Embedded C interview questions On C or C++

1. Using the #define statement, how would you declare a manifest constant that
returns the number of seconds in a year? Disregard leap years in your answer.
#define SECONDS_PER_YEAR
(60 * 60 * 24 * 365)UL
The purpose of this question is to test the following:
· Basic knowledge of the #define syntax (for example, no semi-colon at the end, the need
to parenthesize, and so on)
· An understanding that the pre-processor will evaluate constant expressions for you. Thus,
it is clearer, and penalty-free, to spell out how you are calculating the number of seconds
in a year, rather than actually doing the calculation yourself
· A realization that the expression will overflow an integer argument on a 16-bit machinehence
the need for the L, telling the compiler to treat the variable as a Long
· As a bonus, if you modified the expression with a UL (indicating unsigned long), then you
are off to a great start. And remember, first impressions count!

2. Write the "standard" MIN macro-that is, a macro that takes two arguments and
returns the smaller of the two arguments.
#define MIN(A,B)
((A)
<
= (B) ? (A) : (B))
The purpose of this question is to test the following:
· Basic knowledge of the #define directive as used in macros. This is important because until the inline operator becomes part of standard C, macros are the only portable way of
generating inline code. Inline code is often necessary in embedded systems in order to
achieve the required performance level
· Knowledge of the ternary conditional operator. This operator exists in C because it allows
the compiler to produce more optimal code than an if-then-else sequence. Given that
performance is normally an issue in embedded systems, knowledge and use of this
construct is important
· Understanding of the need to very carefully parenthesize arguments to macros
· I also use this question to start a discussion on the side effects of macros, for example,
what happens when you write code such as:
least = MIN(*p++, b);

3. What is the purpose of the preprocessor directive #error?
Either you know the answer to this, or you don't. If you don't, see Reference 1. This question is
useful for differentiating between normal folks and the nerds. Only the nerds actually read the
appendices of C textbooks to find out about such things. Of course, if you aren't looking for a
nerd, the candidate better hope she doesn't know the answer.

4. Infinite loops often arise in embedded systems. How does you code an infinite loop
in C?
There are several solutions to this question. My preferred solution is:
while(1)
{
}
Many programmers seem to prefer:
for(;;)
{
}
This construct puzzles me because the syntax doesn't exactly spell out what's going on. Thus, if
a candidate gives this as a solution, I'll use it as an opportunity to explore their rationale for
doing so. If their answer is basically, "I was taught to do it this way and I haven't thought about
it since," it tells me something (bad) about them.
A third solution is to use a goto :
goto Loop;
Candidates who propose this are either assembly language programmers (which is probably
good), or else they are closet BASIC/FORTRAN programmers looking to get into a new field.

5. Using the variable a, give definitions for the following:
a) An integer
b) A pointer to an integer
c) A pointer to a pointer to an integer
d) An array of 10 integers
e) An array of 10 pointers to integers
f) A pointer to an array of 10 integers
g) A pointer to a function that takes an integer as an argument and returns an integer
h) An array of ten pointers to functions that take an integer argument and return an
integer
The answers are:
a) int a; // An integer
b) int *a; // A pointer to an integer
c) int **a; // A pointer to a pointer to an integer
d) int a[10]; // An array of 10 integers
e) int *a[10]; // An array of 10 pointers to integers
f) int (*a)[10]; // A pointer to an array of 10 integers
g) int (*a)(int); // A pointer to a function a that takes an integer argument and
returns an integer
h) int (*a[10])(int); // An array of 10 pointers to functions that take an integer
argument and return an integer

People often claim that a couple of these are the sorts of thing that one looks up in textbooksand
I agree. While writing this article, I consulted textbooks to ensure the syntax was correct.
However, I expect to be asked this question (or something close to it) when I'm being
interviewed. Consequently, I make sure I know the answers, at least for the few hours of the
interview. Candidates who don't know all the answers (or at least most of them) are simply
unprepared for the interview. If they can't be prepared for the interview, what will they be
prepared for?

6. What are the uses of the keyword static?
This simple question is rarely answered completely. Static has three distinct uses in C:
· A variable declared static within the body of a function maintains its value between
function invocations
· A variable declared static within a module, (but outside the body of a function) is
accessible by all functions within that module. It is not accessible by functions within any
other module. That is, it is a localized global

· Functions declared static within a module may only be called by other functions within
that module. That is, the scope of the function is localized to the module within which it is
declared
Most candidates get the first part correct. A reasonable number get the second part correct,
while a pitiful number understand the third answer. This is a serious weakness in a candidate,
since he obviously doesn't understand the importance and benefits of localizing the scope of
both data and code.

7. What does the keyword const mean?
As soon as the interviewee says "const means constant," I know I'm dealing with an amateur.
Dan Saks has exhaustively covered const in the last year, such that every reader of ESP should
be extremely familiar with what const can and cannot do for you. If you haven't been reading
that column, suffice it to say that const means "read-only." Although this answer doesn't really
do the subject justice, I'd accept it as a correct answer. (If you want the detailed answer, read
Saks' columns-carefully!)
If the candidate gets the answer correct, I'll ask him these supplemental questions:
What do the following declarations mean?
const int a;
int const a;
const int *a;
int * const a;
int const * a const;
The first two mean the same thing, namely a is a const (read-only) integer. The third means a is
a pointer to a const integer (that is, the integer isn't modifiable, but the pointer is). The fourth
declares a to be a const pointer to an integer (that is, the integer pointed to by a is modifiable,
but the pointer is not). The final declaration declares a to be a const pointer to a const integer
(that is, neither the integer pointed to by a, nor the pointer itself may be modified). If the
candidate correctly answers these questions, I'll be impressed. Incidentally, you might wonder
why I put so much emphasis on const, since it is easy to write a correctly functioning program
without ever using it. I have several reasons:
· The use of const conveys some very useful information to someone reading your code. In
effect, declaring a parameter const tells the user about its intended usage. If you spend a
lot of time cleaning up the mess left by other people, you'll quickly learn to appreciate
this extra piece of information. (Of course, programmers who use const , rarely leave a
mess for others to clean up.)
· const has the potential for generating tighter code by giving the optimizer some
additional information
· Code that uses const liberally is inherently protected by the compiler against inadvertent
coding constructs that result in parameters being changed that should not be. In short,they tend to have fewer bugs

8. What does the keyword volatile mean? Give three different examples of its use.
A volatile variable is one that can change unexpectedly. Consequently, the compiler can make
no assumptions about the value of the variable. In particular, the optimizer must be careful to
reload the variable every time it is used instead of holding a copy in a register. Examples of
volatile variables are:
· Hardware registers in peripherals (for example, status registers)
· Non-automatic variables referenced within an interrupt service routine
· Variables shared by multiple tasks in a multi-threaded application
Candidates who don't know the answer to this question aren't hired. I consider this the most
fundamental question that distinguishes between a C programmer and an embedded systems
programmer. Embedded folks deal with hardware, interrupts, RTOSes, and the like. All of these
require volatile variables. Failure to understand the concept of volatile will lead to disaster.
On the (dubious) assumption that the interviewee gets this question correct, I like to probe a
little deeper to see if they really understand the full significance of volatile . In particular, I'll ask
them the following additional questions:
· Can a parameter be both const and volatile ? Explain.
· Can a pointer be volatile ? Explain.
· What's wrong with the following function?:
int square(volatile int *ptr)
{
return *ptr * *ptr;
}
The answers are as follows:
· Yes. An example is a read-only status register. It is volatile because it can change
unexpectedly. It is const because the program should not attempt to modify it
· Yes, although this is not very common. An example is when an interrupt service routine
modifies a pointer to a buffer
· This one is wicked. The intent of the code is to return the square of the value pointed to
by *ptr . However, since *ptr points to a volatile parameter, the compiler will generate
code that looks something like this:
int square(volatile int *ptr)
{
int a,b;
a = *ptr;
b = *ptr;
return a * b;
}
Because it's possible for the value of *ptr to change unexpectedly, it is possible for a and b to be
different. Consequently, this code could return a number that is not a square! The correct way to
code this is:
long square(volatile int *ptr)
{
int a;
a = *ptr;
return a * a;
}

9. Embedded systems always require the user to manipulate bits in registers or
variables. Given an integer variable a, write two code fragments. The first should set
bit 3 of a. The second should clear bit 3 of a. In both cases, the remaining bits should
be unmodified.
These are the three basic responses to this question:
· No idea. The interviewee cannot have done any embedded systems work
· Use bit fields. Bit fields are right up there with trigraphs as the most brain-dead portion
of C. Bit fields are inherently non-portable across compilers, and as such guarantee that
your code is not reusable. I recently had the misfortune to look at a driver written by
Infineon for one of their more complex communications chips. It used bit fields and was
completely useless because my compiler implemented the bit fields the other way
around. The moral: never let a non-embedded person anywhere near a real piece of
hardware!
· Use #defines and bit masks. This is a highly portable method and is the one that should
be used. My optimal solution to this problem would be:
#define BIT3 (0x1
<
<
3)
static int a;
void set_bit3(void) {
a |= BIT3;
}
void clear_bit3(void) {
a &= ~BIT3;
}
Some people prefer to define a mask together with manifest constants for the set and clear
values. This is also acceptable. The element that I'm looking for is the use of manifest constants,
together with the |= and &= ~ constructs

10. Embedded systems are often characterized by requiring the programmer to access
a specific memory location. On a certain project it is required to set an integer
variable at the absolute address 0x67a9 to the value 0xaa55. The compiler is a pure
ANSI compiler. Write code to accomplish this task.
This problem tests whether you know that it is legal to typecast an integer to a pointer in order
to access an absolute location. The exact syntax varies depending upon one's style. However, I
would typically be looking for something like this:
int *ptr;
ptr = (int *)0x67a9;
*ptr = 0xaa55;
A more obscure approach is:
*(int * const)(0x67a9) = 0xaa55;
Even if your taste runs more to the second solution, I suggest the first solution when you are in
an interview situation.

11. Interrupts are an important part of embedded systems. Consequently, many
compiler vendors offer an extension to standard C to support interrupts. Typically, this
new keyword is __interrupt. The following code uses __interrupt to define an
interrupt service routine (ISR). Comment on the code.
__interrupt double compute_area
(double
radius)
{
double area = PI * radius *
radius;
printf("\nArea = %f", area);
return area;
}
This function has so much wrong with it, it's hard to know where to start:
· ISRs cannot return a value. If you don't understand this, you aren't hired
· ISRs cannot be passed parameters. See the first item for your employment prospects if
you missed this
· On many processors/compilers, floating-point operations are not necessarily re-entrant.
In some cases one needs to stack additional registers. In other cases, one simply cannot
do floating point in an ISR. Furthermore, given that a general rule of thumb is that ISRs
should be short and sweet, one wonders about the wisdom of doing floating-point math
here
· In a vein similar to the third point, printf() often has problems with reentrancy and
performance. If you missed points three and four, I wouldn't be too hard on you.
Needless to say, if you got these last two points, your employment prospects are looking
better and better

12. What does the following code output and why?
void foo(void)
{
unsigned int a = 6;
int b = -20;
(a+b > 6) ? puts("> 6") :
puts("
<
= 6");
}
This question tests whether you understand the integer promotion rules in C-an area that I find
is very poorly understood by many developers. Anyway, the answer is that this outputs "> 6."
The reason for this is that expressions involving signed and unsigned types have all operands
promoted to unsigned types. Thus �comes a very large positive integer and the expression
evaluates to greater than 6. This is a very important point in embedded systems where unsigned
data types should be used frequently (see Reference 2). If you get this one wrong, you are
perilously close to not getting the job.
13. Comment on the following code fragment.
unsigned int zero = 0;
unsigned int compzero = 0xFFFF;
/*1's complement of zero */
On machines where an int is not 16 bits, this will be incorrect. It should be coded:
unsigned int compzero = ~0;
This question really gets to whether the candidate understands the importance of word length on a computer. In my experience, good embedded programmers are critically aware of the
underlying hardware and its limitations, whereas computer programmers tend to dismiss the
hardware as a necessary annoyance.
By this stage, candidates are either completely demoralized-or they're on a roll and having a
good time. If it's obvious that the candidate isn't very good, then the test is terminated at this
point. However, if the candidate is doing well, then I throw in these supplemental questions.
These questions are hard, and I expect that only the very best candidates will do well on them.
In posing these questions, I'm looking more at the way the candidate tackles the problems,
rather than the answers. Anyway, have fun.

14. Although not as common as in non-embedded computers, embedded systems do
still dynamically allocate memory from the heap. What are the problems with dynamic
memory allocation in embedded systems?
Here, I expect the user to mention memory fragmentation, problems with garbage collection,
variable execution time, and so on. This topic has been covered extensively in ESP , mainly by
P.J. Plauger. His explanations are far more insightful than anything I could offer here, so go and
read those back issues! Having lulled the candidate into a sense of false security, I then offer up
this tidbit:
What does the following code fragment output and why?
char *ptr;
if ((ptr = (char *)malloc(0)) ==
NULL)
else
puts("Got a null pointer");
puts("Got a valid pointer");
This is a fun question. I stumbled across this only recently when a colleague of mine
inadvertently passed a value of 0 to malloc and got back a valid pointer! That is, the above code
will output "Got a valid pointer." I use this to start a discussion on whether the interviewee
thinks this is the correct thing for the library routine to do. Getting the right answer here is not
nearly as important as the way you approach the problem and the rationale for your decision.

15. Typedef is frequently used in C to declare synonyms for pre-existing data types. It
is also possible to use the preprocessor to do something similar. For instance,
consider the following code fragment:
#define dPS struct s *
typedef struct s * tPS;
The intent in both cases is to define dPS and tPS to be pointers to structure s. Which
method, if any, is preferred and why?
This is a very subtle question, and anyone who gets it right (for the right reason) is to be
congratulated or condemned ("get a life" springs to mind). The answer is the typedef is
preferred. Consider the declarations:
dPS p1,p2;
tPS p3,p4;
The first expands to:
struct s * p1, p2;
which defines p1 to be a pointer to the structure and p2 to be an actual structure, which is
probably not what you wanted. The second example correctly defines p3 and p4 to be pointers.

16. C allows some appalling constructs. Is this construct legal, and if so what does
this code do?
int a = 5, b = 7, c;
c = a+++b;
This question is intended to be a lighthearted end to the quiz, as, believe it or not, this is
perfectly legal syntax. The question is how does the compiler treat it? Those poor compiler
writers actually debated this issue, and came up with the "maximum munch" rule, which
stipulates that the compiler should bite off as big (and legal) a chunk as it can. Hence, this code
is treated as:
c = a++ + b;
Thus, after this code is executed, a = 6, b = 7, and c = 12.
If you knew the answer, or guessed correctly, well done. If you didn't know the answer then I
wouldn't consider this to be a problem. I find the greatest benefit of this question is that it is
good for stimulating questions on coding styles, the value of code reviews, and the benefits of
using lint.