memcpy() and uint8_t

Hello folks..

In the following code, I m trying to copy struct element to array...

Code:

#include<iostream>
#include<stdint.h>
#include<string>

using namespace std;
struct test
{
	struct header
	{
		int a;
		int b;
	};
	uint8_t c;
	uint8_t d;
	uint8_t e;
	uint32_t f;
};
struct test2
{
	uint8_t n;
	uint8_t m;
	uint8_t p;
};


int main()
{
	test headd;
	test2 data;
        uint8_t* d;
	int x, y,i;

	x= sizeof(test);
	y= sizeof(test::header);
	i= sizeof(data);

	x=x+y+i;
	d = new uint8_t[x+10];
	cout<<"x ="<<x<<endl;

/** Assign **/
	headd.c = 4;
	headd.d = 5;

	data.n=8;
	data.m=7;
	data.p=9;

/** copy 2 elements of struct 'headd' and struct 'data' on to memory block pointed by 'd' **/

	memcpy(d,&headd.c,8); 
	cout<<"d="<<*d<<endl; 
	cout<<"d2="<<*(d+1)<<endl;

	memcpy(d+8,&data, 8);
	cout<<"d="<<*d<<endl;
	cout<<"d2="<<*(d+2)<<endl;

	return 0;
}

OUTPUT:
x =19
d=
d2=
d=
d2=


I ve replaced uint8_t and uint32_t by datatype 'int' and it works well... i can display contents of 'd'
but not for uint8_t and uint32_t .
Please suggest me how can i check whether data successfully copied to destination?


Thank You

 

I'm not sure what you are doing, but it might be a packing issue. By default, MS compilers line-up data in structure to an 8 byte boundary (makes it easier on the memory controller). You can override this behavior with the #pragma pack option:

Code:

[COLOR="Blue"]#pragma pack(push,1)[/COLOR]
struct test
{
	struct header
	{
		int a;
		int b;
	};
	uint8_t c;
	uint8_t d;
	uint8_t e;
	uint32_t f;
};
struct test2
{
	uint8_t n;
	uint8_t m;
	uint8_t p;
};
[COLOR="Blue"]#pragma pack(pop)[/COLOR]

 

I want to copy struct test (except struct header) and struct test2 on to array pointed by 'd'

I 'd put printf() to display *d instead of cout... its working

Thanks for early reply