c++ interview questions

Top c++ frequently asked interview questions

Do I cast the result of malloc?

In this question, someone suggested in a comment that I should not cast the results of malloc, i.e:

int *sieve = malloc(sizeof(int)*length);

rather than:

int *sieve = (int *)malloc(sizeof(int)*length);

Why would this be the case?

Source: (StackOverflow)

What does "static" mean in a C program?

I've seen the word static used in different places in C code; is this like a static function/class in C# (where the implementation is shared across objects)?

Source: (StackOverflow)

How do I achieve the theoretical maximum of 4 FLOPs per cycle?

How can the theoretical peak performance of 4 floating point operations (double precision) per cycle be achieved on a modern x86-64 Intel CPU?

As far as I understand it take three cycles for an SSE add and five cycles for a mul to complete on most of the modern Intel CPUs (see for example Agner Fog's 'Instruction Tables' ). Due to pipelining one can get a throughput of one add per cycle if the algorithm has at least three independent summations. Since that is true for packed addpd as well as the scalar addsd versions and SSE registers can contain two double's the throughput can be as much as two flops per cycle.

Furthermore, it seems (although I've not seen any proper documentation on this) add's and mul's can be executed in parallel giving a theoretical max throughput of four flops per cycle.

However, I've not been able to replicate that performance with a simple C/C++ programme. My best attempt resulted in about 2.7 flops/cycle. If anyone can contribute a simple C/C++ or assembler programme which demonstrates peak performance that'd be greatly appreciated.

My attempt:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <sys/time.h>

double stoptime(void) {
   struct timeval t;
   return (double) t.tv_sec + t.tv_usec/1000000.0;

double addmul(double add, double mul, int ops){
   // Need to initialise differently otherwise compiler might optimise away
   double sum1=0.1, sum2=-0.1, sum3=0.2, sum4=-0.2, sum5=0.0;
   double mul1=1.0, mul2= 1.1, mul3=1.2, mul4= 1.3, mul5=1.4;
   int loops=ops/10;          // We have 10 floating point operations inside the loop
   double expected = 5.0*add*loops + (sum1+sum2+sum3+sum4+sum5)
               + pow(mul,loops)*(mul1+mul2+mul3+mul4+mul5);

   for (int i=0; i<loops; i++) {
      mul1*=mul; mul2*=mul; mul3*=mul; mul4*=mul; mul5*=mul;
      sum1+=add; sum2+=add; sum3+=add; sum4+=add; sum5+=add;
   return  sum1+sum2+sum3+sum4+sum5+mul1+mul2+mul3+mul4+mul5 - expected;

int main(int argc, char** argv) {
   if (argc != 2) {
      printf("usage: %s <num>\n", argv[0]);
      printf("number of operations: <num> millions\n");
   int n = atoi(argv[1]) * 1000000;
   if (n<=0)

   double x = M_PI;
   double y = 1.0 + 1e-8;
   double t = stoptime();
   x = addmul(x, y, n);
   t = stoptime() - t;
   printf("addmul:\t %.3f s, %.3f Gflops, res=%f\n", t, (double)n/t/1e9, x);
   return EXIT_SUCCESS;

Compiled with

g++ -O2 -march=native addmul.cpp ; ./a.out 1000

produces the following output on an Intel Core i5-750, 2.66 GHz.

addmul:  0.270 s, 3.707 Gflops, res=1.326463

That is, just about 1.4 flops per cycle. Looking at the assembler code with g++ -S -O2 -march=native -masm=intel addmul.cpp the main loop seems kind of optimal to me:

inc    eax
mulsd    xmm8, xmm3
mulsd    xmm7, xmm3
mulsd    xmm6, xmm3
mulsd    xmm5, xmm3
mulsd    xmm1, xmm3
addsd    xmm13, xmm2
addsd    xmm12, xmm2
addsd    xmm11, xmm2
addsd    xmm10, xmm2
addsd    xmm9, xmm2
cmp    eax, ebx
jne    .L4

Changing the scalar versions with packed versions (addpd and mulpd) would double the flop count without changing the execution time and so I'd get just short of 2.8 flops per cycle. Is there a simple example which achieves four flops per cycle?

Nice little programme by Mysticial; here are my results (run just for a few seconds though):

  • gcc -O2 -march=nocona: 5.6 Gflops out of 10.66 Gflops (2.1 flops/cycle)
  • cl /O2, openmp removed: 10.1 Gflops out of 10.66 Gflops (3.8 flops/cycle)

It all seems a bit complex, but my conclusions so far:

  • gcc -O2 changes the order of independent floating point operations with the aim of alternating addpd and mulpd's if possible. Same applies to gcc-4.6.2 -O2 -march=core2.

  • gcc -O2 -march=nocona seems to keep the order of floating point operations as defined in the C++ source.

  • cl /O2, the 64-bit compiler from the SDK for Windows 7 does loop-unrolling automatically and seems to try and arrange operations so that groups of three addpd's alternate with three mulpd's (well, at least on my system and for my simple programme).

  • My Core i5 750 (Nahelem architecture) doesn't like alternating add's and mul's and seems unable to run both operations in parallel. However, if grouped in 3's it suddenly works like magic.

  • Other architectures (possibly Sandy Bridge and others) appear to be able to execute add/mul in parallel without problems if they alternate in the assembly code.

  • Although difficult to admit, but on my system cl /O2 does a much better job at low-level optimising operations for my system and achieves close to peak performance for the little C++ example above. I measured between 1.85-2.01 flops/cycle (have used clock() in Windows which is not that precise. I guess, need to use a better timer - thanks Mackie Messer).

  • The best I managed with gcc was to manually loop unroll and arrange additions and multiplications in groups of three. With g++ -O2 -march=nocona addmul_unroll.cpp I get at best 0.207s, 4.825 Gflops which corresponds to 1.8 flops/cycle which I'm quite happy with now.

In the C++ code I've replaced the for loop with

   for (int i=0; i<loops/3; i++) {
       mul1*=mul; mul2*=mul; mul3*=mul;
       sum1+=add; sum2+=add; sum3+=add;
       mul4*=mul; mul5*=mul; mul1*=mul;
       sum4+=add; sum5+=add; sum1+=add;

       mul2*=mul; mul3*=mul; mul4*=mul;
       sum2+=add; sum3+=add; sum4+=add;
       mul5*=mul; mul1*=mul; mul2*=mul;
       sum5+=add; sum1+=add; sum2+=add;

       mul3*=mul; mul4*=mul; mul5*=mul;
       sum3+=add; sum4+=add; sum5+=add;

And the assembly now looks like

mulsd    xmm8, xmm3
mulsd    xmm7, xmm3
mulsd    xmm6, xmm3
addsd    xmm13, xmm2
addsd    xmm12, xmm2
addsd    xmm11, xmm2
mulsd    xmm5, xmm3
mulsd    xmm1, xmm3
mulsd    xmm8, xmm3
addsd    xmm10, xmm2
addsd    xmm9, xmm2
addsd    xmm13, xmm2

Source: (StackOverflow)

What is the name of the "-->" operator?

After reading Hidden Features and Dark Corners of C++/STL on comp.lang.c++.moderated, I was completely surprised that the following snippet compiled and worked in both Visual Studio 2008 and G++ 4.4.

Here's the code:

#include <stdio.h>
int main()
    int x = 10;
    while (x --> 0) // x goes to 0
        printf("%d ", x);

I'd assume this is C, since it works in GCC as well. Where is this defined in the standard, and where has it come from?

Source: (StackOverflow)

Why does the C preprocessor interpret the word "linux" as the constant "1"?

Why does the C preprocessor in GCC interpret the word linux (small letters) as the constant 1?


#include <stdio.h>
int main(void)
    int linux = 5;
    return 0;

Result of $ gcc -E test.c (stop after the preprocessing stage):

int main(void)
    int 1 = 5;
    return 0;

Which -of course- yields an error.

(BTW: There is no #define linux in the stdio.h file.)

Source: (StackOverflow)

What is the strict aliasing rule?

When asking about common undefined behavior in C, souls more enlightened than I referred to the strict aliasing rule.
What are they talking about?

Source: (StackOverflow)

Which is faster: while(1) or while(2)?

This was an interview question asked by a senior manager.

Which is faster?

while(1) {
    // Some code


while(2) {
    //Some code

I said that both have the same execution speed, as the expression inside while should finally evaluate to true or false. In this case, both evaluate to true and there are no extra conditional instructions inside the while condition. So, both will have the same speed of execution and I prefer while (1).

But the interviewer said confidently: "Check your basics. while(1) is faster than while(2)." (He was not testing my confidence)

Is this true?

See also: Is "for(;;)" faster than "while (TRUE)"? If not, why do people use it?

Source: (StackOverflow)

Can code that is valid in both C and C++ produce different behavior when compiled in each language?

C and C++ have many differences, and not all valid C code is valid C++ code.
(By "valid" I mean standard code with defined behavior, i.e. not implementation-specific/undefined/etc.)

Is there any scenario in which a piece of code valid in both C and C++ would produce different behavior when compiled with a standard compiler in each language?

To make it a reasonable/useful comparison (I'm trying to learn something practically useful, not to try to find obvious loopholes in the question), let's assume:

  • Nothing preprocessor-related (which means no hacks with #ifdef __cplusplus, pragmas, etc.)
  • Anything implementation-defined is the same in both languages (e.g. numeric limits, etc.)
  • We're comparing reasonably recent versions of each standard (e.g. say, C++98 and C90 or later)
    If the versions matter, then please mention which versions of each produce different behavior.

Source: (StackOverflow)

Why is one loop so much slower than two loops?

Suppose a1, b1, c1, and d1 point to heap memory and my numerical code has the following core loop.

const int n=100000;

for(int j=0;j<n;j++){
    a1[j] += b1[j];
    c1[j] += d1[j];

This loop is executed 10,000 times via another outer for loop. To speed it up, I changed the code to:

for(int j=0;j<n;j++){
    a1[j] += b1[j];
for(int j=0;j<n;j++){
    c1[j] += d1[j];

Compiled on MS Visual C++ 10.0 with full optimization and SSE2 enabled for 32-bit on a Intel Core 2 Duo (x64), the first example takes 5.5 seconds and the double-loop example takes only 1.9 seconds. My question is: (Please refer to the my rephrased question at the bottom)

PS: I am not sure, if this helps:

Disassembly for the first loop basically looks like this (this block is repeated about five times in the full program):

movsd       xmm0,mmword ptr [edx+18h]
addsd       xmm0,mmword ptr [ecx+20h]
movsd       mmword ptr [ecx+20h],xmm0
movsd       xmm0,mmword ptr [esi+10h]
addsd       xmm0,mmword ptr [eax+30h]
movsd       mmword ptr [eax+30h],xmm0
movsd       xmm0,mmword ptr [edx+20h]
addsd       xmm0,mmword ptr [ecx+28h]
movsd       mmword ptr [ecx+28h],xmm0
movsd       xmm0,mmword ptr [esi+18h]
addsd       xmm0,mmword ptr [eax+38h]

Each loop of the double loop example produces this code (the following block is repeated about three times):

addsd       xmm0,mmword ptr [eax+28h]
movsd       mmword ptr [eax+28h],xmm0
movsd       xmm0,mmword ptr [ecx+20h]
addsd       xmm0,mmword ptr [eax+30h]
movsd       mmword ptr [eax+30h],xmm0
movsd       xmm0,mmword ptr [ecx+28h]
addsd       xmm0,mmword ptr [eax+38h]
movsd       mmword ptr [eax+38h],xmm0
movsd       xmm0,mmword ptr [ecx+30h]
addsd       xmm0,mmword ptr [eax+40h]
movsd       mmword ptr [eax+40h],xmm0

EDIT: The question turned out to be of no relevance, as the behavior severely depends on the sizes of the arrays (n) and the CPU cache. So if there is further interest, I rephrase the question:

Could you provide some solid insight into the details that lead to the different cache behaviors as illustrated by the five regions on the following graph?

It might also be interesting to point out the differences between CPU/cache architectures, by providing a similar graph for these CPUs.

PPS: The full code is at It uses TBB Tick_Count for higher resolution timing, which can be disabled by not defining the TBB_TIMING Macro.

(It shows FLOP/s for different values of n.)

enter image description here

Source: (StackOverflow)

Obfuscated C Code Contest 2006. Please explain sykes2.c

How does this C program work?


It compiles as it is (tested on gcc 4.6.3). It prints the time when compiled. On my system:

    !!  !!!!!!              !!  !!!!!!              !!  !!!!!! 
    !!  !!  !!              !!      !!              !!  !!  !! 
    !!  !!  !!              !!      !!              !!  !!  !! 
    !!  !!!!!!    !!        !!      !!    !!        !!  !!!!!! 
    !!      !!              !!      !!              !!  !!  !! 
    !!      !!              !!      !!              !!  !!  !! 
    !!  !!!!!!              !!      !!              !!  !!!!!!

Source: sykes2 - A clock in one line, sykes2 author hints

Some hints: No compile warnings per default. Compiled with -Wall, the following warnings are emitted:

sykes2.c:1:1: warning: return type defaults to ‘int’ [-Wreturn-type]
sykes2.c: In function ‘main’:
sykes2.c:1:14: warning: value computed is not used [-Wunused-value]
sykes2.c:1:1: warning: implicit declaration of function ‘putchar’ [-Wimplicit-function-declaration]
sykes2.c:1:1: warning: suggest parentheses around arithmetic in operand of ‘|’ [-Wparentheses]
sykes2.c:1:1: warning: suggest parentheses around arithmetic in operand of ‘|’ [-Wparentheses]
sykes2.c:1:1: warning: control reaches end of non-void function [-Wreturn-type]

Source: (StackOverflow)

Why doesn't GCC optimize a*a*a*a*a*a to (a*a*a)*(a*a*a)?

I am doing some numerical optimization on a scientific application. One thing I noticed is that GCC will optimize the call pow(a,2) by compiling it into a*a, but the call pow(a,6) is not optimized and will actually call the library function pow, which greatly slows down the performance. (In contrast, Intel C++ Compiler, executable icc, will eliminate the library call for pow(a,6).)

What I am curious about is that when I replaced pow(a,6) with a*a*a*a*a*a using GCC 4.5.1 and options "-O3 -lm -funroll-loops -msse4", it uses 5 mulsd instructions:

movapd  %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm14, %xmm13

while if I write (a*a*a)*(a*a*a), it will produce

movapd  %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm14, %xmm13
mulsd   %xmm13, %xmm13

which reduces the number of multiply instructions to 3. icc has similar behavior.

Why do compilers not recognize this optimization trick?

Source: (StackOverflow)

Why are these constructs (using ++) undefined behavior?

int main(int argc, char ** argv)
   int i = 0;
   i = i++ + ++i;
   printf("%d\n", i); // 3

   i = 1;
   i = (i++);
   printf("%d\n", i); // 2 Should be 1, no ?

   volatile int u = 0;
   u = u++ + ++u;
   printf("%d\n", u); // 1

   u = 1;
   u = (u++);
   printf("%d\n", u); // 2 Should also be one, no ?

   register int v = 0;
   v = v++ + ++v;
   printf("%d\n", v); // 3 (Should be the same as u ?)

Source: (StackOverflow)

With C arrays, why is it the case that a[5] == 5[a]?

As Joel points out in Stack Overflow podcast #34, in C Programming Language (aka: K & R), there is mention of this property of arrays in C: a[5] == 5[a]

Joel says that it's because of pointer arithmetic but I still don't understand. Why does a[5] == 5[a]?

Source: (StackOverflow)

What is the difference between const int*, const int * const, and int const *?

I always mess up how to use const int*, const int * const, and int const * correctly. Is there a set of rules defining what you can and cannot do?

I want to know all the do's and all don'ts in terms of assignments, passing to the functions, etc.

Source: (StackOverflow)

Speed comparison with Project Euler: C vs Python vs Erlang vs Haskell

I have taken Problem #12 from Project Euler as a programming exercise and to compare my (surely not optimal) implementations in C, Python, Erlang and Haskell. In order to get some higher execution times, I search for the first triangle number with more than 1000 divisors instead of 500 as stated in the original problem.

The result is the following:


lorenzo@enzo:~/erlang$ gcc -lm -o euler12.bin euler12.c
lorenzo@enzo:~/erlang$ time ./euler12.bin

real    0m11.074s
user    0m11.070s
sys 0m0.000s


lorenzo@enzo:~/erlang$ time ./ 

real    1m16.632s
user    1m16.370s
sys 0m0.250s

python with pypy:

lorenzo@enzo:~/Downloads/pypy-c-jit-43780-b590cf6de419-linux64/bin$ time ./pypy /home/lorenzo/erlang/ 

real    0m13.082s
user    0m13.050s
sys 0m0.020s


lorenzo@enzo:~/erlang$ erlc euler12.erl 
lorenzo@enzo:~/erlang$ time erl -s euler12 solve
Erlang R13B03 (erts-5.7.4) [source] [64-bit] [smp:4:4] [rq:4] [async-threads:0] [hipe] [kernel-poll:false]

Eshell V5.7.4  (abort with ^G)
1> 842161320

real    0m48.259s
user    0m48.070s
sys 0m0.020s


lorenzo@enzo:~/erlang$ ghc euler12.hs -o euler12.hsx
[1 of 1] Compiling Main             ( euler12.hs, euler12.o )
Linking euler12.hsx ...
lorenzo@enzo:~/erlang$ time ./euler12.hsx 

real    2m37.326s
user    2m37.240s
sys 0m0.080s


  • C: 100%
  • python: 692% (118% with pypy)
  • erlang: 436% (135% thanks to RichardC)
  • haskell: 1421%

I suppose that C has a big advantage as it uses long for the calculations and not arbitrary length integers as the other three. Also it doesn't need to load a runtime first (Do the others?).

Question 1: Do Erlang, Python and Haskell lose speed due to using arbitrary length integers or don't they as long as the values are less than MAXINT?

Question 2: Why is Haskell so slow? Is there a compiler flag that turns off the brakes or is it my implementation? (The latter is quite probable as Haskell is a book with seven seals to me.)

Question 3: Can you offer me some hints how to optimize these implementations without changing the way I determine the factors? Optimization in any way: nicer, faster, more "native" to the language.


Question 4: Do my functional implementations permit LCO (last call optimization, a.k.a tail recursion elimination) and hence avoid adding unnecessary frames onto the call stack?

I really tried to implement the same algorithm as similar as possible in the four languages, although I have to admit that my Haskell and Erlang knowledge is very limited.

Source codes used:

#include <stdio.h>
#include <math.h>

int factorCount (long n)
    double square = sqrt (n);
    int isquare = (int) square;
    int count = isquare == square ? -1 : 0;
    long candidate;
    for (candidate = 1; candidate <= isquare; candidate ++)
        if (0 == n % candidate) count += 2;
    return count;

int main ()
    long triangle = 1;
    int index = 1;
    while (factorCount (triangle) < 1001)
        index ++;
        triangle += index;
    printf ("%ld\n", triangle);

#! /usr/bin/env python3.2

import math

def factorCount (n):
    square = math.sqrt (n)
    isquare = int (square)
    count = -1 if isquare == square else 0
    for candidate in range (1, isquare + 1):
        if not n % candidate: count += 2
    return count

triangle = 1
index = 1
while factorCount (triangle) < 1001:
    index += 1
    triangle += index

print (triangle)

-module (euler12).
-compile (export_all).

factorCount (Number) -> factorCount (Number, math:sqrt (Number), 1, 0).

factorCount (_, Sqrt, Candidate, Count) when Candidate > Sqrt -> Count;

factorCount (_, Sqrt, Candidate, Count) when Candidate == Sqrt -> Count + 1;

factorCount (Number, Sqrt, Candidate, Count) ->
    case Number rem Candidate of
        0 -> factorCount (Number, Sqrt, Candidate + 1, Count + 2);
        _ -> factorCount (Number, Sqrt, Candidate + 1, Count)

nextTriangle (Index, Triangle) ->
    Count = factorCount (Triangle),
        Count > 1000 -> Triangle;
        true -> nextTriangle (Index + 1, Triangle + Index + 1)  

solve () ->
    io:format ("~p~n", [nextTriangle (1, 1) ] ),
    halt (0).

factorCount number = factorCount' number isquare 1 0 - (fromEnum $ square == fromIntegral isquare)
    where square = sqrt $ fromIntegral number
          isquare = floor square

factorCount' number sqrt candidate count
    | fromIntegral candidate > sqrt = count
    | number `mod` candidate == 0 = factorCount' number sqrt (candidate + 1) (count + 2)
    | otherwise = factorCount' number sqrt (candidate + 1) count

nextTriangle index triangle
    | factorCount triangle > 1000 = triangle
    | otherwise = nextTriangle (index + 1) (triangle + index + 1)

main = print $ nextTriangle 1 1

Source: (StackOverflow)