Serial.print uint64_t in Arduino

Question

I use Arduino UNO (Arduino 1.8.3).

This is my code:

void setup() { 
    Serial.begin(115200);
}
void loop() { 
    uint64_t num = 9223372036854775807; 
    Serial.print(num); 
    delay(1000);
}

This is error message:

exit status 1

call of overloaded 'print(uint64_t&)' is ambiguous

How do I solve this error?

Please edit your question to include the *full* and *complete* error output. Including possible informational notes. Preferably just copy it all and paste it into your question (as text), without any modifications. And please take some time to [read about how to ask good questions](http://stackoverflow.com/help/how-to-ask). — Some programmer dude, Aug 31 '17 at 06:22
It's a helpful question, at least in its current form. It's exactly what I was looking for. — Robert M., Mar 30 '23 at 19:19

score 3 · Answer 1 · answered Sep 06 '17 at 17:40

This is how I would do it. It's messy but it gets the job done.

Output from serial console: 9223372036854775807

void print_uint64_t(uint64_t num) {

  char rev[128]; 
  char *p = rev+1;

  while (num > 0) {
    *p++ = '0' + ( num % 10);
    num/= 10;
  }
  p--;
  /*Print the number which is now in reverse*/
  while (p > rev) {
    Serial.print(*p--);
  }
}

Something Something · Answer 2 · 2023-03-30T22:26:52.883

1

A bit better than printing char by char as suggested somewhere else is to actually build the buffer in reverse and then print it at once.

#include <stdint.h>
char* str( uint64_t num ) {
  static char buf[22];
  char* p = &buf[sizeof(buf)-1];
  *p = '\0';
  do {
    *--p = '0' + (num%10);
    num /= 10;
  } while ( num > 0 );
  return p;
}

char* str( int64_t num ) {
    if ( num>=0 ) return str((uint64_t)num);
    char* p = str((uint64_t)(-num));
    *--p = '-';
    return p;    
}

void setup() {
  int64_t num = -92233720368547758L; 
  Serial.print( str(num) );
}

edited Mar 30 '23 at 22:26

answered Feb 15 '23 at 04:04

Something Something

3,999
1
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@MooingDuck the question would have been closed anyway. Gotta love SO. So toxic. – Something Something Feb 16 '23 at 23:45
1

@njuffa That is an excellent solution! Bad thing they were about to delete the question – Something Something Feb 17 '23 at 04:14
1

I wonder how much awesome content like this is suppressed by these jealous admins of StackOverflow. The level of toxicity on the C++ tag in this website is impressive. – Something Something Feb 17 '23 at 05:53
I agree. Why is that? The C++ forum on stackoverflow is quite pleasant and helpful. – Robert M. Mar 30 '23 at 19:16
1

I wonder if you have some idea how to do this with a signed integer, int64_t? – Robert M. Mar 30 '23 at 19:38
1

Yes that is simple. Will update the answer when I’m back in the office – Something Something Mar 30 '23 at 22:03
1

This is nice, thank you. I don't know why I was testing the leftmost bit instead of just testing for >=0 lol. – Robert M. Mar 30 '23 at 23:35

score 1 · Answer 3 · answered Mar 30 '23 at 21:04

1

If it's helpful to anyone, I modified this a bit to work with an int64_t (I changed the name of the function, sorry).

char* getInt64Str( int64_t num ) {
  int64_t numAbs = abs(num);
  static char buf[22];
  char* p = &buf[sizeof(buf)-1];
  *p = '\0';
  do {
    *--p = '0' + (numAbs%10);
    numAbs /= 10;
  } while ( numAbs > 0 );
  if( (byte)((num & bit(64))>>63) == 1 ){
    *--p = '-';
  }
  return p;
}

answered Mar 30 '23 at 21:04

Robert M.

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5
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I have added the `int64_t` implementation on the question above. No need to modify the original. – Something Something Mar 30 '23 at 22:27

Serial.print uint64_t in Arduino

3 Answers3