16,471
社区成员
发帖
与我相关
我的任务
分享无符号整型溢出问题
C++标准有规定无符号整型溢出要变成0吗
如果一直把一个无符号整型加下去, 直到0xFFFFFFFF, 会不会有某些编译器不把它变回0而一直停在0xFFFFFFFF
如果一直把一个无符号整型加下去, 直到0xFFFFFFFF, 会不会有某些编译器不把它变回0而一直停在0xFFFFFFFF
...全文
请发表友善的回复…
发表回复
yjukh 2010-01-12
- 打赏
- 举报
发现回帖的都有小星星啊
red-fly 2010-01-12
- 打赏
- 举报
应该是变为0的，可以把它看作是有符号位的情形来计算，当达到最大值的时候，它自然就变成0了。好像是有这样的说法：只所以用内码来存储数字，就是为了让符号位也参加计算，这样方便指令的实现
marrco2005 2010-01-12
- 打赏
- 举报
C++ 标准要求必须变为0
3.9.1(4): Unsigned integers, declared unsigned, shall obey the laws of
arithmetic modulo 2^n where n is the number of bits in the value
representation of that particular size of integer
Footnote: This implies that unsigned arithmetic does not overflow
because a result that cannot be represented by the resulting unsigned
integer type is reduced modulo the number that is one greater than the
largest value that can be represented by the resulting unsigned
integer
type.
3.9.1(4): Unsigned integers, declared unsigned, shall obey the laws of
arithmetic modulo 2^n where n is the number of bits in the value
representation of that particular size of integer
Footnote: This implies that unsigned arithmetic does not overflow
because a result that cannot be represented by the resulting unsigned
integer type is reduced modulo the number that is one greater than the
largest value that can be represented by the resulting unsigned
integer
type.
bragi523 2010-01-11
- 打赏
- 举报
是啊
因为你超出上限了啊
因为你超出上限了啊
快乐鹦鹉 2010-01-11
- 打赏
- 举报
这个得有人把编译器遍历一遍......
