从二进制文件读取32位整数 [英] read 32bit integer from binary file

问题描述

  00000000：0000 0803 0000 ea60 0000 001c 0000 001c 
 00000010： 0000 0000 0000 0000 0000 0000 0000 0000 
  

左栏是地址。

我试图读取 0000 0803 （= 2051）如下

  ifstream if; 
 if.open（file）; 
 uint32_t a; 
如果>>一个; 
  

正如预期的那样...它没有工作:-(

a 在执行后只是0。 >。

为什么这些都不起作用，我该如何实现目标？

你有两个问题：

$ ol
保证你读取你想要的字节数（不少于，不是）

我推荐这个语法：

uint32_t a;

inFILE.read（reinterpret_cast< char *>（& a） ）;


确保你用正确的字节顺序来解释这些字节。

问：如果你在PC上，你的CPU可能是小尾数。你知道吗你的数据流也是小端，或者是大端？

如果数据是大端，我会考虑标准的网络功能来容纳字节顺序： ntohl（）等： http://www.retran.com/beej/htonsman.html

ALSO：

遵循Hcorg和Daniel Jour的建议：不要忘记open mode参数，不要忘记检查file open错误。 >

My binary file looks like this.

00000000: 0000 0803 0000 ea60 0000 001c 0000 001c
00000010: 0000 0000 0000 0000 0000 0000 0000 0000

left column is address.

I just tried to read 0000 0803(=2051) as follows

ifstream if;
if.open("file");
uint32_t a;
if >> a;

As expected...It did not work :-(
a was just 0 after execution.
I tried long, int, unsigned int, unsigned long. All failed.

Why these are not working and how can I achieve the goal?

解决方案

You have two issues:

1. Insuring you read the bytes you intend (no fewer, no more) from the stream.

   I'd recommend this syntax:

   uint32_t a;

   inFILE.read(reinterpret_cast<char *>(&a), sizeof(a));

2. Insure you're interpreting those bytes with the correct byte order.

   Q: If you're on a PC, your CPU is probably little endian. Do you know if your data stream is also little-endian, or is it big endian?

   If the data is big-endian, I'd consider the standard networking functions to accomodate byte order: ntohl(), etc: http://www.retran.com/beej/htonsman.html

ALSO:

Follow Hcorg's and Daniel Jour's advice: don't forget about the "open mode" parameter, and don't forget to check for "file open" errors.

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