Reading 16-bit integers from binary file c++

问题

I'm not sure if I'm doing this right so I want to check my code. it works but I'm not sure its working right. I need it to read the binary file, and store the 16 bit integers in an array of ints that is the exact size needed. I tried to do sizeof(storage[i]) so I could see if I was storing 16 bits but it says 32 (I'm guessing because int automatically allocates 4 bytes?

        void q1run(question q){
        int end;
        std::string input = q.programInput;
        std::ifstream inputFile (input.c_str(), ios::in | ios::binary);                     //Open File
        if(inputFile.good()){                                       //Make sure file is open before trying to work with it
                                                                    //Begin Working with information
           cout << "In File:  \t" << input << endl;
           inputFile.seekg(0,ios::end);
           end=inputFile.tellg();
           int numberOfInts=end/2;
           int storage[numberOfInts];
           inputFile.clear();
           inputFile.seekg(0);
           int test = 0;


           while(inputFile.tellg()!=end){       
               inputFile.read((char*)&storage[test], sizeof(2));
               cout << "Currently at position" << inputFile.tellg() << endl;
               test++;
           }

           for(int i=0;i<numberOfInts;i++){
               cout << storage[i] << endl;
           }
       }else{
           cout << "Could not open file!!!" << endl;
      }
 }

EDIT:::::::::::::::::::::::::::::::::::::::::::::;

I changed the read statement to:

      inputFile.read((char*)&storage[test], sizeof(2));

and the array to type short. now Its pretty well working except the output is a little strange:

      In File:        data02b.bin
      8
      Currently at position4
      Currently at position8
      10000
      10002
      10003
      0

I'm not sure what's in the .bin file, but I'm guessing the 0 shouldn't be there. lol

回答1:

Yes, int is 4 bytes (on 32-bit x86 platform).

You have two problems:

As Alec Teal correctly mentioned in comment, you have your storage declared as int, which means 4 bytes. Not a problem, really - your data will fit.
Actual problem: your line which is reading the file: inputFile.read((char*)&storage[test], sizeof(2)); is actually reading 4 bytes, because 2 is integer, so sizeof(2) is 4. You don't need sizeof there.

回答2:

Use: int16_t in <cstdint>. (guaranteed 16bits)

Shorts and ints can be of various sizes depending on the architecture.

回答3:

Storage is declared as an 'array' of "int"s, sizeof(int)=4

This shouldn't be a problem though, you can fit 16 bit values in 32 bit spaces, you probably meant short storage[...

Additionally for the sake of full-disclosure, sizes are defined in terms of sizeof(char) as a monotonically increasing sequence.

4 is by far the most common though, hence the assumption. (Limits.h will clarify)

回答4:

One way to store 16 bits integers is to use the type short or unsigned short.

You used sizeof(2) which is equal to 4 because 2 is of type int so the the way to read 16 is to make storage of type short and the read:

short storage[numberOfInts];
....    
inputFile.read((char*)&storage[test], sizeof(short));

You can find here a table containing the sizes of all the types.

来源：https://stackoverflow.com/questions/20526734/reading-16-bit-integers-from-binary-file-c

标签

c++

binary

sizeof