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How to measure inserting time in seconds?

I tried to use:

struct timeval t1,t2;

I checked time before inserting input: 

gettimeofday(&t1,NULL);

and the same after getting the input:

gettimeofday(&t2,NULL);
double elapsedTime=(t2.tv_sec - t1.tv_sec)*10000.0;

but It's not accurate at all !!

I need better way to measure seconds in Inserting time , and to know the differences in seconds in inserting each character. 

// trying to insert 5 chars
for(i=0; i<=4 ; i++)
{        
    gettimeofday(&t1,NULL);    //get time before getting char
    c=getchar();
    gettimeofday(&t2,NULL);    //get time after geting char             
    elapsedTime=(t1.tv_sec - t2.tv_sec)*10000.0;
    printf("\n char number %d his elapsed time =%d\n",i,elapsedTime);
}

I need to know the seconds rates, on inserting "clicking" chars as input , and to calculate elapsedTime in seconds :

output should be like: 

time between inserting first character and the second is : 0.002 seconds
time between..........second character and the third is:  1.008 seconds
jww
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reham.abass
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  • Your subtraction in the loop code snippet is back-to-front, because `t2` is later than `t1` and therefore bigger. – Weather Vane Jun 21 '19 at 18:43
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    "How to measure inserting time in seconds?" Don't multiply by `10000`, and consider the other `struct` member `suseconds_t tv_usec; /* microseconds */` as well. – Weather Vane Jun 21 '19 at 18:49
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    The keyboard input is buffered. You only receive the characters ***after*** you press the **Enter** key. (Note: buffering the input allows the user to edit the input before pressing Enter.) – user3386109 Jun 21 '19 at 19:01
  • @WeatherVane good idea , thank you , I tried it I began to get different results, but I got also negative numbers , how it could be negative useconds ?, (still the t2 after the t1 and I think it counts also the commands in the loop ) – reham.abass Jun 21 '19 at 19:10
  • You need to put the terminal into non-canonical mode. [Here's some code](https://stackoverflow.com/a/29723614/3386109) that works on MacOS. It should work on Linux, too. – user3386109 Jun 21 '19 at 19:16
  • @user3386109 I think the same , waiting until "Enter" being pressed is a problem for me . how can I overcome the Enter and really count useconds it really takes to user pressing each character (thinking and pressing character) – reham.abass Jun 21 '19 at 19:16

2 Answers2

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elapsedTime=(t1.tv_sec - t2.tv_sec)*10000.0;

You're only taking tv_sec into account. the actual timeval struct has a tv_sec and also a tv_usec part which both of them are integers that don't hold fractions (the exact type is not guaranteed though)

tv_sec holds the seconds and tv_usec holds microseconds.

it's also guaranteed that tv_usec is always less than 1000000 that means you only need to calculate the difference of them individually.

and you're also doing t1-t2 that you should change it to t2-t1 as t2 is the latest timeval. this is why you're getting negative times.

elapsedTime=((double) (t2.tv_sec - t1.tv_sec))+((double) (t2.tv_usec - t1.tv_usec) / 1000000.0);

which returns time in : "sec.usec" format that should be precise enough :)

note that you need to declare elapsedTime as double instead of int, and also replace the second "$d" in the printf with "%f".

you also need to consider that keyboard input is buffered which means you're blocked in getchar() until Enter is pressed, then the buffer is fed to getchar, one char at a call. you cad use just Enter as the input character to test the accuracy of the code, but to use it with actual chars, you need to use the unbuffered input.

GNU docs: 21.2 Elapsed Time

Solid State
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On my system using clock() counts in millisconds, maybe that's accurate enough for your purposes:

#include "time.h"

clock_t c1=clock();
... doing stuff ...
clock_t c2=clock();
printf("diff %i clocks with CLOCKS_PER_SEC = %i\n",c2-c1,CLOCKS_PER_SEC);