Why would you cast a struct as a char* when using fread in c?

Question

Lets say I have a "Passenger" struct, which has a field for a name.

If I do (like the syntax of my book shows):

fread(&passenger, sizeof(Passenger), 1, in_fp);
printf("%s", (*passenger).first_name)

I get a segmentation fault, but if I do:

fread( (char *)passenger, sizeof(Passenger), 1, in_fp);
printf("%s", (*passenger).first_name)

the name read from the file will be printed out.

Answer 1

Looks to me as if 'passenger' is a pointer. If you take &passenger, you are passing the address of the pointer to fread. When you cast it, you are telling fread to treat it as a pointer to a char buffer instead of a pointer to a Passenger.

Answer 2

You probably have a pointer to a Passenger, not a Passenger:

fread(passenger, sizeof(Passenger), 1, in_fp); printf("%s", (*passenger).first_name)

Will most likely do what you want.

Answer 3

In the early days of C language, C had no void * type (it appeared later, was borrowed from C++ actually) and type char * was used as the generic "raw memory" pointer type instead. So, even to this day you might see this habitual rudimentary use of type char * as the generic pointer type in the code, and see other pointer types explicitly converted to and from char * in generic pointer context. I'd guess that the code you quoted does this for that specific reason. (Unless it was you who put this char * cast there. In that case I can only ask "Why?".)

In modern C the first parameter of fread has void * type, meaning that no cast is necessary. This

fread(passenger, sizeof *passenger, 1, in_fp);

whill work just as well.

Your &passenger version makes no sense, since apparently the original intent was to read data into the location passenger points to, not into the passenger pointer object itself.