Is there any cost to convert from pointer to ref in the following type of code?
void bar(Obj&);
void foo(Obj* o) {
bar(*o);
}
On one hand, the value of the pointer is the address of Obj, so the compiler could just re-interpret the value as a reference. However, if o is a null pointer, then that deref would crash so, presumably, there is a memory read there. The compiler doesn't have context on whether the pointer is valid or not.
I understand that in some cases, the compiler could elide reads for conversions, when it has enough context near the conversion.
I found this other question, but that answer is for a different scenario: Cost of converting a pointer to a vector into a reference
In practice, there is no cost to this operation. To visualize this, you could compile the following code with optimizations enabled:
struct Obj;
void bar(Obj&);
void foo(Obj* o) {
bar(*o);
}
GCC compiles this to:
foo(Obj*):
jmp bar(Obj&)
At the ABI level, pointers and references are both passed as the memory address of the object they are referring to, so they're interchangeable without any operation.
However, if
oisnullptr, then that deref would crash so, presumably, there is a memory read there.
No, that doesn't necessarily happen.
If o is a null pointer, dereferencing it is undefined behavior.
That doesn't mean that the program has to crash; it just means that anything can happen in that scenario.
In this case, it results in a reference being bound to something that's not an actual object, which is meant to be impossible. This will probably break some code further down the line.
I understand that, in some cases, the compiler could elide reads for conversions, when it has enough context near the conversion.
There is neither a read nor a conversion in this code, at least not in the C++ standard sense. Dereferencing the pointer does not read from the pointed-to object, and binding references is not considered a conversion.