I'm getting error C2664 on some I/O code

Question

void BinaryTree::InitializeFromFile(string Filename){
ifstream inFile;
inFile.open(Filename, fstream::binary);
if(inFile.fail()){
    cout<<"Error in opening file "<<Filename;
    return;
}
 for(int i=0;i<=255;i++) Freq[i]=0;
  char c;
  inFile.get(c);
  while(!inFile.eof()){
    Freq[c] ++;
    inFile.get(c);
  }
}  



HuffmanTree.cpp(293) : error C2664: 'void std::basic_ifstream<_Elem,_Traits>::
open(const wchar_t *,std::ios_base::openmode,int)' : cannot convert parameter 1 
from 'std::string' to 'const wchar_t *'
1>    with
1>    [
1>        _Elem=char,
1>        _Traits=std::char_traits<char>
1>    ]
1>    No user-defined-conversion operator available that can perform this 
      conversion, or the operator cannot be called

Line 293 is inFile.open(Filename, fstream::binary);

score 4 · Accepted Answer · answered Feb 26 '10 at 18:42

4

Use Filename.c_str() instead - open() doesn't take a std::string as a parameter for the filename.

answered Feb 26 '10 at 18:42

Timo Geusch

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score 3 · Answer 2 · answered Feb 26 '10 at 18:42

3

use Filename.c_str() instead of Filename in the call of ifstream::open

answered Feb 26 '10 at 18:42

f4.

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score 2 · Answer 3 · answered Feb 26 '10 at 18:43

2

Somewhat perplexingly, ifstream::open takes a C-string, not a C++ std::string. Change the line to:

inFile.open(Filename.c_str(), fstream::binary);

I have no idea why the designers of the C++ standard library made this choice, but there you go.

answered Feb 26 '10 at 18:43

Tyler McHenry

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Because in general the C++ standard library design is to not force users to use features they don't want to use, such as std::string. – MSN Feb 26 '10 at 20:23
Fair enough, but to not even provide an overload that accepts `std::string` still strikes me as a bit weird. – Tyler McHenry Feb 26 '10 at 21:12

score 1 · Answer 4 · answered Feb 26 '10 at 18:43

1

The ifstream ctor expects a const char *. Use Filename.c_str().

answered Feb 26 '10 at 18:43

dirkgently

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I'm getting error C2664 on some I/O code

4 Answers4