x = malloc(n * sizeof (int)); cannot convert void to int

Question

I found a code that computes the execution time of merge sort. I'm getting an error that says "Cannot convert void to int" . And I'm getting the error from the part of this code that says x = malloc(n * sizeof (int));

    x = malloc(n * sizeof (int));

Answer 1

Since this question is tagged with C++, I recommend a completely different solution: Embrace the power of C++ and use a std::vector .

std::vector<int> descriptiveNameHere(n);

This approach frees you from having to manually manage the memory allocation. When the vector is allowed to go out of scope the storage will automatically be freed.

Documentation on std::vector

Answer 2

C++ compilers are more strict than C compilers. Therefore not all C code will compile using a C++ compiler, so you shouldn't use a C++ compiler if you are not using any C++ feature.

malloc returns a void * so you must cast it to the corresponding type, provided that you are using a C compiler:

Type * x = (Type *) malloc (sizeof(Type));

But if you must use C++ and malloc, then like mentioned in the comments, the C-style cast should be a last resort and you should rather use static_cast

Type * x = static_cast<Type *>(malloc(sizeof(Type)));

Answer 3

You should cast the malloc like this.

x = (int *)malloc(sizeof(int) * n)

because malloc just return void pointer