◆ memadd_BE_inplace()

bool Chusov::Memory::memadd_BE_inplace	(	void *	pResult,
		const void *	pAddend,
		std::size_t	cb
	)

noexcept

Replaces an integer given by a vector of bytes in the Big-Endian order with the result of its addition with another integer.

Parameters

[in,out]	pResult	A pointer to a vector holding an integer to increase by the second integer pointed to by `pAddend`.
[in]	pAddend	A pointer to a byte vector with the second integer. Although the pointer is qualified with `const`, `pAddend` may address the same integer as `pResult`. Other than that, the pointers may not overlap.
	cb	A byte size of the input integers as well as the result written to `pResult` on output.

Returns: A value of a carry bit which results from the addition.

The addition is carried out modulo b^cb, where b = 2^CHAR_BIT is a number of values that a byte can represent.

All of the integers are specified with vectors in the Big-Endian byte order. That is, if n = cb - 1, and the first and the second integers are respectively specified with vectors \(V_X = \{x_0,\dots, x_n\}\) and \(V_Y = \{y_0,\dots, y_n\}\), where \(x_0\) and \(y_0\) are the first bytes of the vectors, then the respective integers are defined as \(X=\sum_{i=0}^{n}x_{n-i}\cdot b^i\) and \(Y=\sum_{i=0}^{n}y_{n-i}\cdot b^i\). The result of the addition is also written to pResult in the Big-Endian byte order.

To add up numbers given in the Little-Endian byte order use memadd_LE_inplace.

To right the sum into a separate buffer leaving the original operands unmodified, use memadd_BE_copy or memadd_BE.

See also: memadd_LE_inplace;
memadd_BE_copy;
memadd_BE;
memincr_BE;
memsub_BE;
memmul_BE;
memfmadd_BE;
memdiv_BE.