how does function memcpy work? Any convertion occured?

steven_yu · 06-13-2007, 04:33 AM

See following small program

Code:

#include  <stdio.h>
#include  <unistd.h>


int  main()
{
	unsigned  int  a=0x12345678;
	int  i;
		
	memset(str,0,sizeof(str));    	
	memcpy(str,&a,sizeof(a));	
	printf("a = Ox");
	for(i=0;i<4;++i)
  		printf("%x",str[i]);
	printf("\n");

	return   0;
}

when I run it, the result is 0x78563412, it is reserved of orginal variable a, it is suprised to me and I really want to know why does it happen?

If I use the following method, the result is expected.

Code:

#include  <stdio.h>
#include  <unistd.h>

int  main()
{
	unsigned  int  a=0x12345678;
	unsigned  char   str[4];
	int 	i;

	memset(str,0,sizeof(str));

	for(i=3;i>=0;--i)
		str[3-i] = (a >> (8 * i) ) & 0xff;
	
	printf("a = Ox");
	for(i=0;i<4;++i)
  		printf("%x",str[i]);
	printf("\n");
        return   0;

}

Guttorm · 06-13-2007, 04:58 AM

Hi

"memcpy" just copies memory. What you see here, is the problem of little endian versus big endian in the CPU. Your first program's output is "undefined". The C standard doesn't specify how the compiler stores types like "int" in memory (just like float or double).

It's a common mistake - very often programs just make a binary dump when saving to file, since it's fast and easy. But it becomes a problem when you try the program with a different CPU.

nautilus · 06-13-2007, 04:59 AM

Your first program doesn't even compile... You have not declared str... Please repost...

steven_yu · 06-14-2007, 01:18 AM

Quote:

Originally Posted by nautilus

Your first program doesn't even compile... You have not declared str... Please repost...

Thanks.
Oh, My guard, it's my mistake, I copy file error.

Code:

#include  <stdio.h>
#include  <unistd.h>


int  main()
{
	unsigned  int  a=0x12345678;
	int  i;
		
	memset(str,0,sizeof(str));    	
	memcpy(str,&a,sizeof(a));	
	printf("a = Ox");
	for(i=0;i<4;++i)
  		printf("%x",str[i]);
	printf("\n");

	return   0;
}

steven_yu · 06-14-2007, 01:28 AM

Quote:

Originally Posted by Guttorm

Hi

"memcpy" just copies memory. What you see here, is the problem of little endian versus big endian in the CPU. Your first program's output is "undefined". The C standard doesn't specify how the compiler stores types like "int" in memory (just like float or double).

It's a common mistake - very often programs just make a binary dump when saving to file, since it's fast and easy. But it becomes a problem when you try the program with a different CPU.

Thanks very much.

When CPU put data in memory, it will convert it according to its type, such as little-endian or big-endian. My CPU is Intel 386, little-endian, so result is so.

Writing careful code.

jlliagre · 06-14-2007, 11:04 AM

Quote:

Originally Posted by steven_yu

When CPU put data in memory, it will convert it according to its type

Strictly speaking, it doesn't really convert the data.

It just stores multi-byte data in one way or another, just like a telephone and a numeric keypad lay out the digits in different ways.

Conversion would mean some data processing to happen while here the native (i.e. faster) way of storing data for a given CPU is simply used.

atulsvasu · 06-19-2007, 06:25 AM

Big Endian and Small Endian conventions.

bigearsbilly · 06-19-2007, 08:25 AM

have you tried:

man memcpy

always the first place to look.

srp · 06-25-2007, 12:04 PM

The code you shown - reads one byte at a time and prints that byte. But if you read 'int' (i.e 4-byte value) from that location, it will print correct value.
See the modified program below:

Code:

#include  <stdio.h>
#include  <unistd.h>


int  main()
{
	unsigned  int  a=0x12345678;
	int  i;
	unsigned  char str[4];
		
	memset(str,0,sizeof(str));    	
	memcpy(str,&a,sizeof(a));	
	printf("a = Ox");
	for(i=0;i<4;++i)
  	    printf("%x",str[i]); /* prints one byte at a time*/
	printf("\n");
        /* This prints correct 4-byte value */  
	printf("str = 0x%x", *(int *)str);

	return   0;
}

Subhash

Guttorm · 06-25-2007, 01:33 PM

man swab

I forgot to add - but it's just confusing