Thread: Some basic C questions

I was going through a few C quizzes and was quite baffled by the amount of questions i could answer correctly. Never thought C was such a confusing language !

Well, anyway... here are some doubts i've had..

1. Can a pointer be typecasted ? Why or why not ?

2. I've heard that pointer can hold the address of a function. What does that mean ? Where is a function stored ?

3. What does #define f(g,g2) g##g2 mean ?

4.

Code:

Output of this program ?
int i= 5;
printf("%d",i++ + ++i)
Answer : cannot be predicted.

Why ?

5.

Code:

Output of this ?
int i=5;
printf("%d %d %d %d %d",i++,i--,++i,--i,i);

With those questions, I suggest you provide you thoughts on what the answers are. People will then be more inclined to give you pointers (no pun intended) to confirm or help you improve your understanding. But if you don't do that, it will be assumed you are just asking us to do homework for you.

Originally Posted by grumpy

With those questions, I suggest you provide you thoughts on what the answers are. People will then be more inclined to give you pointers (no pun intended) to confirm or help you improve your understanding. But if you don't do that, it will be assumed you are just asking us to do homework for you.

Thats not homework.. I asked these questions because i have no clue as to what the answer is. If i knew so, i wouldnt be asking.

But then, to elaborate on them more..

1. Can a pointer be typecasted ? Why or why not ?

Absolutely no clue about this.

2. I've heard that pointer can hold the address of a function. What does that mean ? Where is a function stored ?

According to my knowledge, when a function is called, the function details are stored in an activation frame which is pushed into the system stack.
I've never heard of function arguments being pushed into anything. So, what exactly is goin on here ?

3. What does #define f(g,g2) g##g2 mean ?

What does ## between g and g2 means ??

4.

Output of this program ?
int i= 5;
printf("%d",i++ + ++i)

Answer : cannot be predicted.

Why ? Shouldnt this yield 5 + 6 = 11 ?

5.

Output of this ?
int i=5;
printf("%d %d %d %d %d",i++,i--,++i,--i,i);

According to the answer to the above question, the function arguments are pushed from last to first and evaluated while being pushed. Please elaborate on this.

Thanks a lot for the help.

Originally Posted by Salem

1. Yes. But the result isn't necessarily valid (eg. a char pointer to an int pointer). Only to void* and back to the SAME type is allowed for sure.

2. Yes. Look at the declaration of qsort in stdlib.h

3. Read about token concatenation operator in the C preprocessor section of your manual / book

4. http://c-faq.com/expr/index.html
5. http://c-faq.com/expr/index.html

Originally Posted by matsp

1. Do you understand what a typecast is? If not, you need to read your C programming literature a bit more. Then the answer will probably be apparent.

2. You are not at all answering the question - what you replied was how a call-stack works, which has extremely little to do with the function pointers - about as much as tyres have to do with the air-fuel mixture of a petrol engine.

3. Read your book (or search the web - this one should definitely be searchable).

4. Because the order of i++, ++i and + is not defined. If i++ and ++i are performed before the addition of the result of the two, then you get 6 + 7 = 13, if ++i is done before addition and i++ after, then you get 11 or 12 (depending on if the value of the first instance of i is before or after ++i). [And technically, the compiler is "allowed" to come up with any answer, such as 3128357912 as the answer, since this is "undefined behaviour", and thus "anything can happen" - but it's likely that the answer is in the range 10-13, depending on which compiler, compiler optimization level, etc].

5. In this case, the comma is separating the statements, so the answer is predictable, and you can then figure out (from the order of evaluation) what the result would be.

--
Mats