How to create a temporary value initialized T * in Standard-C++

Question

How to create a temporary value-initialized T* in standard C++?

void foo( int );
void bar( int * );

int main()
{
    foo( int() );  // works. a temporary int - value initialized.
    bar( ??? );    // how to create a temporary int *?
}

Just out of curiousity.

I think it's designed so that you can't create a temporary pointer, because doing such a thing doesn't make much sense. What would it point to? If nothing, juts use `nullptr` — BWG, Jan 29 '15 at 01:38
Does `bar()` dereference the pointer, or is it "safe" to pass a pointer value that is illegal (i.e. points to unallocated memory)? — Jens, Jan 29 '15 at 01:55
@MarsonMao As I wrote: "Just out of curiousity." ... and because I were at a very uncreative state at the time asking that question and couldn't come up with an answer myself. — Swordfish, Oct 14 '18 at 16:43
@Swordfish nice to know it ;) and it's really an old comment! — Marson Mao, Oct 16 '18 at 03:05
@MarsonMao I'm intrigued to delete the question, though, since it STILL attracts downvoters for no appearant reason :/ — Swordfish, Oct 16 '18 at 03:07
@Swordfish i see...lol but fortunately it has only 4 down-votes :p — Marson Mao, Oct 16 '18 at 06:45
@Swordfish dont worry, u got other good answers to compensate this one. just leave it as a history of ... life :p — Marson Mao, Oct 17 '18 at 02:17

score 4 · Accepted Answer · edited May 23 '17 at 10:34

4

The easiest is to use curly braces:

 bar({});

Or a using statement:

using p = int*;
bar( p() );    // how to create a temporary int *?

sehe just reminded me of the stupidly obvious answer of nullptr, 0, and NULL.

bar(nullptr);

And I'm sure there's many more ways.

GCC lets you use compound literals, but technically this isn't allowed

 bar((int*){});

http://coliru.stacked-crooked.com/a/7a65dcb135a87ada

edited May 23 '17 at 10:34

Community

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answered Jan 29 '15 at 01:38

Mooing Duck

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that's easier with `nullptr` – sehe Jan 29 '15 at 01:38
Yes, there's the ancient `NULL`, and the decrepit `0`. Aside from millions of variations on those themes. – Deduplicator Jan 29 '15 at 01:58
using p = T*; is technically the same as a type alias introduced by typedef T* p; i guess? – Swordfish Jan 29 '15 at 02:18
1

@Swordfish: It's strictly equivalent. Still wondering about your post, why do those contortions? – Deduplicator Jan 29 '15 at 02:35
@Deduplicator: Somehow I had it in my head he wanted default-initialization, but the question clearly says value-initialization, so I added `0` and `NULL`. – Mooing Duck Jan 29 '15 at 17:42

score 0 · Answer 2 · answered Jan 29 '15 at 01:45

Just for kicks, you can try a typedef:

#include <iostream>
void foo( int ) {}
typedef int* PtrInt;
void bar( PtrInt p ) 
{
   std::cout << "The value of p is " << p;
}

int main()
{
    foo( int() );  
    bar( PtrInt() );  
}

Live Example: http://ideone.com/sjOMlj

score 0 · Answer 3 · answered Jan 29 '15 at 01:48

Why not simply use something like this:

int i=0;
bar(&i);  // safe to dereference in bar()

Or are you looking for an inliner? If so, you can use some frown-worthy casting, but then bar() shouldn't actually dereference that pointer:

bar((int*)0); // or use nullptr if your C++ compiler is more recent

How to create a temporary value initialized T * in Standard-C++

3 Answers3