An int is 2 bytes but Serial.print with HEX or BIN formatting outputs 4 bytes:
int x = 0x9876;
Serial.println(x, HEX);
// output is FFFF9876
Why?
(and what is a good way to print out only 2 bytes)
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\$\begingroup\$ This question would be a better fit for Stack Overflow possibly with the "embedded" tag. \$\endgroup\$embedded.kyle– embedded.kyle2012-11-27 21:43:29 +00:00Commented Nov 27, 2012 at 21:43
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\$\begingroup\$ It's a fine line. See here: meta.electronics.stackexchange.com/questions/2457/… \$\endgroup\$embedded.kyle– embedded.kyle2012-11-29 13:12:29 +00:00Commented Nov 29, 2012 at 13:12
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2 Answers
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The Arduino print / println function casts the int to a long, which is 4 bytes long for Arduinos. See here: https://github.com/arduino/Arduino/blob/master/hardware/arduino/cores/arduino/Print.cpp
To have more control over printing check out the C++ sprintf function. For example,
int x = 0x9876;
char buf[9];
sprintf(buf, "%04x", x);
Serial.println(buf);
Will print it out correctly.
sprintf - http://www.cplusplus.com/reference/clibrary/cstdio/sprintf/
format string reference - http://www.cplusplus.com/reference/clibrary/cstdio/printf/
Some implementations of the Print class, include printf (Like Adafruit's, see here). In that case, you just do Serial.printf("%04X", x).
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\$\begingroup\$ Based on the source you linked, it seems that println( (unsigned) x ) will also work (casting to unsigned long rather than long). \$\endgroup\$greggo– greggo2013-03-23 17:56:12 +00:00Commented Mar 23, 2013 at 17:56
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\$\begingroup\$ Why did you allocated 9 bytes for
bufvariable whenxfits under 5 bytes? \$\endgroup\$Binar Web– Binar Web2020-01-21 16:30:17 +00:00Commented Jan 21, 2020 at 16:30
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This is because the data-type 'int' in 'C' language, as per ANSI C standards, is a 32-bit numeric type, and thus you see 4 bytes.
If your variable 'x' can only take 2 byte values, then declare it as a 'short' variable, instead of 'int'. This is because 'short' type as per the language standards is a 16-bit numeric type.
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1\$\begingroup\$ Arduino ints are 16 bits. According to answers here stackoverflow.com/questions/589575/size-of-int-long-etc the C standard says ints have to be able to represent from -32767 to 32767, which is possible in 16 bits. \$\endgroup\$geometrikal– geometrikal2012-11-24 05:32:02 +00:00Commented Nov 24, 2012 at 5:32
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\$\begingroup\$ Correct ints are 16, the problem as pointed out by geometrikal is that println converts all to longs (32) and println( int ) converts to long. println( unsigned ) will convert to unsigned long, and should do what the OP is expecting. \$\endgroup\$greggo– greggo2013-03-23 18:00:04 +00:00Commented Mar 23, 2013 at 18:00