How Bit Endianness Affects Bitwise Shifts and File IO in C

Let L and B be two machines. L order its bits from LSB
(Least Significant Bit) to MSB (Most Significant Bit) while B order
from MSB to LSB. Or, in other words, L uses Little Endian while
B uses Big Endian bit – not to be confused with byte – ordering.

Problem 1 SOLVED:

We are writing the following code which we want to be portable:

#include <stdio.h>

int main()
{
    unsigned char a = 1;
    a <<= 1;

    printf("a = %d\n", (int) a);

    return 0;
}

on L, it will print 2, but what happens on B? Will it shift the
1 out and print 0?.

SOLUTION: The C99 definition at 6.5.7 says it that, at least on
unsigned integer types, << and >> will multiply and divide by 2
respectively.

Problem 2:

We are writing the following code which we want to be portable:

READ program:

/* program READ */
#include <stdio.h>

int main()
{
    FILE* fp;
    unsigned char a;

    fp = fopen("data.dat", "rb");
    fread(&a, 1, 1, fp);
    fclose(fp);

    return 0;
}

and WRITE program:

/* program WRITE */
#include <stdio.h>

int main()
{
    FILE* fp;
    unsigned char a = 1;

    fp = fopen("data.dat", "wb");
    fwrite(&a, 1, 1, fp);
    fclose(fp);

    return 0;
}

what happens if we run WRITE on L, move the data file to B and
run READ there? And if we run WRITE on B and then READ on L?

Sorry if this is a FAQ. I googled for hours without luck.