如何确定我是否过度括号？ [英] How can I determine if I'm overparenthesizing?

问题描述

我目前正在尝试避免在C语言中使用指针算术来编写模拟器。

I'm currently trying to avoid the pointer arithmetic workings in C to write an emulator.

通常，如果您添加 1 到C中的指针，则改为添加指向对象的大小。但是，我试图使用位和字节，所以这是不希望的。

Usually, if you add 1 to a pointer in C, you add the size of the pointed to object instead. However, I am trying to work with bits and bytes, so this is undesired.

我想知道在此示例中是否使用了太多的括号：

I was wondering if I was using too many parentheses in this example:

*(int16_t *)(((intptr_t)bc)+sp)

如果不是，那么是否等同于此？ ：

And if not, then is it equivalent to this? :

*(int16_t *)((intptr_t)bc+sp)

sp 是我的模拟器的页面对齐堆栈地址（通过 mmap （未设置 MAP_FIXED ）。这是 intptr_t 类型。

sp is a page-aligned stack address for my emulator (obtained via. mmap without MAP_FIXED set). It is an intptr_t type.

bc 是 int16_t * 类型的名称。这是两个 int8_t 的组合的指针。

bc is the name of an int16_t * type. It's a pointer to a combination of two int8_t's.

推荐答案

（（（（intptr_t）bc）+ sp）等同于（（intptr_t）bc + sp）。

但这整个避免指针算术方法是不可移植的。

But this whole "avoid the pointer arithmetic" approach is not portable.

至少3个关注点：

• 指针，转换为

• Pointers, converted to an integer, do not certainly maintain the math properties needed.

// possible outcome
uint16_t x[2];
printf("%llx\n", (unsigned long long) (intptr_t) &x[0]); // --> abcd0000
printf("%llx\n", (unsigned long long) (intptr_t) &x[1]); // --> abcd0010

作为整数的差可能是16和不是2的希望-即使相差2更常见。

The difference as integers may be 16 and not the hoped for 2 -- even if a difference of 2 is more common.

• 进一步，其中 *（int16_t *）（（intptr_t）bc + sp），如果 sp 是奇数，则（int16_t *） 可能由于对齐限制而失败。

• Further, with *(int16_t *)((intptr_t)bc+sp), if sp is odd, (int16_t *) can fail due to alignment restrictions.

也出现了抗锯齿问题。 @Andrew Henle

Anti-aliasing issues occur too. @Andrew Henle

尽管整数有各种陷阱，但还是避免了指针算术-祝你好运。

Such avoidance of pointer arithmetic though integers has various pitfalls - good luck.

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