Will the following program execute? C++

Will the following program execute?
void main()
{
void *vptr = (void *) malloc(sizeof(void));
vptr++;
}


Answer1
It will throw an error, as arithmetic operations cannot be performed on void pointers.

Answer2
It will not build as sizeof cannot be applied to void* ( error “Unknown size” )

Answer3
How can it execute if it won’t even compile? It needs to be int main, not void main. Also, cannot increment a void *.

Answer4
According to gcc compiler it won’t show any error, simply it executes. but in general we can’t do arthematic operation on void, and gives size of void as 1

Answer5
The program compiles in GNU C while giving a warning for “void main”. The program runs without a crash. sizeof(void) is “1? hence when vptr++, the address is incremented by 1.

Answer6
Regarding arguments about GCC, be aware that this is a C++ question, not C. So gcc will compile and execute, g++ cannot. g++ complains that the return type cannot be void and the argument of sizeof() cannot be void. It also reports that ISO C++ forbids incrementing a pointer of type ‘void*’.

Answer7
in C++
voidp.c: In function `int main()’:
voidp.c:4: error: invalid application of `sizeof’ to a void type
voidp.c:4: error: `malloc’ undeclared (first use this function)
voidp.c:4: error: (Each undeclared identifier is reported only once for each function it appears in.)
voidp.c:6: error: ISO C++ forbids incrementing a pointer of type `void*’

But in c, it work without problems