compile error about void*

[George2]

Hello everyone,


What is wrong with the code, I just want to allocate an array of 100 void* pointers. :-)

int main()
{
    void** p;
 
    p = new (void*) [100];
 
    return 0;
}
 

>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C2143: syntax error : missing ';' before '['
1>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C3409: empty attribute block is not allowed
1>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C2143: syntax error : missing ']' before 'constant'
1>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C2143: syntax error : missing ';' before 'constant'
1>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C2143: syntax error : missing ';' before ']'
1>d:\visual studio 2008\projects\t est_void1\test_ void1\main.cpp( 5) : error C2143: syntax error : missing ';' before ']'


thanks in advance,
George

[Savage]

You just remove those '()' around void*,and it should compile.

p = new void* [100];

[weaknessforcats]

Use of void* deprecated in C++.

[George2]

Thanks Savage,


Why adding () does not work, which C++ rule is violated?

You just remove those '()' around void*,and it should compile.

p = new void* [100];

regards,
George

[George2]

Why weaknessforcats ?


Any links for this topic and using what to replace void*?

Use of void* deprecated in C++.

regards,
George

[Savage]

Thanks Savage,


Why adding () does not work, which C++ rule is violated?




regards,
George

p = new (void*) [100];

this is interpreted by compiler as:

p=(new unknown_type) (void*)[100];

because of new operator precedence.

[weaknessforcats]

p = new (void*) [100];

C++ ain't C.

In C++ new(void*) is a function named new that has a void* arguiment.

Why weaknessforcats ?


Any links for this topic and using what to replace void*?

Quote:
Originally Posted by weaknessforcats
Use of void* deprecated in C++.

Function overloading.

[Savage]

C++ ain't C.

In C++ new(void*) is a function named new that has a void* arguiment.



Function overloading.

What about this then:

void foo();
void (**p)();
 
p = new (void (*[3])());
p[0] = f;
 
//etc...

Is this interpreted as a overloaded function called new that takes a pointer to the array of generic function pointers,or does it allocate memory for p?

[George2]

Thanks Savage,


I have tried your code can compile. What does the following statement mean?

p = new (void (*[3])());

what is the type of p? Could you interpret the meaning of right side of assignment and left side of assignment please?

What about this then:

void foo();
void (**p)();
 
p = new (void (*[3])());
p[0] = f;
 
//etc...

Is this interpreted as a overloaded function called new that takes a pointer to the array of generic function pointers,or does it allocate memory for p?

regards,
George

[George2]

Thanks Savage,


Could you explain why the code will be explained to

p=(new unknown_type) (void*)[100];

please?

What is unknown_type comes from? What is new operator precedence?

p = new (void*) [100];

this is interpreted by compiler as:

p=(new unknown_type) (void*)[100];

because of new operator precedence.

regards,
George