Question

I want to assign a char* to char* , if I use strcpy I get several run time memory problem so I fix it by simple assignment using = operator.
Can any one explain what should prepare before using strcpy to avoid memory issues.

EDIT:

int Function(const char* K,char* FileIN,char* FileOut,int DC)
{   
    char *fic_in,*n_fic,*fic_out,*fic_dec;

    unsigned char k[17];

        fic_in = (char*)malloc(60*sizeof(char));

        strcpy((char*)k,K);

        //strcpy(fic_in,FileIN);  //I remove this one
        fic_in=FileIN;            //and replace it by this
      ...

Answer 1

const char *a = "string literal";
const char *b;

a is a pointer that points to a string literal. b is a pointer that doesn’t point anywhere in particular (it is uninitialized).

b = a;

b now points to the same memory that a points to.

char c[100];

c is now an array of 100 chars. That area of memory is completely separate from everything else so far.

strcpy(c, a);

The contents of c (to be specific, the first 15 bytes) now hold the same values as the memory pointed to by a. So now there are two regions of memory that contain the same string data.

So as you can see, assigning pointers has pretty much nothing in common with strcpy. If you want to assign a char* to a char*, then you shouldn’t be anywhere near strcpy.

It’s essential that you read a book about C++, but the basic requirement for strcpy is that the destination pointer must point to a region of memory with enough space for the string that the source pointer points to. Probably your “run time memory problems” were because you didn’t ensure that, which is why you need to read a book.