how is _Bool implemented?

What is it that makes value in a variable of type _Bool have the value 1, even when we assign a value greater than 1 to it?

For example:

#include <stdio.h>

int main(void) {
    _Bool tmp = 10;

    printf("%x, %lu\n", tmp, sizeof(tmp));
    return 0;
}

This would print 1, 1. I am trying to understand what makes a variable of size byte act as a single bit and when assigned a value greater than 1 which has its LSB 0 still get converted to 1.

What is it that makes value in a variable of type _Bool 1 , even when we assign a value greater than 1 to it. The compiler does.

For example on ARM (arm-none-eabi-gcc):

#include "stdio.h"
#include "stdbool.h"

int main()
{

_Bool tmp = 10;

printf("%x , %lu", tmp, sizeof(tmp));


return 0;

}

compiles to:

.LC0:
        .ascii  "%x , %lu\000"
main:
        stmfd   sp!, {fp, lr}
        add     fp, sp, #4
        sub     sp, sp, #8
        mov     r3, #1
        strb    r3, [fp, #-5]
        ldrb    r3, [fp, #-5]   @ zero_extendqisi2
        mov     r2, #1
        mov     r1, r3
        ldr     r0, .L3
        bl      printf
        mov     r3, #0
        mov     r0, r3
        sub     sp, fp, #4
        ldmfd   sp!, {fp, lr}
        bx      lr
.L3:
        .word   .LC0

you can see in the instruction mov r3, #1 that the compiler directly converts the initialisation value 10 to 1 as specified by the standard .