Dynamic char* Allocation
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Thread: Dynamic char* Allocation

  1. #1
    Join Date
    Jun 1999
    Location
    Wichita, KS
    Posts
    33

    Dynamic char* Allocation

    What is wrong with the following code? The code crashes on the delete statment (true condition) every time I run the code. Thanks.

    bool ReadWriteString(CFile* fileIn, CStdioFile* fileOut, CString strOut, unsigned short nLength)
    {
    UINT nBytesRead;
    char *pch = new char[nLength];

    pch[nLength] = NULL;

    nBytesRead = fileIn->Read(pch, nLength);

    if (nBytesRead == nLength)
    {
    // Write Label
    fileOut->WriteString(strOut + "\n");
    // Write value
    fileOut->WriteString(pch);
    fileOut->WriteString("\n");
    delete [] pch;
    return true;
    }
    else
    {
    delete [] pch;
    return false;
    }
    }


    -Albert

  2. #2
    Join Date
    Apr 1999
    Posts
    27,422

    Re: Dynamic char* Allocation

    You are writing out of bounds:

    UINT nBytesRead;
    char *pch = new char[nLength];

    pch[nLength] = NULL; <<--Out of bounds!!




    First, arrays are 0-based, therefoer pch[nLength] is over the limit by 1. Also, the value that should be written is '\0', not NULL. I'm sure you got a warning on the type mismatch.
    Here is the corrected code (at least for the first part)

    bool ReadWriteString(CFile* fileIn, CStdioFile* fileOut, CString strOut, unsigned short nLength)
    {
    UINT nBytesRead;
    char *pch = new char[nLength + 1];
    pch[nLength] = '\0';
    nBytesRead = fileIn->Read(pch, nLength);
    ...



    Regards,

    Paul McKenzie


  3. #3
    Join Date
    May 1999
    Location
    Seattle, WA USA
    Posts
    423

    Re: Dynamic char* Allocation

    correct about the out-of-bounds thing, but it really doesn't matter if he assigns NULL or '\0' or 0 to pch[nLength] , they all have the same value of 0x00, which is what the NULL Byte is.

    Strousstrup, that guy that created c++, tells us that "A constant expression that evaluates to 0 is converted to a pointer, commonly called the NULL pointer".

    --michael


  4. #4
    Join Date
    Jun 1999
    Location
    Wichita, KS
    Posts
    33

    Re: Dynamic char* Allocation

    Thanks for the help. I noticed in the code sample you posted that you delimited '\0' with the single-quote character. What is the reason for this? Is there any difference between using double and single quotes?

    -Albert

  5. #5
    Join Date
    May 1999
    Location
    Seattle, WA USA
    Posts
    423

    Re: Dynamic char* Allocation

    the '\0' one byte (or two if you are using double-byte characters) that represents NULL.

    "\0" is a string (2 chars or 4 if you are using double-byte chars), so assigning it to a char would give an error that it cannot convert form a char[2] to a char.

    --michael


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