Assign values to a dereferenced void pointer

Question

I'm doing a class in c++ that supports any kind of variable to help me in a future project. The thing is, when I try to assign a value to a void* variable, I get the error: void* is not a pointer-to-object type. Here is the code:

int main (void) {
    void* a;
    int x;
    a = malloc(sizeof(int));
    x = 120;
    ((int)(*a)) = x;
    printf("%d",((int)*a));
    free(a);
    system("pause");
    return 0;
}

I see it like I am trying to assign the value of x in the memory block reserved for a. I want that the value stored in x be stored in the memory block of a. Can any1 help me?

score 9 · Answer 1 · answered Aug 16 '11 at 16:13

9

You probably want to do:

void* a;
int x;
a = malloc(sizeof(int));
x = 120;
*(int*)a = x;
printf("%d", * (int*)a);
free(a);

answered Aug 16 '11 at 16:13

karlphillip

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score 4 · Answer 2 · answered Aug 16 '11 at 16:11

You're making it too complicated. Just cast a to be an int *, or better yet declare a as int * to start with.

int x = 120;
int * a ;
// now just cast the result from malloc as int*
if((a = (int *)malloc(sizeof(int))) == NULL){ //ALWAYS check
   exit(-1);
}
*a = x;

There are slicker ways to do similar things in C++, but you appear to be primarily caught up in pointer syntax.

Puppy · Answer 3 · 2011-08-16T16:17:26.087

3

Firstly, you must cast a to the appropriate pointer type and then de-reference it. Secondly, use a template- that's what they're for. Thirdly, malloc and free is baaaad, OK? Don't. If you think that this is good code, you need serious remedial C++. Fourthly, this is already done- boost::any serves this purpose.

edited Aug 16 '11 at 16:17

answered Aug 16 '11 at 16:10

Puppy

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11

This is why I hate "C/C++". – Puppy Aug 16 '11 at 16:12
1

@Charlie: yeah, but he didn't say "C ∩ C++" nor "C ∪ C++". – R. Martinho Fernandes Aug 16 '11 at 16:17
No, but he then uses stuff that's only C. – Charlie Martin Aug 16 '11 at 16:19
@Charlie: That doesn't make it C. He clearly states that he's going to put it in a class, in a class in C++, and he only tags C++. I think it's pretty clear what language he's discussing. – Puppy Aug 16 '11 at 16:22
it doesn't really answered my question.. malloc and free can be easily replaced. boost::any ins't even a C++ standard class. This problem has only the purpose of knowledge. Its important to know how things work before use a real class to understand whats going on on my program.. – Ruy Aug 17 '11 at 00:06

score 3 · Answer 4 · answered Aug 16 '11 at 16:11

3

You need the cast as *(int*)(a) = x;.

answered Aug 16 '11 at 16:11

Asha

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I think, this might break strict aliasing rule as `(int*)(a)` is not yet a lvalue. – iammilind Aug 16 '11 at 16:14
@iammilind can you pls elaborate? what do you mean by `break strict aliasing rule`? – user2635088 Nov 28 '16 at 06:34
@user, refer this [What is the strict aliasing rule?](http://stackoverflow.com/questions/98650/what-is-the-strict-aliasing-rule). – iammilind Nov 28 '16 at 06:58

iammilind · Answer 5 · 2011-08-16T16:20:36.237

0

There are other elegant ways to assign a variable of different types. For your current question you can follow the standard way as,

int* const a = new int;
//...
*a = 120;
//...
delete a;

Or just for this problem specifically,

int a = 120; // simple

[Note: In your current solution if you do,

*((int*)(a)) = x;

it might be breaking strict aliasing rule.]

edited Aug 16 '11 at 16:20

answered Aug 16 '11 at 16:13

iammilind

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Well, that completely avoids his question about malloc and void *. – Charlie Martin Aug 16 '11 at 16:20
@Charlie, it's evident that whatever OP is doing is not correct and seemingly doesn't need `malloc/void*`. – iammilind Aug 16 '11 at 16:25

Assign values to a dereferenced void pointer

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