CODEWITHSUNDEEP
c++
cyboashu
cyboashu

Reputation: 10433

How to return void* from function?

I am trying to

When I am doing this directly, like passing &x to the conversion method it returns correct value, but through method call I am getting weird number.

here is the code...

void *GetVoidPointer(int a)
{   
    return &a;
}

int ReadValueFromVoidPointer(void *ptr)
{   
    int *i = static_cast<int*>(ptr);
    return *i;
}

int main()
{
    int x = 10;

    //Through method call to get void* : This doesn't work
    int i= ReadValueFromVoidPointer(GetVoidPointer(x));
    printf("%d\n", i);

    //Direct call  : This works
    int y= ReadValueFromVoidPointer(&x);
    printf("%d\n", y);

    return 0;
}

I am very new to C++ so might be doing something very silly. Please advise.

Upvotes: 1

Views: 149

Answers (2)

cocool97
cocool97

Reputation: 1251

When compiling your code you receive the following error : 

test.cpp: In function ‘void* GetVoidPointer(int)’:
test.cpp:4:26: warning: address of local variable ‘a’ returned [-Wreturn-local-addr]
 void *GetVoidPointer(int a)

It means that your are returning the adress of a local variable (a), which is a problem as this variable is removed at the end of the function call.
Therefore you have to pass a reference to it (I'll let you look for some detailed explanations about it...).
So you just have to change the GetVoidPointer() by the following and everything is OK : void *GetVoidPointer(int& a).
The result values are 10 and 10.
I'm using the compile g++ in version 8.3.0 with no added options.

Upvotes: 4

Allen Zhu
Allen Zhu

Reputation: 870

In function GetVoidPointer, the local param 'a' is temporary. When GetVoidPointer returned, 'a' will be destroyed by the system. So, you will not get the correct value. You could use 'C++ reference'.

Upvotes: 0

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