Hi Vardan,
Eus said...
> It looks like that you have not zeroed your `a'.
> `a[0]='\0';' is not enough.
> Either you do `char a[BUFFER_SIZE] = {0};' when defining `a' or, better, you
> do `memset(a, 0, BUFFER_SIZE);'
For strcpy, it should not be necessary to zero the 'a'.
For a strcat, just a[0]='\0'; should be sufficient to make 'a' strcat-able.
For C, I like Eus' suggestion: char a[1024] = { 0 };
For C++, you can omit the first element: char a[1024] = { };
However, for the problem as presented, is necessary to actually allocate the
buffer for 'a'.
These won't work:
char* a = NULL; // a does not actually refer to any space.
strcpy(a,x("sample text","default text")); // ERROR!
char* a; // a has a garbage pointer
strcpy(a,x("sample text","default text")); // ERROR!
char a[5]; // Not enough space.
strcpy(a,x("sample text","default text")); // ERROR!
- - - - - -
These will work:
char a[1024]; // Overkill, but will work. 'a' is on the stack.
strcpy(a,x("sample text","default text"));
char* a;
a = malloc(1024); // Overkill, but will work. Note#1
strcpy(a,x("sample text","default text"));
char* temp = x("sample text","default text");
char a = malloc(strlen(temp)+1); // Exact allocation. Note#1.
strcpy(a, temp);
Note#1: Be very mindful of the ownership & lifespan of 'a' which was
allocated off the heap. Remember to free(a) when done with it.
HTH,
--Eljay
