从int16_t到int的对话能否导致实现定义的行为? [英] Can a int16_t to int conversation result in implementation-defined behavior?

问题描述

在C99标准的第7.18.1.1节第1段中:

In section 7.18.1.1 paragraph 1 of the C99 standard:

typedef名称 intN_t 指定宽度为 N 且没有填充的带符号整数类型 位和两个补码表示形式.

The typedef name intN_t designates a signed integer type with width N, no padding bits, and a two’s complement representation.

根据C99标准，要求精确宽度的带符号整数类型具有二进制补码表示形式.例如，这意味着int8_t的最小值为-128，而不是一个人的补语 的最小值为-127.

According to the C99 standard, exact-width signed integer types are required to have a two's complement representation. This means, for example, int8_t has a minimum value of -128 as opposed to the one's complement minimum value of -127.

第6.2.6.2节第2段允许实现决定将符号位解释为符号和大小，二进制补码还是一个人的补语:

Section 6.2.6.2 paragraph 2 allows the implementation to decide whether to interpret a sign bit as sign and magnitude, two's complement, or one's complement:

如果符号位为1，则值应为 已通过以下方式之一进行了修改:
—取反了符号位0的对应值(符号和幅度)；
—符号位的值为-(2 ^N )(二进制补码)；
—符号位的值为-(2 ^N -1)(一个人的补语).

If the sign bit is one, the value shall be modified in one of the following ways:
— the corresponding value with sign bit 0 is negated (sign and magnitude);
— the sign bit has the value -(2^N) (two’s complement);
— the sign bit has the value -(2^N - 1) (ones’ complement).

方法之间的区别很重要，因为二进制补码(-128)中整数的最小值可能超出 one补码表示的值范围>(-127至127).

The distinct between the methods is important because the minimum value of an integer in two's complement (-128) can be outside the range of values representable in ones' complement (-127 to 127).

假设一个实现将int类型定义为具有ones' complement表示形式，而int16_t类型具有two's complement表示形式(由C99标准保证).

Suppose an implementation defines the int types as having ones' complement representation, while the int16_t type has two's complement representation as guaranteed by the C99 standard.

int16_t foo = -32768;
int bar = foo;

在这种情况下，由于foo持有的值超出了bar可以表示的值范围，从int16_t到int的转换是否会引起实现定义的行为?

In this case, would the conversion from int16_t to int cause implementation-defined behavior since the value held by foo is outside the range of values representable by bar?

推荐答案

是.

具体来说，该转换将产生实现定义的结果. (对于-32768以外的任何值，其结果和行为都将得到很好的定义.)或者该转换可能会产生一个实现定义的信号，但是我不知道执行该操作的实现.

Specifically, the conversion would yield an implementation-defined result. (For any value other than -32768, the result and the behavior would be well defined.) Or the conversion could raise an implementation-defined signal, but I don't know of any implementations that do that.

有关转换规则的参考: N1570 6.3.1.3p3 :

Reference for the conversion rules: N1570 6.3.1.3p3:

否则，将对新类型进行签名，并且无法表示该值 在里面;结果要么是实现定义的，要么是 实施定义的信号被引发.

Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

这只能在以下情况下发生:

This can only happen if:

• int是16位(更确切地说，具有15个值位，1个符号位和0个或更多填充位)
• int使用补码或正负号和幅度
• 该实现还支持二进制补码(否则将不会定义int16_t).

• int is 16-bits (more precisely, has 15 value bits, 1 sign bit, and 0 or more padding bits)
• int uses one's-complement or sign-and magnitude
• The implementation also supports two's-complement (otherwise it just won't define int16_t).

我很惊讶地看到一个符合这些标准的实现.它必须同时支持二进制补码和一个或两个数的补码，并且必须为类型int选择后者中的一个. (也许非二进制补码实现可能在软件中支持二进制补码，只是为了能够定义int16_t.)

I'd be surprised to see an implementation that meets these criteria. It would have to support both two's-complement and either one's complement or sign-and-magnitude, and it would have to chose one of the latter for type int. (Perhaps a non-two's-complement implementation might support two's-complement in software, just for the sake of being able to define int16_t.)

如果您担心这种可能性，则可以考虑将其添加到您的头文件之一:

If you're concerned about this possibility, you might consider adding this to one of your header files:

#include <limits.h>
#include <stdint.h>

#if !defined(INT16_MIN)
#error "int16_t is not defined"
#elif INT_MIN > INT16_MIN
#error "Sorry, I just can't cope with this weird implementation"
#endif

#error不太可能在任何理智的现实世界实现中触发.

The #errors are not likely to trigger on any sane real-world implementation.

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