在-255到255范围内未定义带符号的char溢出吗？ [英] Is signed char overflow undefined within the range -255 to 255?

问题描述

在C99模式下，根据GCC，以下代码是否为未定义行为：

Is the following code undefined behavior according to GCC in C99 mode:

signed char c = CHAR_MAX; // assume CHAR_MAX < INT_MAX
c = c + 1;
printf("%d", c);

推荐答案

签名字符溢出的确会导致未定义的行为，但这不是发布的代码中发生的情况。

signed char overflow does cause undefined behavior, but that is not what happens in the posted code.

With c = c +1 ，整数提升在加法之前执行，因此表达式中的 c 被提升为 int 在右边。由于 128 小于 INT_MAX ，因此这种添加不会发生意外。请注意， char 通常比 int 窄，但在罕见的系统 char 和 int 的宽度可以相同。无论哪种情况，在算术表达式中， char 都被提升为 int 。

With c = c + 1, the integer promotions are performed before the addition, so c is promoted to int in the expression on the right. Since 128 is less than INT_MAX, this addition occurs without incident. Note that char is typically narrower than int, but on rare systems char and int may be the same width. In either case a char is promoted to int in arithmetic expressions.

然后对 c 进行赋值时，如果普通 char 是 unsigned 在所讨论的系统上，相加的结果小于 UCHAR_MAX （必须至少为255），并且该值在转换并分配给 c 。

When the assignment to c is then made, if plain char is unsigned on the system in question, the result of the addition is less than UCHAR_MAX (which must be at least 255) and this value remains unchanged in the conversion and assignment to c.

如果不是普通的 char 是带符号的，加法的结果在赋值前会转换为带符号的字符的值。在这里，如果加法的结果不能用带符号的字符表示，则转换是实现定义的，或者引发实现定义的信号。 SCHAR_MAX 必须至少为127，如果是这种情况，那么当纯 char 是签名的。

If instead plain char is signed, the result of the addition is converted to a signed char value before assignment. Here, if the result of the addition can't be represented in a signed char the conversion "is implementation-defined, or an implementation-defined signal is raised," according to §6.3.1.3/3 of the Standard. SCHAR_MAX must be at least 127, and if this is the case then the behavior is implementation-defined for the values in the posted code when plain char is signed.

该行为不是未定义代码的行为，而是实施定义。

The behavior is not undefined for the code in question, but is implementation-defined.

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