Assume the following code in C:

long int a = 262143;

which in binary will be 111111111111111111 (18 bits). If an atmega328p register can hold 8 bits, how is the above represented in the register?

Assume the following code in C:

long int a = 262143;

which in binary will be 111111111111111111 (18 bits). If an atmega328p register can hold 8 bits, how is the above represented in the register?

Compiled Assembly:

    .file   "a.c"
__SP_H__ = 0x3e
__SP_L__ = 0x3d
__SREG__ = 0x3f
__tmp_reg__ = 0
__zero_reg__ = 1
    .text
.global main
    .type   main, @function
main:
    push r28
    push r29
    rcall .
    rcall .
    in r28,__SP_L__
    in r29,__SP_H__
/* prologue: function */
/* frame size = 4 */
/* stack size = 6 */
.L__stack_usage = 6
    ldi r24,lo8(-1)
    ldi r25,lo8(-1)
    ldi r26,lo8(3)
    ldi r27,0
    std Y+1,r24
    std Y+2,r25
    std Y+3,r26
    std Y+4,r27
    ldi r24,0
    ldi r25,0
/* epilogue start */
    pop __tmp_reg__
    pop __tmp_reg__
    pop __tmp_reg__
    pop __tmp_reg__
    pop r29
    pop r28
    ret
    .size   main, .-main
    .ident  "GCC: (GNU) 8.2.0"