Lifetime of pointers value

Projects Programming
21

This is different:

void loop() {
  char * local_string = "Will this string survive until second call to print_buffer()?"; 
  print_buffer();
  buffer[1] = local_string;
}

Now you are not asking for a local array to be allocated but a local pointer. The compiler generates a much shorter loop:

void loop() {
  char * local_string = "Will this string survive until second call to print_buffer()?"; 
  print_buffer();
  ea:	0e 94 5f 00 	call	0xbe	; 0xbe <_Z12print_bufferv>
  buffer[1] = local_string;
  ee:	80 e0       	ldi	r24, 0x00	; 0
  f0:	91 e0       	ldi	r25, 0x01	; 1
  f2:	90 93 6f 01 	sts	0x016F, r25
  f6:	80 93 6e 01 	sts	0x016E, r24
}
  fa:	08 95       	ret

This is actually a bit dodgy because you have converted a const pointer to non-const. I think there is a warning about that. Ah yes, if you turn on verbose compiling:

sketch_sep20a.cpp: In function 'void loop()':
sketch_sep20a.cpp:17: warning: deprecated conversion from string constant to 'char*'