从 C++ 中的二进制文件中读取 32 位整数？
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Read 32-bit integer from binary file in C++?
提问by plhn
My binary file looks like this.
我的二进制文件看起来像这样。
00000000: 0000 0803 0000 ea60 0000 001c 0000 001c
00000010: 0000 0000 0000 0000 0000 0000 0000 0000
left column is address.
左栏是地址。
I just tried to read 0000 0803(=2051) as follows
我只是尝试阅读0000 0803（= 2051）如下
ifstream if;
if.open("file");
uint32_t a;
if >> a;
As expected...It did not work :-( awas just 0 after execution.
I tried long, int, unsigned int, unsigned long. All failed.
正如预期的那样......它没有用 :-(a执行后只是 0。
我试过了long, int, unsigned int, unsigned long。一切都失败了。
Why these are not working and how can I achieve the goal?
为什么这些不起作用，我怎样才能实现目标？
回答by paulsm4
You have two issues:
你有两个问题：
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));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
确保您从流中读取您想要的字节（不多也不少）。
我推荐这种语法：
uint32_t a;inFILE.read(reinterpret_cast<char *>(&a), sizeof(a));确保您使用正确的字节顺序解释这些字节。
问：如果您在 PC 上，您的 CPU 可能是小端。你知道你的数据流是小端还是大端？
如果数据是大端数据，我会考虑使用标准网络功能来适应字节顺序：
ntohl()等：http: //www.retran.com/beej/htonsman.html
ALSO:
还：
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.
遵循 Hcorg 和 Daniel Jour 的建议：不要忘记“打开模式”参数，不要忘记检查“文件打开”错误。
回答by VolAnd
Open file in binary mode and then use read()method, something like:
以二进制模式打开文件，然后使用read()方法，例如：
uint32_t a;
ifstream file ("file", ios::in | ios::binary);
if (file.is_open())
{
file.read ((char*)&a, sizeof(a));
}
