hi, im having trouble with this:
"write a function that accepts two strings. Use malloc()
function to allocate enough memory to hold the two strings after they have bgeen linked. return a pointer to this new string"
when i try to allocate the memory:
" pointer = malloc( 10 * sizeof(char)); "
it say that i cannot "convert void types into char *"
and how can i link the two strings toghether?
thnks 4 ur help
Type cast it, or use a C compiler instead of a C++ compiler. Additionally, make sure you're including the right header for malloc.
Quzah.
Hope is the first step on the road to disappointment.
try this
pointer = (char *)malloc(sizeof(char));
l
malloc() always returns a pointer to the object of void data type by default.In C compilers(especially turbo c)this warning shouldn't
come but this is a problem with c++ compilers.
Always type cast the return value of malloc() like
CharPtr = (char*)malloc(10 * sizeof(char));
>In C compilers(especially turbo c)this warning shouldn't
>come but this is a problem with c++ compilers.
I think you mean:
In C compilers (especially turbo c) this warning DOESN'T come...
Isn't it?
> Always type cast the return value of malloc() like
Never cast the result of malloc (not in any ANSI compiler at least)
If you need to cast the result of malloc, and you're using C++, then you should be using the new operator anyway.
If you need to cast the result of malloc, and you're using C, then it means you haven't included stdlib.h. The side effect of this is that C initially thinks that malloc returns an int (by the implicit declaration rules). On machines where ints and pointers are different sizes, this results in seriously incorrect code.
On the off-chance that you're using some pre-ANSI fossil, then you still might need the cast.
http://www.eskimo.com/~scs/C-faq/q7.7.html
thanks for help guys
