
            if square.act:
                if arrow_pos == square.center:
                    square.run()

如果达到eof或读取8行，此处std::vector<string> getLines(const string& inputFile){ int counter; std::vector<string> lines; ifstream file(inputFile); std::string str; for(counter = 0; getline(file, str) && counter < 8; counter++){ lines.push_back(str); } return lines; }将退出循环。

在一个文件中混合多个文件行

1 个答案: