Tag: memcpy

struct – setup and memory locations

When you setup a struct within c++, it is kinder like having a array of data and if you want to you can access the internal parts by using some memory pointer location fun!.

Lets say that you have a struct of

struct intvalue {
    int a;
    int b;
};

Just to say that since I am using int(eger) values so that is what I am incrementing by in the pointer arithmetic which is why I am casting the pointer to a integer value.

So lets say that we create a variable of intvalue and setup the values as below

intvalue testvalue;
testvalue.a = 4;
testvalue.b = 5;

We can then pull out the value of a or b, but using memcpy and just outputting to a int(eger) variable as below, the key is the ((int*)&testvalue)+1, this will first convert the testvalue variable to a pointer to memory location and then (int*) casts that pointer to a int pointer, because internally that is what it is, and then just add 1 to it, which points to the second value ( in this case the value of b which is 5)

    int avalue;
    // convert to a int pointer type and then add one to it (to the next array element as such).
    memcpy(&avalue, ((int*)&testvalue)+1,sizeof(int));
    cout << "a value (or is it the b value :) ) " << avalue << endl;

The output would be

a value (or is it the b value :) ) 5

because I am pointing to the second value which is b and thus 5 :).

Of course if you just wanted the value of first int (the value of a in this case) you do not add the 1 to the memory location, for example

    memcpy(&avalue, ((int*)&testvalue),sizeof(int));

this time I am just converting the testvalue (casting) to a int pointer and thus pointing to the start of the struct and that is where the int a variable is living :).

memcpy – copy the memory contents from one place to another

The memcpy function, does not really care about the parameters passed in, if they are the same types or of different types. Just that they are variables already declared (and of course can hold the data being passed from one place to another, because otherwise there may be a slight overlap if the copied value is larger in memory size to the copier).

So the memcpy is just a binary data copy and nothing else, there is no type checking.

Here is a example in c++ code that will copy a newly created struct(ure) from one place to another.

void* memcpy (void *copied_to, const void *copier, size_t num);

where the return is also the copied_to value, since we are not altering the copier then a const(ant) can be applied and the last parameter is num which is the size of the data you want to copy.

#include <stdio.h>
#include <string.h>
 
// diferent type of object, e.g. not a int,char
typedef struct 
{
    int bob;
} insertType;
 
int main()
{
  // create two variables of insertType with default values for bob of 10 and 20
  insertType t1 = { 10 };
  insertType t2 = { 20 };
 
  // output the declared values
  printf("t1 = %d\nt2 = %d\n", t1.bob, t2.bob);
 
  // do a memory copy of t2 into t1 only copy as big as the variable size.
  memcpy(&t1, &t2, sizeof(insertType));
 
  // output the new values
  printf("t1 = %d\nt2 = %d\n", t1.bob, t2.bob);
  return 0;
}

here is the output

t1 = 10
t2 = 20
t1 = 20
t2 = 20

also note that the size_t is a unsigned integral type (normal of int, of the base operating system)