Re: Returning Char array/pointer? Continuing of thread I am confused
with these concepts.
rockdale wrote:
long ReadFileToByteArray(char const* aSrcFile, char*& aDestArr);
Forgive my ignorance, can you show me how to call this function? like
this?
-----------------------------------
char* myByte;
long lngSize = 0;
lngSize = ReadFileToByteArray("myFileName",&myByte);
//process data in myByte
Do I need to delete myByte here? but I never new or malloc memory to
myByte.
------------------------------------
Also, should I do delete inside the function ReadFileToByteArray.
Suppose I am using
char* aDestArr = new char[lngFileSize]; inside the function. I think I
can not, since I am passing the reference of the array into the func.
--------------------------------------
2 reasons I do not want using std::string. In my appl I still need to
convert the std::string to char array and I want to know more about
this kind of concepts as I am always confused.
Rockdale:
1. When a function uses C++ pass by reference you call it the same as pass by
value. In your case:
char* myByte;
long lngSize = 0;
lngSize = ReadFileToByteArray("myFileName", myByte);
2. Whne you pass by reference, you are indeed calling malloc (or new) on the
original pointer (that is what pass by reference does for you). In C the same
effct can be obtained by passing as char**.
You use malloc() in the function; therefore the caller should use free(). You
cannot call free() in the function or the caller cannot retrieve the
information. Again, in a C++ program you should use new[]/delete[] rather than
malloc/free.
3. To convert a std::string to a const char* array, use the c_str() method.
--
David Wilkinson
Visual C++ MVP
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