\(\newcommand{\W}[1]{ \; #1 \; }\) \(\newcommand{\R}[1]{ {\rm #1} }\) \(\newcommand{\B}[1]{ {\bf #1} }\) \(\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }\) \(\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }\) \(\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }\) \(\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }\)
compound_assign¶
View page sourceAD Compound Assignment Operators¶
Syntax¶
Purpose¶
Performs compound assignment operations
where either x has type
AD < Base > .
Op¶
The operator Op is one of the following
Op |
Meaning |
|
x is assigned x plus y |
|
x is assigned x minus y |
|
x is assigned x times y |
|
x is assigned x divided by y |
Base¶
The type Base is determined by the operand x .
x¶
The operand x has the following prototype
AD< Base > & x
y¶
The operand y has the following prototype
constType & y
where Type is
VecAD < Base >:: reference ,
AD < Base > ,
Base , or
double .
Result¶
The result of this assignment can be used as a reference to x . For example, if z has the following type
AD< Base > z
then the syntax
z = x += y
will compute x plus y and then assign this value to both x and z .
Operation Sequence¶
This is an atomic_base AD of Base operation and hence it is part of the current AD of Base operation sequence .
Example¶
The following files contain examples and tests of these functions. Each test returns true if it succeeds and false otherwise.
Derivative¶
If \(f\) and \(g\) are Base functions