以十六进制字符串形式读取MIDI文件：缺少一些信息

Question

I have a MIDI file which I am trying to read as a hex string: in particular I want to input a MIDI file and to have the hex string available for use. I have the following:

ostringstream ss;
char * memblock;
unsigned char x;
std::string hexFile;

ifstream file ("row.mid", ios::binary);
ofstream output;
output.open("output.txt");

while(file >> x){
    ss << hex << setw(2) << setfill('0') << (int) x;
}

hexFile = ss.str();
cout << hexFile; 

When I output hexFile, I get the following (note the white space near the end):

4d546864000000060001000400f04d54726b0000001300ff58040402180800ff5103 27c000ff2f00

When I view the MIDI in a hex editor, it reads as follows:

4d546864000000060001000400f04d54726b0000001300ff58040402180800ff5103 0927c000ff2f00

The latter is definitely correct, as confirmed by track size (around the white space I manually inserted, the correct one has a 09 the former lacks).

What could have caused this 09 to go missing in my code?

Answer 1

By default ifstream skips whitespace.
All you need to do is tell it not to.

ifstream file ("row.mid", ios::binary);
file.unsetf(ios::skipws); //add this line to not skip whitespace

Answer 2

09 is an ANSII code for tabulation character. The default ofstream mode is for text, and that's why the 09 byte is written as the actual tabulation. Try to set ios::binary also for the output file, and it should be fine.

output.open("output.txt", ios::binary);

Answer 3

在声明ifstream file()之后添加以下行似乎可以解决问题：

file >> std::noskipws;