Why did the language designers of C do type equivalance like this?

Why did the language designers of C do type equivalance like this?

高跟鞋的旋律 发布于 2021-11-25 字数 388 浏览 709 回复 5 原文

I'm learning C and I'm reading about type equivalence.

I'm curious, does anyone have an opinion why they used structural equivalence for arrays and pointers but they used declaration equivalence for structs and unions?

Why the disparity there? What is the benefit of doing declaration equivalence with structs/unions and structural equivalence for everything else?

如果你对这篇文章有疑问,欢迎到本站 社区 发帖提问或使用手Q扫描下方二维码加群参与讨论,获取更多帮助。



需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。


叹倦 2022-06-07 5 楼

Don't underrate Dennis Ritchie. Every statically typed language should have a way to create an abstract type which it is impossible for a user to forge. For this you need a type construct or declaration construct that is generative, i.e., every instance of the construct generates a fresh type, distinct from any other. Such a construct is vital if you want to keep other people's mitts off your data. (For plenty of examples, see Dave Hanson's book C Interfaces and Implementations.)

So here values p1 and p2 have different types but the same representation:

struct { float x, y } p1;
struct { float x, y } p2;

Why pick struct to be generative? Because it's general enough to allow a wide range of convenient representations. union is a bit of a stretch, but I suspect it's "collateral design"; in C's type system, union behaves as much list struct as possible, which simplifies the compiler.

Incidentally, "declaration equivalence" is a term I have never heard before. 25 years ago terms like "name equivalence", "structural equivalence", and "occurence equivalence" were popular. Today type systems are much more formal, and equivalence is typically defined by logical rules rather than informal English. When it is helpful to resort to informal English, I usually find that the idea of "generativity" has more explanatory power than inventing a new name for the equivalence rules of each new language.

柠栀 2022-06-07 4 楼

Structural equivalence is difficult to check. Pointers and arrays are pretty simple, but structs and union types are more complex. Testing for structural equivalence is very difficult on these complex types, but easier for pointers and arrays.


Originally I had written an answer that dealt with value equivalence instead of type equivalnce, so it wasn't really n answer to this question. I did receive a few upvotes on it, though, so I decided to keep in here.

What I know about [value equivalence], is that pointers and arrays always have a pretty simple layout in memory. This which makes it easy to do a simple byte-for-byte comparison.

For structs and union, this memory layout isn't necessarily this simple. You could for example, have a struct with an int (32 bit) and a double (64 bit). Such a struct requires 128 bits of memory, 32 of which aren't actually relevant for comparison. So, byte-forbyte comparison is out of the question. So, declaration equivalence is just easier to implement.

反话 2022-06-07 3 楼

In a lot of ways c is a clean expression of assembly idiom from the 1970s. Arrays on these processors are implemented directly using pointers, and c simply copies that fact.

浅暮の光 2022-06-07 2 楼

I'm sure it was because it looked like a good idea at the time.

Don't try to overthink the design of C. Kernighan & Ritchie were designing a system implementation language that might be good for other things, and wound up with decisions they regretted later (operator precedence being the best documented). Some of the issues were cleaned up by the Standards Committee, and others were too deeply ingrained.

As Uri points out in his answer, type equivalence is a difficult problem, and is therefore one of those likely to have been punted by K&R in a desire to get a working compiler soon rather than a clean language design later.

温柔少女心 2022-06-07 1 楼

I'm sure others will present C specific information, but I'll mention that type-equivalence is one of the classic main problems in programming languages theory. Determining whether two types are actually equivalent is a much trickier problem than it may seem.

For example, here are some overview slides from an academic course just to give you a taste of the headache.