Allocating char array using malloc

Yes, it’s a matter of style, because you’d expect sizeof(char) to always be one.

On the other hand, it’s very much an idiom to use sizeof(foo) when doing a malloc, and most importantly it makes the code self documenting.

Also better for maintenance, perhaps. If you were switching from char to wchar, you’d switch to

wchar *p = malloc( sizeof(wchar) * ( len + 1 ) );

without much thought. Whereas converting the statement char *p = malloc( len + 1 ); would require more thought. It’s all about reducing mental overhead.

And as @Nyan suggests in a comment, you could also do

type *p = malloc( sizeof(*p) * ( len + 1 ) );

for zero-terminated strings and

type *p = malloc( sizeof(*p) * len ) );

for ordinary buffers.