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gcc C/C++ assume no pointer aliasing

开发者 https://www.devze.com 2023-04-02 12:40 出处:网络
Having recently read that the main reason why fo开发者_运维技巧rtran is faster than c/c++ in numerical computations is because there is no pointer aliasing.

Having recently read that the main reason why fo开发者_运维技巧rtran is faster than c/c++ in numerical computations is because there is no pointer aliasing.

Apparently, using restrict or __restrict__ keywords allows on a case by case basis to indicate the absence of pointer aliasing for a given memory element.

The icc compiler apparently has an option -fno-alias which allows one to globally assume that no aliasing is present. On gcc there is -fno-strict-aliasing, which applies only to a subset of all the aliasing situations.

Is there an option present in gcc, or are there some cases where no aliasing is assumed, when using certain optimization flags?


GCC has the option -fstrict-aliasing which enables aliasing optimizations globally and expects you to ensure that nothing gets illegally aliased. This optimization is enabled for -O2 and -O3 I believe.

C++ has well defined aliasing rules, though, and standard-compliant code will not conflict with strict aliasing. In particular this means you're not allowed to access one variable through a pointer to a different type:

float f;
int * p = reinterpret_cast<int*>(&f);  // uh-oh
*p = 0x3FF00000;                       // breaks strict aliasing

The crucial exception to this rule is that you can always access any variable through a pointer to char. (This is necessary for serializing through IO operations.)

The aliasing rules do not help the compiler to know whether any pointers of the same type alias each other. Consider this:

void add(float * a, float * b, float * c) { *c = *a + *b; }

Here the compiler cannot know whether c points to different memory than a or b and has to be careful. I think this is where restrict makes a difference, essentially by promising that float * restrict c means that nobody aliases c.

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