i was trying to test some timers in contikiOS, i used printf() to output the value of a variable but the printed value changed acording to how i printed the message.
What do i mean?, well i print two variables, RTIMER_SECOND is a long unsigned int and CLOCK_SECOND is an integer. The variations bellow are what i tried and the output is bellow that code:
printf("1->RTIMER_SECOND=%lu CLOCK_SECOND=%d\n", RTIMER_SECOND, CLOCK_SECOND);
printf("2->RTIMER_SECOND=%d CLOCK_SECOND=%d\n\n", RTIMER_SECOND, CLOCK_SECOND);
printf("3->CLOCK_SECOND=%d RTIMER_SECOND=%lu\n", CLOCK_SECOND,RTIMER_SECOND);
printf("4->CLOCK_SECOND=%d RTIMER_SECOND=%d\n\n", CLOCK_SECOND,RTIMER_SECOND);
printf("Ticks per second: %lu\n", RTIMER_SECOND);
printf("second per second: %d\n", CLOCK_SECOND);
The output is this:
1->RTIMER_SECOND=15625 CLOCK_SECOND=128
2->RTIMER_SECOND=15625 CLOCK_SECOND=0
3->CLOCK_SECOND=128 RTIMER_SECOND=15625
4->CLOCK_SECOND=128 RTIMER_SECOND=15625
Ticks per second: 15625
second per second: 128
My question is this why does the 0 appear in the 2nd print? is the undefined behaviour when printing the %d instead of %lu of the RTIMER_SECOND affecting the next variable? PS: the 4th is the reverse of the 2nd and the value of the CLOCK_SECOND is not 0 there.
printf() will be implemented by the OS. I don't know the details of the implementation.
My guess is that the variables get passed to printf() as a sequence of bytes on the stack. The specifier string tells the function how many bytes to consume when printing the output.
A long unsigned integer is 4 bytes, probably in little-endian order. Since 15625 is less than 2^16, then last two bytes of RTIMER_SECOND are both 0. When the string tells the function to only printf 2 bytes for RTIMER_SECOND, the next two 0 bytes are interpreted as the two bytes of CLOCK_SECOND.
To test this hypothesis, try passing in a value for RTIMER_SECOND that takes more than 2 bytes, i.e., more than 65536. In that case, the 4th line will be messed up, too.