7

I am trying to binary serialize the data of vector. In this sample below I serialize to a string, and then deserialize back to a vector, but do not get the same data I started with. Why is this the case?

vector<size_t> v;
v.push_back(1);
v.push_back(2);
v.push_back(3);

string s((char*)(&v[0]), 3 * sizeof(size_t));

vector<size_t> w(3);
strncpy((char*)(&w[0]), s.c_str(), 3 * sizeof(size_t));

for (size_t i = 0; i < w.size(); ++i) {
    cout << w[i] << endl;
}

I expect to get the output 

1  
2
3

but instead get the output 

1
0
0

(on gcc-4.5.1)

Mooing Duck
  • 64,318
  • 19
  • 100
  • 158
typedef
  • 1,800
  • 3
  • 11
  • 19

4 Answers4

4

The error is in the call to strncpy. From the linked page:

If the length of src is less than n, strncpy() pads the remainder of dest with null bytes.

So, after the first 0 byte in the serialized data is found the remainder of w's data array is padded with 0s.

To fix this, use a for loop, or std::copy

std::copy( &s[0], 
           &s[0] + v.size() * sizeof(size_t), 
           reinterpret_cast<char *>(w.data()) );

IMO, instead of using std::string as a buffer, just use a char array to hold the serialized data.

Example on ideone

Praetorian
  • 106,671
  • 19
  • 240
  • 328
  • Good catch on the `strncpy`, I never knew about that "feature" and was baffled as to why his code didn't work. – Mooing Duck Jul 05 '12 at 23:48
  • Thank you for the explanation. Why std::copy is better than memcpy? – typedef Jul 06 '12 at 00:06
  • Since you're copying an array of integers both will work the same. But let's say the vector contained an object that managed some resource. memcpy would perform a bitwise copy of the object, which is most likely not the way you want it copied. std::copy, on the other hand, will invoke the assignment operator ensuring the object is copied correctly. – Praetorian Jul 06 '12 at 00:15
2

strncpy is a giant pile of fail. It will terminate early on your input because the size_t have some zero bytes, which it interprets as the NULL terminator, leaving them as default-constructed 0. If you ran this test on a BE machine, all would be 0. Use std::copy.

Puppy
  • 144,682
  • 38
  • 256
  • 465
-1

To serialize this vector into a string, You first want to convert each of the elements of of this vector from an int into a string containing the same the ascii representation of that number, this operation can be called serialization of an int to string.

So for example, assuming an integer is 10 digits we can

// create temporary string to hold each element
char intAsString[10 + 1];

then convert the integer to a string

sprintf(intAsString, "%d", v[0]);

or

itoa( v[0], intAsString, 10 /*decimal number*/ );

You can also make use of the ostringstream and the << operator

if you look at the memory contents of intAsString and v[0], they are very different, the first contains the ascii letters that represent the value of v[0] in the decimal number system(base 10) while v[0] contains the binary representation of the number(because that's how computers store numbers).

Diaa Sami
  • 3,249
  • 24
  • 31
  • 1
    It seems clear to me he wants binary serialization, not text serialization. Also, `sprintf` and `itoa` in C++ code? – Mooing Duck Jul 05 '12 at 23:26
  • well he made a string to hold the output so that's why I assumed he wants text serialization. Thanks for the comment though :) – Diaa Sami Jul 05 '12 at 23:32
  • No, the string is copied to a vector, and the vector holds the output. The string is merely a binary buffer. – Mooing Duck Jul 05 '12 at 23:33
-1

The safest way would be to just loop through the vector and store the values individually into a char array of size 3*sizeof(size_t). That way you don't have a dependency on the internal structure of the vector class implementation.

Deepan
  • 26
  • 3