我在下面的问题中需要帮助。 我有一个知道在32位上进行计算的cpu（unsigned 整数（ 写一个函数，得到2个64位数字并返回它们的总和。 我从以下结构开始。 typedef struct { unsigned int low; unsigned int high; } 64bits; 并定义nums 64bits num1.num2; 1.我知道无符号整数的极限是~64000 的情况从0开始溢出它。？ 2.有人可以为这个问题添加解决方案吗？
Hi, I need help in the following question . I have a cpu that knows to do the computations on 32 bits(unsigned integer( write a function that gets 2 64 bits numbers and return their sum. I start with the following structure. typedef struct{ unsigned int low; unsigned int high; }64bits; and define the nums 64bits num1.num2; 1.I know that the limit of unsigned integer is ~64000 what happend in case of overflow it start again from 0.? 2.could someone add solution for the question?
推荐答案5月13日18:17，sarahh< cohen ... @ gmailwrote： On 13 May, 18:17, sarahh <cohen...@gmailwrote: 您好，我在以下问题中需要帮助。 我有一个知道在32位上进行计算的cpu（无符号 整数（ 写一个得到2 6的函数） 4位数字并返回它们的总和。 我从以下结构开始。 typedef struct { unsigned int low; unsigned int high; } 64位; 并定义nums 64bits num1.num2; Hi, I need help in the following question . I have a cpu that knows to do the computations on 32 bits(unsigned integer( write a function that gets 2 64 bits numbers and return their sum. I start with the following structure. typedef struct{ unsigned int low; unsigned int high; }64bits; and define the nums 64bits num1.num2;
您是否正在使用C89编译器 表示您没有未签名的长期 可用吗？
Are you working on a C89 compiler which means you don''t have unsigned long long available ?
1.我知道无符号整数的限制是〜64000发生的事情 溢出的情况它从0开始。？ 1.I know that the limit of unsigned integer is ~64000 what happend in case of overflow it start again from 0.?UINT_MAX保证至少为65535.它可以更多，实际上标准没有指定 上限。是的，它会在到达时徘徊 UINT_MAX + 1
UINT_MAX is guaranteed to be at least 65535. It could be a lot more , in fact the standard doesn''t specify an upper bound. And yes it will wrap around upon reaching UINT_MAX + 1
2.有人可以为问题添加解决方案吗？ 2.could someone add solution for the question?这是家庭作业吗？
Is it homework ?
sarahh写道： sarahh writes: 1.我知道无符号整数的极限是~64000 1.I know that the limit of unsigned integer is ~64000
....在主机上，unsigned int是32位的就像你的问题，是的....
....on a host where unsigned int is 32-bit like on your question, yes....
如果发生溢出，它会从0开始再次发生什么？ what happend in case of overflow it start again from 0.?是的。使用32位无符号int，0xffffffffU（所有位= 1）+ 1U == 0. 无符号算术定义为模数（最大值+ 1）算术。 （严格来说，这不是所谓的C语言溢出 - 正是因为它定义明确。溢出是发生在 $ b上的事情$ b签名整数，其结果未定义。）
Yes. With 32-bit unsigned int, 0xffffffffU (all bits = 1) + 1U == 0. Unsigned arithmetic is defined as modulo-(max value + 1) arithmetic. (Strictly speaking this is not what is called overflow in the C language - precisely because it is well-defined. "overflow" is what happens to signed integers, where the result is not defined.)
2.是否有人为问题添加解决方案？ 2.could someone add solution for the question?家庭作业的重点是边做边学。但是如果你需要一个提示： 想想你如何手工添加两位数字，并认为 的低位和高位是一个数字的两位数。 （在基础0x100000000 而不是基数10，但是嘿......）你需要以某种方式检查 是否有来自添加低的进位数字。 - Hallvard
The point of homework is to learn by doing. But if you need a hint: Think of how you do addition of two-digit numbers by hand, and think of low and high as the two digits of a number. (In base 0x100000000 instead of base 10, but what the hey...) You need to check somehow whether there was carry from adding the "low" digits. -- Hallvard
在13×？×？×？，20 ：45，Hallvard B Furuseth< hbfurus ... @ usit.uio.no> 写道： On 13 ×?×?×?, 20:45, Hallvard B Furuseth <h.b.furus...@usit.uio.no> wrote: sarahh写道： sarahh writes: 1.我知道主机上无符号整数的限制为~64000 1.I know that the limit of unsigned integer is ~64000
...其中unsigned int在你的问题上是32位的，是的....
...on a host where unsigned int is 32-bit like on your question, yes....
如果发生溢出，它会从0开始再次发生什么？ what happend in case of overflow it start again from 0.?是的。使用32位无符号int，0xffffffffU（所有位= 1）+ 1U == 0 .. 无符号算术定义为模数（最大值+ 1）算术。 （严格来说，这不是所谓的C语言溢出 - 正是因为它定义明确。溢出是发生在 签名的整数，其结果未定义。）
Yes. With 32-bit unsigned int, 0xffffffffU (all bits = 1) + 1U == 0.. Unsigned arithmetic is defined as modulo-(max value + 1) arithmetic. (Strictly speaking this is not what is called overflow in the C language - precisely because it is well-defined. "overflow" is what happens to signed integers, where the result is not defined.)
2.是否可以为此问题添加解决方案？ 2.could someone add solution for the question?家庭作业的重点是边做边学。但是如果你需要一个提示： 想想你如何手工添加两位数字，并认为 的低位和高位是一个数字的两位数。 （在基础0x100000000 而不是基数10，但是嘿......）你需要以某种方式检查 是否有来自添加低的进位数字。 - Hallvard
The point of homework is to learn by doing. But if you need a hint: Think of how you do addition of two-digit numbers by hand, and think of low and high as the two digits of a number. (In base 0x100000000 instead of base 10, but what the hey...) You need to check somehow whether there was carry from adding the "low" digits. -- Hallvard
谢谢，知道我更了解它怎么样没有签名，如果我加上max + 1，会发生什么？b / b？
Thanks,know I understand it better what about not unsigned, what happend if I add to max+1 ?
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