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Why are global and static variables initialized to their default values?

开发者 https://www.devze.com 2022-12-16 16:50 出处:网络
In C/C++, why are globals and static variables initialized to default values? Why not leave it with 开发者_如何学Cjust garbage values? Are there any special

In C/C++, why are globals and static variables initialized to default values?

Why not leave it with 开发者_如何学Cjust garbage values? Are there any special reasons for this?


  1. Security: leaving memory alone would leak information from other processes or the kernel.

  2. Efficiency: the values are useless until initialized to something, and it's more efficient to zero them in a block with unrolled loops. The OS can even zero freelist pages when the system is otherwise idle, rather than when some client or user is waiting for the program to start.

  3. Reproducibility: leaving the values alone would make program behavior non-repeatable, making bugs really hard to find.

  4. Elegance: it's cleaner if programs can start from 0 without having to clutter the code with default initializers.

One might then wonder why the auto storage class does start as garbage. The answer is two-fold:

  1. It doesn't, in a sense. The very first stack frame page at each level (i.e., every new page added to the stack) does receive zero values. The "garbage", or "uninitialized" values that subsequent function instances at the same stack level see are really the previous values left by other method instances of your own program and its library.

  2. There might be a quadratic (or whatever) runtime performance penalty associated with initializing auto (function locals) to anything. A function might not use any or all of a large array, say, on any given call, and it could be invoked thousands or millions of times. The initialization of statics and globals, OTOH, only needs to happen once.


Because with the proper cooperation of the OS, 0 initializing statics and globals can be implemented with no runtime overhead.


Section 6.7.8 Initialization of C99 standard (n1256) answers this question:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then:

— if it has pointer type, it is initialized to a null pointer;

— if it has arithmetic type, it is initialized to (positive or unsigned) zero;

— if it is an aggregate, every member is initialized (recursively) according to these rules;

— if it is a union, the first named member is initialized (recursively) according to these rules.


Think about it, in the static realm you can't tell always for sure something is indeed initialized, or that main has started. There's also a static init and a dynamic init phase, the static one first right after the dynamic one where order matters.

If you didn't have zeroing out of statics then you would be completely unable to tell in this phase for sure if anything was initialized AT ALL and in short the C++ world would fly apart and basic things like singletons (or any sort of dynamic static init) would simple cease to work.

The answer with the bulletpoints is enthusiastic but a bit silly. Those could all apply to nonstatic allocation but that isn't done (well, sometimes but not usually).


In C, statically-allocated objects without an explicit initializer are initialized to zero (for arithmetic types) or a null pointer (for pointer types). Implementations of C typically represent zero values and null pointer values using a bit pattern consisting solely of zero-valued bits (though this is not required by the C standard). Hence, the bss section typically includes all uninitialized variables declared at file scope (i.e., outside of any function) as well as uninitialized local variables declared with the static keyword.

Source: Wikipedia

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