C++ error in malloc - c++

Question

I am new to C/C++ and trying to test my understanding of pointers. For this I created the following simple program to create an integer array to store 10 numbers and print it out.
#include<iostream>
#include <cstdlib>
// #include <cstdint>
int main(){
int numberOfItems = 10;
int *array;
// int *array = malloc((int*)numberOfItems*sizeof(int));
array = malloc((int*)numberOfItems*sizeof(int));
for(int i=0;i<numberOfItems;i++){
*array = i;
array++;
i++;
}
std::cout<<"The size of the array is : "<<sizeof(array)<<std::endl;
for(int i=0;i<numberOfItems;i++){
std::cout<<*array<<std::endl;
array++;
i++;
}
return 0;
}
I ran the program using the command g++ main.cpp.
But it gives me the following error:
main.cpp:8:47: error: invalid operands of types ‘int*’ and ‘long unsigned int’ to binary ‘operator*’
array = malloc((int*)numberOfItems*sizeof(int));
Tried and Failed
I tried to change the int to uint64_t but still it gave me similar kind of error.
I do not understand what I am doing wrong.

Answer 1

The error is due to the fact that your syntax is incorrect in the following line.
array = malloc((int*)numberOfItems*sizeof(int));
The right syntax would be as follows:
array = (int*) malloc(numberOfItems*sizeof(int));
You need to typecast the return value of malloc as it returns a (void *) pointer. So, you need to specify what type of data you would be pointing when you access the memory.

Answer 2

Let's go through your code step by step:
int *array;
array = malloc((int*)numberOfItems*sizeof(int));
Here you have two problems:
malloc is C, not C++. Yes, malloc is supported in C++, but for backwawrds compatibility with legacy code. Do not use it.
You cast numberOfItems to a pointer, then multiply the pointer with an integer. That's undefined behavior. Your compiler might even do the right thing here, but you cannot know and you cannot trust any result you're getting.
Let's continue:
for(int i=0;i<numberOfItems;i++){
*array = i;
array++;
i++;
}
The for declaration looks fine. You initialize an integer i to 0, and you increment it using the ++ operator as long as it is smaller than numberOfItems, that's good. But:
You increment i again inside your loop. That means you actually increment i by 2, so i goes 0, 2, 4, 6, 8, 10, and your loop is only executed 5 times.
You increment your array pointer. While technically perfectly valid code, this will bite you further down.
So you write the values 0, 2, 4, 6 and 8 into memory at the initial location of your array pointer, and 1*sizeof(int), 2*sizeof(int), 3*sizeof(int), 4*sizeof(int) bytes further down each.
Next problem:
std::cout<<"The size of the array is : "<<sizeof(array)<<std::endl;
This will not print what you meant it to print. array is a pointer type, so sizeof(array) will return the size of a pointer, which is probably either 4 or 8 bytes depending on the bitnes of your application/OS.
And last but not least your last loop:
for(int i=0;i<numberOfItems;i++){
std::cout<<*array<<std::endl;
array++;
i++;
}
You do the same thing with i as in your first loop. This time, this saves your neck, because...
...since you incremented the pointer array already in your first loop, you're now continuing at 5*sizeof(int) bytes further down in the memory. You're therefore not printing the values you set in the first loop (which was probably your intention), but uninitialized values that reside in the second half of the memory originally allocated at array.