问题描述
C99标准的第6.3.1.1节包括：
Section 6.3.1.1 of the C99 standard contains:
以下可在使用
前pression只要一个 INT 或
unsigned int类型可用于：
[...]类型的位字段 _Bool ，
INT ，符号int 或符号。
INT
[...] A bit-field of type _Bool, int, signed int, or unsigned int.
如果一个 INT 可以重新present所有值
原始类型的，这个值是
转换到 INT ;否则，它
被转换为 unsigned int类型。
If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int.
在我看来，这意味着 unsigned int类型位字段都提升到 INT ，除了当无符号位字段的宽度等于 INT 的宽度，在这种情况下，最后一句适用。
It seems to me that this implies that unsigned int bit-fields are promoted to int, except when the width of the unsigned bit-field is equal to the width of int, in which case the last phrase applies.
我有以下程序：
struct S { unsigned f:32; } x = { 28349};
unsigned short us = 0xDC23L;
main(){
int r = (x.f ^ ((short)-87)) >= us;
printf("%d\n", r);
return r;
}
和两个系统执行该程序（ INT 是这两个系统上的32位）。一个系统说，这种程序打印1，另说，它打印0.我的问题是，针对的两个系统，我应该提交错误报告？ （我倾向于提出反对，打印0系统的报告，因为摘录以上）
And two systems to execute this program (int is 32-bit on both systems). One system says this program prints 1, and the other says that it prints 0. My question is, against which of the two systems should I file a bug report? (I am leaning towards filing the report against the system that prints 0, because of the excerpt above)
推荐答案
看来，这种不确定性已经由标准委员会，因为目前的草案明确了那句话检测：
It seems that this ambiguity has already been detected by the standards committee since the current draft clarifies that sentence:
如果int可以重新present的所有值
原始类型（如通过限制
宽度，对于一个位域），值
被转换为int;
这篇关于无符号位字段类型：int或unsigned int的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！