C ++ pwrite（），pread（）数据到文本文件

Question

I have to pwrite() characters to a text file with each character representing 1 byte. Also, I need to write integers to the text file, so 12 has to be one byte also, not 2 bytes (even though two characters).

I am using char *pointer for the characters and integers, but I am getting stuck since the text fill prints jumbled values for the integers (@'s, upside-down ?'s, etc.) Like when I pwrite() pointer[0] = 105; The 105 translates 'i' in the text.txt file (and pread() reads as 'i') Somehow the 105 is lost in translation.

Any ideas how to pwrite()/pread() correctly?

ofstream file; file.open("text.txt");
char *characters = new char;
characters[0] = 105;
cout << pwrite(3, characters, 1, 0);

Also, the 3, is the filedes, which I guess :-P Don't know how to actually find. The text.txt file then has 'i' in it (ASCII 105 I'm assuming). When I pread() then, how will I know if it was originally and 'i' or 105?

Answer 1

Breaking this down in chunks:

"I have to pwrite() characters to a text file with each character representing 1 byte"

By definition, each ASCI character is one byte, and you make no mention of need to to write locale-aware multi-byte characters, or Unicode derivatives, so I'm thinking on this one you're probably covered.

"Also, I need to write integers to the text file, so 12 has to be one byte also, not 2 bytes (even though two characters)"

You're describing a binary write of your integer data. However, keep in mind that "integers" as a numeric representation can be larger than just a number represented by "one byte". If you want to write an integer that can be represented in a single byte, your options are:

1. For signed data, values can range from [-128,127]
2. For unsigned data, values can range from [0, 255]

These are the limitations of an integer value in a single octet.

"I am using char *pointer for the characters and integers, but I am getting stuck since the text fill prints jumbled values for the integers (@'s, upside-down ?'s, etc.)"

The char pointer for characters we covered before, and will likely be fine. The integers will NOT be. Your resulting file per your description will not be a "text" file in the literal sense. It will contain both character data (your char buffers) and binary data (your integers). Please remember an integer within a single byte with a value of 0x01 will be just that, a single octet with the first bit set. A byte representing the ASCI character '1' will have a value of 0x31 (see any ASCI chart ), and value 0xF1 for EBCDIC (don't ask). Using your example, **you cannot write the value 12 in a single byte and have it be displayable "text" (character) data in your file. The single-byte integer value 12 will be represented in your file as a single byte value 0x0C. Trying to view this as "text" will not work; it is not printable ASCI. In fact, the ASCI value of 0x0C is actually a form-feed control character.

Bottom line, if you don't know the difference between ASCI characters and integer bytes, explaining how pwrite() works will do little good but to confuse you more.

"Like when I pwrite() pointer[0] = 105; The 105 translates 'i' in the text.txt file (and pread() reads as 'i') Somehow the 105 is lost in translation"

Refer to the ASCI chart linked several places in this answer. The byte value 105 is, infact, the ASCI value of the character 'i'. The 105 isn't lost; its being displayed as the character it represents.

Finally, pwrite() is a POSIX system call for Linux, BSD, and anyone else that chooses to expose it. It is not part of the C or C++ standards. That said, your first argument for pwrite() should be obtained from a system call, open() . You should never piggyback on a file descriptor assumed to be opened by a different api call unless you go through a supported API to do so. The code in this question does not.