在c ++中读取和写入二进制文件的int [英] Reading and writing int to a binary file in c++

问题描述

我不清楚如何读取长整数工作。如果我说

I am unclear about how reading long integers work. If I say

long int a[1]={666666}
ofstream o("ex",ios::binary);
o.write((char*)a,sizeof(a));

将值存储到文件，并想要读取它们

to store values to a file and want to read them back as it is

long int stor[1];
ifstream i("ex",ios::binary);
i.read((char*)stor,sizeof(stor));

如何使用存储在多个字节字符中的信息显示相同的数字数组？

how will I be able to display the same number as stored using the information stored in multiple bytes of character array?

推荐答案

o.write 不写入字符，如果标记为ios :: binary）。使用char指针是因为char的长度为1个字节。

o.write does not write character, it writes bytes (if flagged with ios::binary). The char-pointer is used because a char has length 1 Byte.

o.write((char*)a,sizeof(a)); 

（char *）a 的 o.write 应该写。然后将 sizeof（a）字节写入文件。没有字符存储，只是字节。

(char*) a is the adress of what o.write should write. Then it writes sizeof(a) bytes to a file. There are no characters stored, just bytes.

如果你在一个十六进制编辑器中打开文件，你会看到这样的如果a int i = 10 ：
0A 00 00 00 （4字节，x64上）。

If you open the file in a Hex-Editor you would see something like this if a is int i = 10: 0A 00 00 00 (4 Byte, on x64).

阅读是类比。

这是一个工作示例：

#include <iostream>
#include <fstream>
#include <string>

using namespace std;


int main (int argc, char* argv[]){
    const char* FILENAM = "a.txt";
    int toStore = 10;
    ofstream o(FILENAM,ios::binary);

    o.write((char*)&toStore,sizeof(toStore));
    o.close();

    int toRestore=0;
    ifstream i(FILENAM,ios::binary);
    i.read((char*)&toRestore,sizeof(toRestore));

    cout << toRestore << endl;


    return 0;
}

这篇关于在c ++中读取和写入二进制文件的int的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！