运行时操作员用C [英] Run time operators in C

问题描述

什么是编译时的定义和使用C运行时操作人员？我知道的sizeof（）是在C编译时运营商，但它是在C运行时操作？

What is the definition of compile time and run time operator in C ? I know sizeof() is a compile time operator in C but which are the run time operators in C ?

推荐答案

对于C;各种优化技术（例如常量传送和常量折叠）意味着每个运营商有可能在编译时（如果情况允许的话）来完成。

For C; various optimisations (e.g. constant propagation and constant folding) mean that every operator can potentially be done at compile time (if the situation permits it).

对于一个简单的例子，考虑这个功能：

For a simple example, consider this function:

int foo(int a, int b) {
    return a+b;
}

这看起来像需要在运行时做加法。

This looks like the addition needs to be done at run time.

现在考虑这个code：

Now consider this code:

int x = foo(1, 2);

现在的编译器可以内联的功能，传播常数，并做加法在编译时。你最终与 INT X = 3; （加上 foo的一个潜在的冗余副本（）函数是能够做到在运行时增加）的。

Now the compiler can "inline" the function, propagate the constants, and do the addition at compile time. You'd end up with int x = 3; (plus a potentially redundant copy of the foo() function which is capable of doing the addition at run time).

还有哪里的优化不能在编译时完成，但可以连接（使用LTO /链接时优化）期间完成情况;其中，运营商不是在编译时或运行时进行评估。一个简单的例子是，如果在另一个目标文件code（编译单元）做了 INT X = foo的（1,2）;

There are also cases where an optimisation can't be done at compile time but can be done during linking (with LTO/Link Time Optimisation); where the operator isn't evaluated at compile time or run time. A simple example would be if code in another object file ("compilation unit") did int x = foo(1, 2);.

另外，在一般情况下，相反也是如此 - 无保证操作者将在当它是可能的编译时进行评估;所以你不能说，运营商始终​​是编译时。对于一个简单的例子，考虑字符串连接（如的char * MyString的=你好，世界; ！） - 这将是法律的编译器生成code，做串联在运行时（即使它是很难想象的一个原因，编译器会想）。

Also, in general, the opposite is also true - nothing guarantees that an operator will be evaluated at compile time when it is possible; so you can't say that an operator is always "compile time". For a simple example of this, consider string concatenation (e.g. char *myString = "Hello " "World!";) - it would be legal for the compiler to generate code that does the concatenation at run time (even though it's hard to think of a reason why a compiler would want to).

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