Time in milliseconds in C

Question

Using the following code    include lt stdio h gt   include lt time h gt  int main         clock t start  stop      int i      start   clock        for i 0  i lt 2000 i                  printf   d    i 1   1 4              printf   n n        stop   clock            double  stop - start    CLOCKS PER SEC      printf   6 3f   start       printf   n n 6 3f   stop       return 0      I get the following output    56789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263264265266267268269270271272273274275276277278279280281282283284285286287288289290291292293294295296297298299300301302303304305306307308309310311312313314315316317318319320321322323324325326327328329330331332333334335336337338339340341342343344345346347348349350351352353354355356357358359360361362363364365366367368369370371372373374375376377378379380381382383384385386387388389390391392393394395396397398399400401402403404405406407408409410411412413414415416417418419420421422423424425426427428429430431432433434435436437438439440441442443444445446447448449450451452453454455456457458459460461462463464465466467468469470471472473474475476477478479480481482483484485486487488489490491492493494495496497498499500501502503504505506507508509510511512513514515516517518519520521522523524525526527528529530531532533534535536537538539540541542543544545546547548549550551552553554555556557558559560561562563564565566567568569570571572573574575576577578579580581582583584585586587588589590591592593594595596597598599600601602603604605606607608609610611612613614615616617618619620621622623624625626627628629630631632633634635636637638639640641642643644645646647648649650651652653654655656657658659660661662663664665666667668669670671672673674675676677678679680681682683684685686687688689690691692693694695696697698699700701702703704705706707708709710711712713714715716717718719720721722723724725726727728729730731732733734735736737738739740741742743744745746747748749750751752753754755756757758759760761762763764765766767768769770771772773774775776777778779780781782783784785786787788789790791792793794795796797798799800801802803804805806807808809810811812813814815816817818819820821822823824825826827828829830831832833834835836837838839840841842843844845846847848849850851852853854855856857858859860861862863864865866867868869870871872873874875876877878879880881882883884885886887888889890891892893894895896897898899900901902903904905906907908909910911912913914915916917918919920921922923924925926927928929930931932933934935936937938939940941942943944945946947948949950951952953954955956957958959960961962963964965966967968969970971972973974975976977978979980981982983984985986987988989990991992993994995996997998999100010011002100310041005100610071008100910101011101210131014101510161017101810191020102110221023102410251026102710281029103010311032103310341035103610371038103910401041104210431044104510461047104810491050105110521053105410551056105710581059106010611062106310641065106610671068106910701071107210731074107510761077107810791080108110821083108410851086108710881089109010911092109310941095109610971098109911001101110211031104110511061107110811091110111111121113111411151116111711181119112011211122112311241125112611271128112911301131113211331134113511361137113811391140114111421143114411451146114711481149115011511152115311541155115611571158115911601161116211631164116511661167116811691170117111721173117411751176117711781179118011811182118311841185118611871188118911901191119211931194119511961197119811991200120112021203120412051206120712081209121012111212121312141215121612171218121912201221122212231224122512261227122812291230123112321233123412351236123712381239124012411242124312441245124612471248124912501251125212531254125512561257125812591260126112621263126412651266126712681269127012711272127312741275127612771278127912801281128212831284128512861287128812891290129112921293129412951296129712981299130013011302130313041305130613071308130913101311131213131314131513161317131813191320132113221323132413251326132713281329133013311332133313341335133613371338133913401341134213431344134513461347134813491350135113521353135413551356135713581359136013611362136313641365136613671368136913701371137213731374137513761377137813791380138113821383138413851386138713881389139013911392139313941395139613971398139914001401140214031404140514061407140814091410141114121413141414151416141714181419142014211422142314241425142614271428142914301431143214331434143514361437143814391440144114421443144414451446144714481449145014511452145314541455145614571458145914601461146214631464146514661467146814691470147114721473147414751476147714781479148014811482148314841485148614871488148914901491149214931494149514961497149814991500150115021503150415051506150715081509151015111512151315141515151615171518151915201521152215231524152515261527152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   2 169    2 169    Start and stop times are the same  Does it mean that the program hardly takes time to complete execution  If 1  is false  then atleast the no of digits beyond the     should differ  which does not happen here  Is my logic correct    Note  I need to calculate the time taken for execution  and hence the above code

User · Accepted Answer

Yes  this program has likely used less than a millsecond  Try using microsecond resolution with timeval  e g   include  lt sys time h gt   struct timeval stop  start  gettimeofday  amp start  NULL     do stuff gettimeofday  amp stop  NULL   printf  quot took  lu us n quot    stop tv sec - start tv sec    1000000   stop tv usec - start tv usec     You can then query the difference  in microseconds  between stop tv usec - start tv usec  Note that this will only work for subsecond times  as tv usec will loop   For the general case use a combination of tv sec and tv usec  Edit 2016-08-19 A more appropriate approach on system with clock gettime support would be  struct timespec start  end  clock gettime CLOCK MONOTONIC RAW   amp start     do stuff clock gettime CLOCK MONOTONIC RAW   amp end    uint64 t delta us    end tv sec - start tv sec    1000000    end tv nsec - start tv nsec    1000

User · Answer

The standard C library provides timespec get  It can tell time up to nanosecond precision  if the system supports  Calling it  however  takes a bit more effort because it involves a struct  Here s a function that just converts the struct to a simple 64-bit integer so you can get time in milliseconds   include  lt stdio h gt   include  lt inttypes h gt   include  lt time h gt   int64 t millis         struct timespec now      timespec get  amp now  TIME UTC       return   int64 t  now tv sec    1000     int64 t  now tv nsec    1000000     int main void        printf  quot Unix timestamp with millisecond precision    quot  PRId64  quot  n quot   millis        Unlike clock  this function returns a Unix timestamp so it will correctly account for the time spent in blocking functions  such as sleep

User · Answer

This code snippet can be used for displaying time in seconds milliseconds and microseconds     include  lt sys time h gt   struct timeval start  stop  double secs   0   gettimeofday  amp start  NULL       Do stuff  here  gettimeofday  amp stop  NULL   secs    double  stop tv usec - start tv usec    1000000    double  stop tv sec - start tv sec   printf  time taken  f n  secs

User · Answer

Here is what I write to get the timestamp in millionseconds    include lt sys time h gt   long long timeInMilliseconds void        struct timeval tv       gettimeofday  amp tv NULL       return    long long tv tv sec  1000   tv tv usec 1000

User · Answer

A couple of things might affect the results you re seeing    You re treating clock t as a floating-point type  I don t think it is  You might be expecting  1 4  to do something else than compute the bitwise XOR of 1 and 4   i e  it s 5  Since the XOR is of constants  it s probably folded by the compiler  meaning it doesn t add a lot of work at runtime  Since the output is buffered  it s just formatting the string and writing it to memory   it completes very quickly indeed    You re not specifying how fast your machine is  but it s not unreasonable for this to run very quickly on modern hardware  no   If you have it  try adding a call to sleep   between the start stop snapshots  Note that sleep   is POSIX though  not standard C

User · Answer

You can use gettimeofday   together with the timedifference msec   function below to calculate the number of milliseconds elapsed between two samples    include  lt sys time h gt   include  lt stdio h gt   float timedifference msec struct timeval t0  struct timeval t1        return  t1 tv sec - t0 tv sec    1000 0f    t1 tv usec - t0 tv usec    1000 0f     int main void       struct timeval t0     struct timeval t1     float elapsed      gettimeofday  amp t0  0             YOUR CODE HERE           gettimeofday  amp t1  0       elapsed   timedifference msec t0  t1       printf  Code executed in  f milliseconds  n   elapsed       return 0      Note that  when using gettimeofday    you need to take seconds into account even if you only care about microsecond differences because tv usec will wrap back to zero every second and you have no way of knowing beforehand at which point within a second each sample is obtained

User · Answer

From man clock      The clock   function returns an approximation of processor time used by the program    So there is no indication you should treat it as milliseconds  Some standards require precise value of CLOCKS PER SEC  so you could rely on it  but I don t think it is advisable  Second thing is that  as  unwind stated  it is not float double  Man times suggests that will be an int  Also note that      this function will return the same value approximately every 72 minutes   And if you are unlucky you might hit the moment it is just about to start counting from zero  thus getting negative or huge value  depending on whether you store the result as signed or unsigned value    This   printf   n n 6 3f   stop     Will most probably print garbage as treating any int as float is really not defined behaviour  and I think this is where most of your problem comes   If you want to make sure you can always do   printf   n n 6 3f    double  stop     Though I would rather go for printing it as long long int at first   printf   n n lldf    long long int  stop

User · Answer

Modern processors are too fast to register the running time  Hence it may return zero  In this case  the time you started and ended is too small and therefore both the times are the same after round of

[c] Time in milliseconds in C

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