将外部变量传递给类c ++

时间:2014-02-22 06:44:11

标签: c++ fifo

我在这里放了一个简化版的代码。我要做的是从输入文件中读取每一行并将其存储在“fifo”类中。然而,在每个商店之前,我尝试打印我之前的fifo值(这是用于调试)。我看到的是,即使在调用fifo.add函数之前,前一个值也被覆盖了。 getline函数本身会自行覆盖fifo。有人可以告诉我这里发生了什么吗?

//fifo.h

#ifndef FIFO_H_
#define FIFO_H_

class fifo {
        private:
                char* instr_arr;
                int head;
                int tail;
        public:
                fifo();
                fifo(int,int);
                void add(char*);
                void print_instruction(void);
};

#endif

//fifo.cpp

#include "fifo.h"
#include <iostream>
//using namespace std;

fifo::fifo() {
        head = 0;
        tail = 0;
        instr_arr = "NULL";
}

fifo::fifo(int h, int t) {
        head = h;
        tail = t;
}

void fifo::add (char *instr) {
        instr_arr = instr;
}

void fifo::print_instruction (void) {
        std::cout << instr_arr << std::endl;
}

//code.cpp

#include <iostream>
using namespace std;
#include <fstream>
using namespace std;
#include "fifo.h"

#define MAX_CHARS_INLINE 250

int main (int argc, char *argv[]) {
        char buf[MAX_CHARS_INLINE];     //Character buffer for storing line
        fifo instruction_fifo;          //FIFO which stores 5 most recent instructions

        const char *filename = argv[1];
        ifstream fin;
        fin.open(filename);

        while(!fin.eof()) {
                std::cout << buf << std::endl;
                fin.getline(buf, MAX_CHARS_INLINE);
                instruction_fifo.print_instruction();  //This instruction prints the line which was read from getline above!! Why??
                instruction_fifo.add(buf);
                }
        return 0;
}

1 个答案:

答案 0 :(得分:1)

您实际上并未将数据存储在FIFO(或由FIFO管理的存储器块中)中,而只存储指针(指向buf)。

当您打印instr_arr时,基本上是打印buf,因为instr_arr指向buf