Reading an array of chars from a binary file into another array of chars dynamically allocated in C++

Question

I have a homework and I have to write an object into a binary file. Then I have to read that object from the binary file. Everything works fine,except a variable of type char*. I read it, but at the end of the text i got some random characters. I think the problem is the string terminator /0 ,but i don't know how to handle it. Can someone help me?

This is the code I used to write in the file:

        size_t n = strlen(input);// input is declared as char* input
        f.write((char*)&n, sizeof(n+1));
        f.write(input, n);

And this is how I tried to read this variable from the binary file:

        size_t n;
        f.read((char*)&n, sizeof(n));
        delete[] buffer;// buffer is also a char*
        buffer = new char[n];
        f.read(buffer, n);

I got the text from variable input, but at the end i got some random characters

sizeof(n+1) doesn't do what you think it does, but I dont think that relates to your problem — Mooing Duck
– Mooing Duck, Commented Dec 28, 2022 at 21:47
The variable was initialized with a text. My code is a lot bigger, I took only this part where I got a problem. @MooingDuck — Gabriel Boşcan
– Gabriel Boşcan, Commented Dec 28, 2022 at 21:51
sizeof(n+1) -- No need to read any further. This is totally wrong. — PaulMcKenzie
– PaulMcKenzie, Commented Dec 28, 2022 at 21:56
f.write((char*)&n, sizeof(n+1)); you broke file by writing extra byte with unknown content — Swift - Friday Pie
– Swift - Friday Pie, Commented Dec 28, 2022 at 23:09

Dúthomhas · Accepted Answer · 2022-12-28 22:31:22Z

1

C-strings must be nul-terminated

C-strings have length elements plus a terminating nul character. You do in fact write the nul to file, but you fail to allocate space for and read it from file. Make sure things match.

size_t n;
f.read( (char *)&n, sizeof n );
char * s = new char[n+1];
f.read( s, n+1 );

UPDATE: You have changed your code (as per commentary below) to not write the nul-terminator to file. Make sure to initialize it as such when you read it:

...
char * buffer = new char[n+1];
f.read( buffer, n );
buffer[n] = '\0';

Endianness nonsense you can totally ignore

As an aside, you are not considering endianness issues with your size_t. I would personally make functions specifically for reading and writing integers in a file-specific endianness.

size_t n = read_integer( f, 4 );
char * s = new char[n+1];
f.read( s, n+1 );

Supposing a little-endian file, we could have:

size_t read_integer( std::istream & f, int n )
{
  size_t result = 0;
  while (n--)
    result = (result << 8) | ((unsigned char)f.get() & 0xFF);
  return result;
}

A corresponding function to write a little-endian value:

std::ostream & write_integer( std::ostream & f, size_t n, size_t value )
{
  while (n--)
  {
    f.put( value & 0xFF );
    value >>= 8;
  }
  return f;
}

(Untested. I might have goofed something.)

Oh, also:

write_integer( f, 4, strlen(s) );
f.write( s, strlen(s)+1 );

edited Dec 28, 2022 at 22:31

answered Dec 28, 2022 at 22:00

Dúthomhas

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7 Comments

Gabriel Boşcan Over a year ago

I change the code like this and now it's working, but i want to see if it's correct like this

size_t n = strlen(input)+1;      // input is declared as char* input   f.write((char*)&n, sizeof(n));   f.write(input, n);

And for reading

size_t n;  f.read((char*)&n, sizeof(n));  delete[] buffer;// buffer is also a char*  buffer = new char[n];  f.read(buffer, n);

Dúthomhas Over a year ago

Yes... but... you then have the same problem you started with: a non-terminated C-string. Re-read the very first part of my answer.

Gabriel Boşcan Over a year ago

ok, so i should initialize like this: buffer=new char[n+1] and in this way i will have enough space for the string terminator?

Dúthomhas Over a year ago

Yes, that is correct.

Gabriel Boşcan Over a year ago

ok, thank you very much for your help. I am sorry that I didn't understand your answear really well. I am still learning:))

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C-strings must be nul-terminated

Endianness nonsense you can totally ignore

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