C / C++ / MFC

George_George wrote:

In the world of Windows and COM, does the codebase transfer also apply?

George_George wrote:

"Specifies where the common language runtime can find an assembly.". We are talking about marshalling in COM, and I do not think they are related?

George_George wrote:

In the world of Windows and COM

George_George wrote:

code base means binary executable file

George_George wrote:

For automation compatible data types, we only need to let the other party know the type, and they could be able to decode/deserial

George_George wrote:

For the Java, I am not sure if we need to decode/deserialize the content for a type, we need to provide not only the type information, but also the binary executable code?

#include <windows.h>

void foo()
{
	int h = 4;
	int w = 7;
	int A[4 * 7];

	for (int i = 0; i < h; i++)
	{
		DebugBreak();
		
		for (int j = 0; j < w; j++)
		{
			A [i*w + j] = -(i * i + j * j);
		}
	}
}

int main()
{
	foo();

	return 0;
}

#include <windows.h>
#include <iostream>

using namespace std;

int A[4 * 7];

void foo()
{
	int h = 4;
	int w = 7;

	for (int i = 0; i < h; i++)
	{
		DebugBreak();
		
		for (int j = 0; j < w; j++)
		{
			A [i*w + j] = -(i * i + j * j);
		}
	}
}

int main()
{
	foo();

	cout << A [0] << endl;

	return 0;
}

George_George wrote:

find there is a new file named main.asm generated, do you mean that file

George_George wrote:

My purpose is to find the related assembly code is optimized in release mode, does turn on the option of Assembler Output impact the actual optimization in release mode?

George_George wrote:

Is that correct?

George_George wrote:

I find there is a new file named main.asm generated, do you mean that file?