假设sizeof(double)> = sizeof(void *)是否安全? [英] Is it safe to assume sizeof(double) >= sizeof(void*)?

问题描述

可以安全地假设sizeof(double)将始终大于或等于sizeof(void*)吗?

Is it safe to assume that sizeof(double) will always be greater than or equal to sizeof(void*)?

在某种情况下，以下内容可移植吗?

To put this in some context, is the following portable?

int x = 100;
double tmp;

union {
  double dbl;
  void* ptr;
} conv;

conv.ptr = (void*)&x;
tmp = conv.dbl;

conv.dbl = tmp;
printf("%d\n", *((int*)conv.ptr));

它确实可以在我测试过的几台机器上运行，但是我发现这很可怕如果sizeof(void*) > sizeof(double)是错误的.

It does work on the few machines that I've tested it on, but I can see this going horribly wrong if sizeof(void*) > sizeof(double).

推荐答案

在当前系统上可以. double在所有当前和将来的系统上都是64位，因为它们与IEEE算术的双精度一致.指针将来不太可能但肯定会更大-可能不是为了更大的地址空间，而是为了携带边界信息.

On current systems yes. double is 64 bits on all current and future systems because they're aligned with IEEE arithmetic's double-precision. It's unlikely but certainly possible that pointers could be larger in the future - probably not for the sake of larger address space, but instead for carrying with them bounding information.

在任何情况下，依靠double和void *之间的任何关系似乎是一个非常糟糕的主意...

In any case it seems like a really bad idea to rely on any relationship between double and void *...

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