Go4Expert - memcpy() and uint8

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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

Re: memcpy() and uint8_t

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:



#pragma pack(push,1)

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;

};

#pragma pack(pop)

Re: memcpy() and uint8_t

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

Quote:

Originally Posted by Gene Poole (Post 63091)

Code:



#pragma pack(push,1)

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;

};

#pragma pack(pop)