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When using PInvoke, I noticed that we need to use IntPtr to refer to Windows handles. I am wondering why not just use int for the handle? My understanding of a handle is that it is just an integer value.

Ani
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user705414
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4 Answers4

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A windows handle is defined as an integer of the native machine pointer size. That's so that they can secretly be a pointer if they need to be. (A handle probably is not a pointer, but it is permitted to be one if the operating system implementers deem it necessary. They typically are not actually pointers for security reasons; it makes it too easy for people to abuse the system if they are actually pointers.)

An int in C# defined as a 32 bit integer, which will be too small on a 64 bit machine. An IntPtr is defined as an integer that can hold a pointer of the machine size. That's why you always use IntPtr when interoperating with handles.

Eric Lippert
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Handles are pointer-sized values.

They're 8 bytes wide on 64-bit platforms.

SLaks
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    I'd say "pointer-*sized*" instead of "pointer" (even though I guess in the strict C/C++ sense, that isn't too accurate, since handles are `typedef`'d as a pointer all right, and `size_t` isn't necessarily the size of a pointer)... but if they're indices into a table, chances are they're used as integers. But yeah, point taken. – user541686 Feb 07 '12 at 00:26
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The size of a handle is dependant on the machine architecture (32/64 bit). IntPtr will take care of that.

Eugen Rieck
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A handle is essentially a system level pointer, and a pointer does not implicitly cast to an int. It is a type by itself. So that's why you have IntPtr in .NET to represent the Handle type.

Tony The Lion
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