A error source for ANSI C (ISO/IEC 9899:TC2; 6.3 Conversions) is the implicit type conversion mechanism. The following code fragment illustrates a bogus usage of the shift operation in realtion with the type of the operands:

void bit_mark(unsigned long *seq_bits, int32_t n)
  int32_t word_offset, word_num;

  word_num    = n / sizeof(unsigned long) * 8;
  word_offset = n % sizeof(unsigned long) * 8;

  seq_bits[word_num] |= 1 << word_offset;

The error source is the type of "1" which is implicit casted to an integer (Section: "If an int can represent all values of the original type the value is converted to an int; [..] These are called the integer promotions.All other types are unchanged by the integer promotions.). seq_bits on the other hand is a array of unsigned long types. On 32 bit architectures this may work, but on 64 architectures like x86_64 where int is 32 bit wide and unsigned long is 64 bit wide the upper 32 bit are left out! The solution is to define 1 as 1UL to promote this type as the general processing type. Another sidenote: character types are inplicit converted to a integer type, this includes both operand to the shift statement. The formal ANSI C definition can be found under: 6.5.7 Bitwise shift operators:

The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand. If the value of the right operand is negative or is greater than or equal to the wide of the promoted left operand, the behavior is undefined.