How to use LocalAlloc and LocalReAlloc correctly - c++

Question

I am going to learn how to use LocalAlloc and also LocalReAlloc of the Win32 API. I written the following code, but it gives me exceptions. I don't know what is wrong with the following code.
#include <Windows.h>
#include <iostream>
namespace code
{
namespace memory
{
void allocation()
{
char* string = reinterpret_cast<char*>(LocalAlloc(LPTR, 6 + 1));
CopyMemory(string, "WINAPI", 6);
std::printf("%s\n", string);
string = reinterpret_cast<char*>(LocalReAlloc(string, 6 + 13 + 1, LMEM_MOVEABLE));
CopyMemory(string + 6, "IS THE BEST", 13);
std::printf("%s\n", string);
delete string;
}
}
}
int main(int argc, char* argv[])
{
code::memory::allocation();
return 0;
}
When I compile the above program, it doesn't give me any error but when I run it, it gives me the exception. The following message is from the exception:
---------------------------
Microsoft Visual C++ Runtime Library
---------------------------
Debug Assertion Failed!
Program: ...Windows\00 Windows API Programming\Debug\52_DynamicMemory.exe
File: minkernel\crts\ucrt\src\appcrt\heap\debug_heap.cpp
Line: 904
Expression: _CrtIsValidHeapPointer(block)
For information on how your program can cause an assertion
failure, see the Visual C++ documentation on asserts.
(Press Retry to debug the application)
---------------------------
Abort Retry Ignore
---------------------------

Answer 1

There are several issues with your code.
A complete lack of error handling.
If LocalReAlloc() fails, you are leaking the memory allocated by LocalAlloc().
The second CopyMemory() is exceeding the bounds of the string literal being copied. And you are not ensuring the reallocated memory is null terminated for the following printf(), were the literal being copied properly.
You are not freeing the allocated memory correctly. You must use LocalFree(), not delete.
Try this instead:
#include <Windows.h>
#include <cstdio>
#include <cstring>
namespace code
{
namespace memory
{
void allocation()
{
char* string = static_cast<char*>(LocalAlloc(LMEM_FIXED, 6 + 1));
if (!string) return;
std::strcpy(string, "WINAPI");
std::printf("%s\n", string);
char* newstring = static_cast<char*>(LocalReAlloc(string, 6 + 12 + 1, LMEM_MOVABLE));
if (!newstring) { LocalFree(string); return; }
string = newstring;
std::strcpy(string + 6, " IS THE BEST");
std::printf("%s\n", string);
LocalFree(string);
}
}
}
int main()
{
code::memory::allocation();
return 0;
}

Answer 2

LocalAlloc Allocates the specified number of bytes from the heap, and if fails, returns value is NULL. To get extended error information, call GetLastError.
If the LocalAlloc function succeeds, allocates at least the amount requested, and to free allocated memory call LocalFree
LocalReAlloc used to re-alloceate memory that allocated either by LocalAlloc or LocalReAlloc, note that if LocalReAlloc fails, the original memory is not freed, and the original handle and pointer are still valid.