Void pointer

[howa]

int i = 10;
void *j = &i;


How to print out the integer "10" ?

Thanks.

 

[Ian Collins]

int i = 10;
void *j = &i;


How to print out the integer "10" ?

#include <iostream>

int main() { std::cout << 10 << std::endl; }

Or where you asking something else?

 

[Ian Collins]

* howa:

#include <iostream>
int main
{
std::cout << "10" << std::endl;
}

Echo?

 

[Alf P. Steinbach]

* howa:

int i = 10;
void *j = &i;


How to print out the integer "10" ?

#include <iostream>
int main
{
std::cout << "10" << std::endl;
}

But I guess you meant how to access the variable "i" given the void*
pointer "j".

Technically the answer is to cast the pointer and dereference it,
*static_cast<int*>(j).

However, as a novice you should not use raw pointers at all, much less
void* pointers: what's the problem you're trying to use void* to solve?


Cheers, & hth.,

- ALf

 

[Alf P. Steinbach]

* Ian Collins:

Echo?

No, just simultanous postings: great minds think alike. ;-)

Cheers,

- Alf

 

[howa]

* howa:



Technically the answer is to cast the pointer and dereference it,
*static_cast<int*>(j).

However, as a novice you should not use raw pointers at all, much less
void* pointers: what's the problem you're trying to use void* to solve?

How does the following one compare with static_cast?

e.g. cout<< *(int*)j;


Howard

 

[Ian Collins]

How does the following one compare with static_cast?

e.g. cout<< *(int*)j;

It's harder to search for in code and more likely to hide errors (the
compiler will reject inappropriate use of static_cast).

 

[howa]

It's harder to search for in code and more likely to hide errors (the
compiler will reject inappropriate use of static_cast).

Thanks.