Assigning variable values "on the fly" within expressions

[Swamp Thing]

TL;DR

This works in JavaScript : ((a=3) +( b=2) )**2 + 1/a +1/b

Is there a term for it? Does it work in may other languages?

I've used this kind of thing often in JavaScript and Mathematica.

z = ((a=3) +( b=2) )**2 + 1/a +1/b

I'm currently learning Julia, and it doesn't seem to be supported.

What is this syntax called?

 

[Nugatory]

“Assignment expression” is one term sometimes used.

I won’t enter into the debate about whether using assignment expressions is a good idea (basically, whether the unnecessary and irrelevant reduction in line count justifies the gratuitous obscuration of the code, interference with the compiler’s ability to optimize, and potential for introducing subtle errors) but I will say that if you’re going to use them…. Be very sure that you understand what your language specification says about order of evaluation.

 

[jtbell]

It works in C++, too.

#include <iostream>
#include <iomanip>

using namespace std;

int main ()
{
    int a, b;
    int c = ((a = 3) + (b = 2)) * 2;  
    cout << a << ' ' << b << ' ' << c << endl;
    return 0;
}

produces the output

3 2 10

 

[pbuk]

It works in C++, too.

But note that the more complicated expression in the OP (((a = 3) + (b = 2)) ** 2 + 1/a + 1/b) will compile but there is no guarantee that the compiler will set a to 3 before it calculates 1/a (the order of these operations is not defined in the standard).

So as somebody not quite said upthread, even if this works it is never a good idea.

 

[pbuk]

Be very sure that you understand what your language specification says about order of evaluation.

Please forgive me for paraphrasing that liberally!

 

[rcgldr]

In APL, variable assignment can be done according to the order of evaluation. In this example, the right Z is declared, but uninitialized, where initialization occurs within the parenthesis. The expression is 2 modulo (Z$\leftarrow$ 1 2 3 4 5) reduce Z (delete all values where Z modulo 2 are zero).

$$ 2|(Z \leftarrow 1 \ 2 \ 3 \ 4 \ 5)/Z $$

returns 1 3 5