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A friend told me that he had read somewhere that giving pointers to an
average programmer is like giving a gun to a 6 year old. Since I got
the wrong answer for the problem below I must be an average
programmer.

I got this problem from cprogramming.com. Does it display 10 or 12 and
why?

void afunction(int *x)
{
x = new int;
*x = 12;
}

void main()
{
int v = 10;
afunction(&v);
cout << "\nv:" << v;

getchar();
}


Thanks,
jvh

jvh wrote:
> void afunction(int *x)
> {
> x = new int;
> *x = 12;
> }
>
> void main()

int main()

> {
> int v = 10;
> afunction(&v);
> cout << "\nv:" << v;

std::cout << "\nv:" << v;
>
> getchar();
> }

It displays 10. C++ passes the address of v in main to afunction by
value. The parameter, x, in afunction is the value of the pointer, and
changing it doesn't change v in main.


I think it'll become clearer if you try adding some output to see the
changing values, maybe like so:

void a(int *x) {
std::cout << "a,x " << x << " a,*x " << *x << std::endl;
x = new int(12); // leaks memory
std::cout << "a,x " << x << " a,*x " << *x << std::endl;
}

void testa() {
int v = 10;
std::cout << "testa,&v " << &v << " testa,v " << v << std::endl;
a(&v);
std::cout << "testa,&v " << &v << " testa,v " << v << std::endl;
}

int main() {
testa();
}

>
> It displays 10. *C++ passes the address of *v in main to afunction by
> value. *The parameter, x, in afunction is the value of the pointer, and
> changing it doesn't change v in main.
>
> I think it'll become clearer if you try adding some output to see the
> changing values, maybe like so:
>
> void a(int *x) {
> * * std::cout << "a,x " << x << " a,*x " << *x << std::endl;
> * * x = new int(12); // leaks memory
> * * std::cout << "a,x " << x << " a,*x " << *x << std::endl;
>
> }
>
> void testa() {
> * * int v = 10;
> * * std::cout << "testa,&v " << &v *<< " testa,v *" << v << std::endl;
> * * a(&v);
> * * std::cout << "testa,&v " << &v *<< " testa,v *" << v << std::endl;
>
> }
>
> int main() {
> * * testa();
>
>

Thanks for the post, I should know better. If x had not been
reassigned then it would display 12. So I think it would be correct to
say that V was passed by reference (by its address) but it's address
was passed by value.

I had put it many display statements and they were telling me what you
told me but not as concisely and I did not see it.

Thanks again,

jvh (AKA an average programmer)

jvh wrote:
>> It displays 10. C++ passes the address of v in main to afunction by
>> value. The parameter, x, in afunction is the value of the pointer,
>> and changing it doesn't change v in main.
>>
>> I think it'll become clearer if you try adding some output to see the
>> changing values, maybe like so:
>>
>> void a(int *x) {
>> std::cout << "a,x " << x << " a,*x " << *x << std::endl;
>> x = new int(12); // leaks memory
>> std::cout << "a,x " << x << " a,*x " << *x << std::endl;
>>
>> }
>>
>> void testa() {
>> int v = 10;
>> std::cout << "testa,&v " << &v << " testa,v " << v << std::endl;
>> a(&v);
>> std::cout << "testa,&v " << &v << " testa,v " << v << std::endl;
>>
>> }
>>
>> int main() {
>> testa();
>>
>>
>
> Thanks for the post, I should know better. If x had not been
> reassigned then it would display 12. So I think it would be correct to
> say that V was passed by reference (by its address) but it's address
> was passed by value.

That's still not right. The following is closer to equivalent:

void a(int * p)
{
int * x = new int;
*x = 12;
}

You probably intended the following instead:

void a(int ** p)
{
*p = new int;
**p = 12;
}

int main()
{
int * p;
a(&p);
...
}

On Dec 30 2011, 7:31*pm, jvh <jvh75...@gmail.com> wrote:
> A friend told me that he had read somewhere that giving pointers to an
> average programmer is like giving a gun to a 6 year old. Since I got
> the wrong answer for the problem below I must be an average
> programmer.
>
> I got this problem from cprogramming.com. Does it display 10 or 12 and
> why?
>
> void afunction(int *x)
> {
> * * * * x = new int;
> * * * * *x = 12;
>
> }
>
> void main()
> {
> * * * * int v = 10;
> * * * * afunction(&v);
> * * * * cout << "\nv:" << v;
>
> * * * * getchar();
>
> }

In addition to what's been said, its sometimes illuminating to
"inline" the function yourself, so you can see what's going on.

void main()
{
int v = 10;
// afunction(&v);
{
int *x = &v; // x points to v
x = new int; // x now points someplace else
*x = 12; // set the value stored "someplace else"
}
cout << "\nv:" << v;
getchar();
}