whats the significance of void * here

cranium2004 · 03-09-2005, 07:13 AM

hello,
I want to know if a fuction say malloc is declared as void *malloc() then whats the significance of void here.
Does void * i used when function has the flexibility to return any type of value by casint that functions with appropriate data type?

trevelluk · 03-09-2005, 07:30 AM

void * means that the function returns a pointer of some unspecified type. You'll need to cast the return value to be a pointer to the correct type for it to be of much use.

Code:

void foo (void) {
    char *mychars;
    int *myints;

    mychars = (char *) malloc (100);
    myints = (int *) malloc (100);
}

itsme86 · 03-09-2005, 08:37 AM

In C, you want to avoid typecasting the return value of malloc(). As long as you #include <stdlib.h> then the compiler will automatically convert the void * pointer. For C++ you'll need to typecast it, but you shouldn't be using malloc() in C++ anyway. You should be using new.

DaFrawg · 03-09-2005, 09:10 AM

Quote:

Originally posted by itsme86 As long as you #include <stdlib.h>

#include <cstdlib> perhaps?

itsme86 · 03-09-2005, 10:26 AM

Quote:

Originally posted by DaFrawg
#include <cstdlib> perhaps?

That's for C++. #include <stdlib.h> is for C which is what I was referring to.

perfect_circle · 03-09-2005, 10:46 AM

Quote:

Originally posted by trevelluk
void * means that the function returns a pointer of some unspecified type. You'll need to cast the return value to be a pointer to the correct type for it to be of much use.

Code:

void foo (void) { char *mychars; int *myints; mychars = (char *) malloc (100); myints = (int *) malloc (100); }

Many people say that now days it's better to not cast the returning value of malloc in ANSI C, and that this is an old habit when coding in traditional C, that should now stop:
http://www.stanford.edu/~blp/writing...lloc-cast.html
http://faq.cprogramming.com/cgi-bin/...&id=1043284351

Quote:

One should not cast the return value of malloc() when writing C.

When stdlib.h has been included, malloc()'s declaration will be visible. Its return type is void*, and void* can be silently converted to other pointer types.

If one should forget to include the declaration, malloc will be assumed to return int.

ints will be converted to pointer types -- but typically that conversion will not be silent. Most, if not all, C compilers will produce a warning to the effect of "you are converting an int into a pointer without explicitly casting -- are you sure you want to?". This alerts one to the fact that one has omitted to include a header.

If you put the cast in, that message will be quashed -- after all, the point of casts is to say to the compiler, "Ignore what you think you know and do what I say" -- and a [potential] problem be left latent in your code.

As you see casting the returning value of malloc may lead to bugs that are really hard to track down.
This is something that is been discussed in this forum again. I did not know that until recently either...