How do we convert 8 byte char array into long since << does not work for type long?
#define word_size 8
long num = 0;
char a[word_size] = "\x88\x99\xaa\x0bb\xcc\xdd\xee\xff";
for (i=0; i < word_size;i++) {
a[(word_size-1) - i] |= (num << (8*(word_size - i - 1))) & 0xFF;
}
printf("%lx\n", num);
The following code is more efficient:
unsigned char[word_size] = ...;
int64_t num = 0;
for ( int i = 0 ; i < sizeof(a) ; i++ )
num = (num << 8) | a[i];
This assumes big endian (highest order byte first) ordering of the bytes in the array. For little endian (as you appear to use) just process it top-down:
for ( int i = sizeof(a) ; --i >= 0 ; )
Note: whether char is signed or unsigned is implementation-dependent, so nail it down to be unsigned, otherwise the logical-or will not work. Better use uint8_t; that is defined to be 8 bits, while char is not.
Note: You should use all-uppercase for constants: WORD_SIZE instead of word_size. That is a commonly accepted standard (quite the only about case for identifiers in C).
unsigned types only for shifting. Also note that long long is not required to have 64 bits and a byte/char (see text) does not need to have 8 bits either. Using the correct types makes your code more self-documenting. Even if that is no production code, you might value that in some years.
<<works fine withlong. In this case,numis always 0, and so you're clearingato all 0s.longis not necessarily 64 bits. If you want to make sure, useint64_tfromstdint.h. That is the proposed way. Also, make sure the array has the proper byte-order.