C malloc function's size parameter

Question

I am reading in a book that the malloc function in C takes the number of 'chunks' of memory you wish to allocate as a parameter and determines how many bytes the chunks are based on what you cast the value returned by malloc to. For example on my system an int is 4 bytes:

int *pointer;

pointer = (int *)malloc(10);

Would allocate 40 bytes because the compiler knows that ints are 4 bytes.

This confuses me for two reasons:

1. I was reading up, and the size parameter is actually the number of bytes you want to allocate and is not related to the sizes of any types.

2. Malloc is a function that returns an address. How does it adjust the size of the memory it allocated based on an external cast of the address it returned from void to a different type? Is it just some compiler magic I am supposed to accept?

I feel like the book is wrong. Any help or clarification is greatly appreciated!

Here is what the book said:

char *string;
string = (char *)malloc(80);

The 80 sets aside 80 chunks of storage. The chunk size is set by the typecast, (char *), which means that malloc() is finding storage for 80 characters of text.

Answer 1

Yes the book is wrong and you are correct please throw away that book.
Also, do let everyone know of the name of the book so we can permanently put it in our never to recommend black list.

Good Read:
What is the Best Practice for malloc?