bool fp[81];
From my understanding fp should use ceil(81/8) bytes because it is in succession.
Am I correct?
How can I prove this?
No, the sizeof your buffer is implementation defined, as per this quote from the Standard:
$5.3.3/1 - "The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is not evaluated, or a parenthesized type-id. The sizeof operator shall not be applied to an expression that has function or incomplete type, or to an enumeration type before all its enumerators have been declared, or to the parenthesized name of such types, or to an lvalue that designates a bit-field. sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1; the result of sizeof applied to any other fundamental type (3.9.1) is implementation-defined. [Note: in particular, sizeof(bool) and sizeof(wchar_t) are implementation-defined.69) ] [Note: See 1.7 for the definition of byte and 3.9 for the definition of object representation.]
Therefore the size you can expect is 81 * X, where X is the size of bool, which is implementation defined.