how does function memcpy work? Any convertion occured?
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"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.
"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.
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;
}
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