"const char *foo means it cannot be modified" -> This is not completly correct. 'const char *foo' means foo is a pointer which points to a constant char (pointer to a constant variable, not a constant pointer variable). That means, the address in foo can be changed(foo is not a constant), but we cannot modify the value contained in the address held by foo(the value is considered as constant).

For example

Code:

const char *foo;

foo="Test1";

cout<<"\n foo 1 : "<<foo<<endl;  // Result will be 'Test1'

foo="Foo is changed now."       

cout<<"\n foo 2 : "<<foo<<endl;  // Result will be 'Foo is changed now'

In your code,

you got error in line 6 because, you are trying to change the value contained in the address held by 'dest'. That is if you remove the '*' it wont generate any errors. But the meaning will be different.

you got error in 7 because the dest is a const char * variable. memcpy does not allow const char * as its first argument (memcpy(void *,const void *,size_t))

The work around you are suggesting is similar to what I have given in my previous post. Only difference is you are asking to assign the address of the const char * variable to a char * variable and change the content of the char * variable. I declared the char * variable first and assigned its address to the const char * variable. After that I am also doing the same thing.

I think my previous post created some miss understandings in you. Sorry if it is so.