是什么在铸造整型常量的编译器？ [英] What does the compiler at casting integer constants?

问题描述

使用下面的宏：

#define MIN_SWORD (signed int) 0x8000

在如下面前pression：

In e.g. the following expression:

signed long s32;
if (s32 < (signed long)MIN_SWORD)...

预计将做如下检查：

is expected to do the following check:

if (s32 < -32768)

一个有些编译器似乎很好地工作。但在其他一些编译器的p21蛋白表达被评价为：

One some compilers it seems to work fine. But on some other compiler the exprssion is evaluated as:

if (s32 < 32768)

我的问题：一个ANSI-C编译器应该如何评估以下EX pression：
（长签字）（签字INT）为0x8000 ？

看来，在一些编译器中投为`（符号的int）不会导致从正不断的0x8000（预期）转换为符号整数的最小负值，如果事后前pression浇铸到签订了更广泛的类型。
换句话说，所评估的恒定不等同于：
-32768L（但32768L）

It seems that on some compilers the cast to `(signed int) does not cause the (expected) conversion from the positive constant 0x8000 to the minimum negative value of a signed int, if afterwards the expression is casted to the wider type of signed long. In other words, the evaluated constant is not equivalent to: -32768L (but 32768L)

这是行为可能是由ANSI-C未定义？

Is this behavior maybe undefined by ANSI-C?

推荐答案

如果一个 INT 是平台上的16位，那么类型为0x8000 是 unsigned int类型（参见6.4.4第5页的标准）。转换为符号int 是实现定义如果该值不能被重新psented $ P $（见6.3.1.3第3页）。所以，你的code的行为是实现定义的。

If an int is 16-bit on your platform, then the type of 0x8000 is unsigned int (see 6.4.4 p.5 of the standard). Converting to a signed int is implementation-defined if the value cannot be represented (see 6.3.1.3 p.3). So the behaviour of your code is implementation-defined.

话虽如此，在实践中，我会一直认为这应该总是做什么你期望。什么编译是什么？

Having said that, in practice, I would've assumed that this should always do what you "expect". What compiler is this?

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