problem in reading binary file (mix of ascii and binary)

Question

My code for reading binary file is:

dataFile.open(fileName.c_str());
ifstream binData("Trafficlog_Data.txt", ios::in | ios::binary);  //binary data file 
if(!binData) { 
  cout << "Cannot open file.\n"; 
  return -1; 
} 
char *memblock;int nBytes =12;
memblock = new char [nBytes+1];
binData.read(memblock,nBytes);
memblock[nBytes+1]='\0';
std::string message;message.assign(memblock,nBytes);
printf("%s\n",message.c_str());

Now i have given a file as input which contain binary and ascii data
RFB 003.003
RFB 003.003
--some binary data--

When I am reading first 12 bytes of file which is "RFB 003.003\\n" but it prints "RFB 003.003=". Can anyone tell where im getting it wrong please. promblem is not with '\\0'. problem is it is not reading "RFB 003.003\\n" . IS it because this file is mix of binary and ascii data

Answer 1

你没有为memblock分配内存：

char *memblock = new char[nBytes+1];

Answer 2

Change:

memblock[nBytes+1]='\0';

to:

memblock[nBytes]='\0';

Let's say you read in six bytes to memblock , that goes into positions 0 through 5 inclusive:

  0   1   2   3   4   5    6    7
+---+---+---+---+---+----+---+---+
| H | E | L | L | O | \n | ? | ? |
+---+---+---+---+---+----+---+---+

(the ? areas still contain whatever rubbish was there before).

You then need to put the null terminator at position 6 rather than position 7 as your code is doing.

By placing the null terminator too far to the "right", you're including that first ? position, which could be holding anything.

---

That's what's causing your specific problem. You also have an issue that you're not allocating space to hold the data you're reading in. You just have a char * but you're not actually initialising it to point to usable memory. That's almost certainly going to cause problems.

Probably the simplest solution is to define it as:

char memblock[nBytes+1];

Although I see you've fixed that in your question now so it's of little consequence. The actual problem (putting the null byte at the wrong location) is covered above.

Answer 3

You're off-by-one: just do memblock[nBytes]='\\0'; The index starts at 0, so if nBytes is 0 you're writing at the first position, if nBytes is 1 you're writing at the second position and so on. By doing nBytes + 1 you actually jumped ahead one position and left one garbage byte at the end of the string.