Reputation: 532
What happens in memory when we pass a value type - which has been stored on the stack - by reference?
A temp value/pointer must be created somewhere to change the origninal value when the method completes. Could someone please explain or point me to the answer - lots of stuff on memory but none seem to answer this. ty
Upvotes: 3
Views: 239
Reputation: 217313
If you have a method like this:
static void Increment(ref int value)
{
value = value + 1;
}
and call it like this:
int value = 5;
Increment(ref value);
then what happens is that, instead of the value 5 being pushed on the stack, the location of the variable value
is pushed on the stack. I.e. the contents of value
are changed directly by Increment
and not after the method completes.
Here's the IL of the method and the method call respectively:
.method private hidebysig static void Increment(int32& 'value') cil managed
{
.maxstack 8
L_0000: nop
L_0001: ldarg.0
L_0002: ldarg.0
L_0003: ldind.i4 // loads the value at the location of 'value'
L_0004: ldc.i4.1
L_0005: add
L_0006: stind.i4 // stores the result at the location of 'value'
L_0007: ret
}
.method private hidebysig static void Main() cil managed
{
.entrypoint
.maxstack 9
.locals init ([0] int32 value) // <-- only one variable declared
L_0000: nop
L_0001: ldc.i4.5
L_0002: stloc.0
L_0003: ldloca.s 'value' // call Increment with the location of 'value'
L_0005: call void Program::Increment(int32&)
L_000a: ret
}
Upvotes: 4
Reputation: 46376
It sounds like you're looking for some details on Boxing and Unboxing, which are the terms used to describe treating a value type as a reference type.
There's lots of articles out there that describe the process, I'll try to find some decent ones--but here's one for starters.
Upvotes: 0