Question

uint8_t a = 0x1234;

Will a == 34? Is this because 0x34 is the first byte?

If so, how do I perform checks to ensure that the value I assign to a does not exceed the size of uint8_t?

Answer 1

jslagle@AR-E642-45QS3R1:~ $ cat test.c

#include <stdio.h>
#include <inttypes.h>

int main (int argc, char **argv) {
  uint8_t a = 0x1234;

  printf("%d %x\n",a,a);
  return 0;
}

---

jslagle@AR-E642-45QS3R1:~ $ gcc -o test test.c
test.c: In function ‘main’:
test.c:5:3: warning: large integer implicitly truncated to unsigned type [-Woverflow]

jslagle@AR-E642-45QS3R1:~ $ ./test
52 34

It does indeed behave like that. GCC gives you a warning if you do it.