Assigning an assignment in C++

Question

I'm reading a piece of code which does this

void *ptr_to_something = (void*)0x23; // I made this up, it's just a pointer to something void *addr; void *dst = (void*)(addr = ptr_to_something); // This is the line that confuses me

it seems to assign a pointer to something to another pointer of the same thing. And that's okay.. but then the result is enclosed in parenthesis, cast to the same thing and somehow reassigned to a third pointer to the same thing.

Is this valid C++ at all? Is it guaranteed that assigning the result of an assignment yields the same assigned object?

Neil · Accepted Answer · 2018-06-25 16:44:25Z

It is valid in C++ to do:

int a, b, c; a = b = c = 1;

So, therefore the result of an assignment is the value of that assignment.

What the cast is for, is a mystery*. Have you tried removing it, and does that generate a warning?

*not a mystery, just perhaps unnecessary.

It's already a void* so just unnecessary. Also note that the assignment operator is on the precidence list and has right-to-left associativity en.cppreference.com/w/cpp/language/operator_precedence

Xirema · Accepted Answer · 2018-06-25 16:50:31Z

This is valid C++. The choice to format the code this way is odd and not advised, but it's perfectly legal.

This code is equivalent to this (what I consider to be better written) code:

void *ptr_to_something = (void*)0x23; void *addr = ptr_to_something; void *dst = addr;

Oh that line was just made up and entirely crap. The address in the real code is calculated elsewhere. Thanks!

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