分配给较大宽度整数时的 C 整数溢出行为
如果我在 C 中执行以下代码：
#include <stdint.h>
uint16_t a = 4000;
uint16_t b = 8000;
int32_t c = a - b;
printf("%d", c);
它会正确打印“-4000”作为结果。 但是，我有点困惑：从另一个整数中减去一个更大的无符号整数时，不应该出现算术溢出吗？ 这里有什么选角规则在起作用？ 这个问题似乎有点菜鸟，所以任何参考资料将不胜感激。
If I execute the following code in C:
#include <stdint.h>
uint16_t a = 4000;
uint16_t b = 8000;
int32_t c = a - b;
printf("%d", c);
It correctly prints '-4000' as the result. However, I'm a little confused: shouldn't there be an arithmetic overflow when subtracting a larger unsigned integer from the other? What casting rules are at play here? This question seems a bit noobish, so any references would be greatly appreciated.
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这个问题其实有些复杂。 算术表达式的操作数使用特定规则进行转换，您可以在 的第 3.2.1.5 节中看到这些规则标准（C89）。 对于您的情况，答案取决于
uint16_t的类型。 如果它小于int，例如short int，则操作数将转换为int并且您得到 -4000，但在 16-位系统，uint16_t可能是unsigned int并且不会自动转换为有符号类型。The issue is actually somewhat complicated. Operands of arithmetic expressions are converted using specific rules that you can see in Section 3.2.1.5 of the Standard (C89). In your case, the answer depends on what the type
uint16_tis. If it is smaller thanint, sayshort int, then the operands are converted tointand you get -4000, but on a 16-bit system,uint16_tcould beunsigned intand conversion to a signed type would not happen automatically.简短的答案是，在减法过程中，这些都被提升为
int。 对于长答案，请查看 C 标准，其中讨论了算术表达式中的整数提升。 标准中的相关语言：细节也在那里，但它们变得非常令人讨厌。
The short answer is that these are all promoted to
intduring the subtraction. For the long answer, look at section 6.3.1.1 of the C standard, where it talks about integer promotions in arithmetic expressions. Relevant language from the standard:The details are in there, too, but they get pretty nasty.
在减法过程中，两个操作数都提升为
int32_t。 如果结果大于int32_t的最大值，您就会看到溢出。Both operands are promoted to
int32_tduring the subtraction. If the result had been larger than the maximum value forint32_tyou would've seen overflow.事实上，存在溢出，但 C 没有告诉你。
当解释为有符号整数时，溢出留下的值恰好为 -4000。 这按照 2 的补码机器上的设计工作。
尝试将结果解释为无符号，您会注意到，当 u1 < > 时，(u1-u2) 的计算结果为一些看似不相关的数字。 u2。
There is, in fact, an overflow but C does not tell you.
The overflow leaves a value that happens to be -4000 when interpreted as a signed integer. This works as designed on 2's complement machines.
Try to interpret the result as unsigned, and you'll notice that (u1-u2) evaluates to some seemingly unrelated number when u1 < u2.