Wrong conversion of time_t to char












1















I have a simple sample of code that i want to convent a time_t to char array in order to concatenate it after, in an other array.
Code:



time_t unixt;

char exm [40];





unixt = RTC.get();

Serial.println(unixt);

dtostrf(unixt, 10, 0, exm);

Serial.println(exm);


The problem that i face is that, the console is printing different the second one every time like this: 



1548258220

1548258200


I tried also with unsigned long with no result either. Can someone point me what's wrong or an other way of doing this?






arduino-mega







share|improve this question





























    1















    I have a simple sample of code that i want to convent a time_t to char array in order to concatenate it after, in an other array.
    Code:



    time_t unixt;
    
char exm [40];
    

    

    
unixt = RTC.get();
    
Serial.println(unixt);
    
dtostrf(unixt, 10, 0, exm);
    
Serial.println(exm);


    The problem that i face is that, the console is printing different the second one every time like this: 



    1548258220
    
1548258200


    I tried also with unsigned long with no result either. Can someone point me what's wrong or an other way of doing this?






    arduino-mega







    share|improve this question



























      1












      1








      1








      I have a simple sample of code that i want to convent a time_t to char array in order to concatenate it after, in an other array.
      Code:



      time_t unixt;
      
char exm [40];
      

      

      
unixt = RTC.get();
      
Serial.println(unixt);
      
dtostrf(unixt, 10, 0, exm);
      
Serial.println(exm);


      The problem that i face is that, the console is printing different the second one every time like this: 



      1548258220
      
1548258200


      I tried also with unsigned long with no result either. Can someone point me what's wrong or an other way of doing this?






      arduino-mega







      share|improve this question
















      I have a simple sample of code that i want to convent a time_t to char array in order to concatenate it after, in an other array.
      Code:



      time_t unixt;
      
char exm [40];
      

      

      
unixt = RTC.get();
      
Serial.println(unixt);
      
dtostrf(unixt, 10, 0, exm);
      
Serial.println(exm);


      The problem that i face is that, the console is printing different the second one every time like this: 



      1548258220
      
1548258200


      I tried also with unsigned long with no result either. Can someone point me what's wrong or an other way of doing this?






      arduino-mega




      arduino-mega






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      ratchet freak

      2,7151511




      2,7151511










      asked 1 hour ago









      kakaviakakavia

      2015




      2015






















          1 Answer
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          Use ultoa() instead of dtostrf(). The former is meant to serialize
          unsigned long integers, while the later is for floating point numbers.
          By using dtostrf(), you are implicitly converting unixt to a
          floating point number. The Arduino Mega only supports single precision
          floating point, so this conversion incurs some loss of precision, hence
          the result you see.






          share|improve this answer
























          • Thanks a lot! That was exactly what i needed.

            – kakavia
            49 mins ago











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          4














          Use ultoa() instead of dtostrf(). The former is meant to serialize
          unsigned long integers, while the later is for floating point numbers.
          By using dtostrf(), you are implicitly converting unixt to a
          floating point number. The Arduino Mega only supports single precision
          floating point, so this conversion incurs some loss of precision, hence
          the result you see.






          share|improve this answer
























          • Thanks a lot! That was exactly what i needed.

            – kakavia
            49 mins ago
















          4














          Use ultoa() instead of dtostrf(). The former is meant to serialize
          unsigned long integers, while the later is for floating point numbers.
          By using dtostrf(), you are implicitly converting unixt to a
          floating point number. The Arduino Mega only supports single precision
          floating point, so this conversion incurs some loss of precision, hence
          the result you see.






          share|improve this answer
























          • Thanks a lot! That was exactly what i needed.

            – kakavia
            49 mins ago














          4












          4








          4







          Use ultoa() instead of dtostrf(). The former is meant to serialize
          unsigned long integers, while the later is for floating point numbers.
          By using dtostrf(), you are implicitly converting unixt to a
          floating point number. The Arduino Mega only supports single precision
          floating point, so this conversion incurs some loss of precision, hence
          the result you see.






          share|improve this answer













          Use ultoa() instead of dtostrf(). The former is meant to serialize
          unsigned long integers, while the later is for floating point numbers.
          By using dtostrf(), you are implicitly converting unixt to a
          floating point number. The Arduino Mega only supports single precision
          floating point, so this conversion incurs some loss of precision, hence
          the result you see.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          Edgar BonetEdgar Bonet

          24.3k22345




          24.3k22345













          • Thanks a lot! That was exactly what i needed.

            – kakavia
            49 mins ago



















          • Thanks a lot! That was exactly what i needed.

            – kakavia
            49 mins ago

















          Thanks a lot! That was exactly what i needed.

          – kakavia
          49 mins ago





          Thanks a lot! That was exactly what i needed.

          – kakavia
          49 mins ago


















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