为什么这个函数没有超过值 2^31？
幂函数（用 C++ 编写）...
long power (long a, long b){
long result=1l;
for (int i = 0;i<b;i++){
result*=a;
}
return result;
}
现在我做了一些输出测试...
cout<<power(2l,2l)<<endl;
cout<<power(2l,4l)<<endl;
cout<<power(2l,31l)<<endl;
cout<<power(2l,32l)<<endl;
cout<<power(2l,61l)<<endl;
输出：
4
16
-2147483648
0
0
嗯，长时间回落到 32 位大小（而不是保持 64 位大小）似乎存在一些问题。我想知道为什么这不起作用，但如果我使用 long long 类型，一切正常。
一些额外信息：
我正在使用 C++ 和编译器 MinGW
我正在运行 64 位操作系统 (Windows 7)
更新：
你们太棒了！没想到竟然会发生这样的事情。
我刚刚使用 sizeof 检查了一些任意 PDT，这就是我发现的...
cout<<sizeof(long)<<" "<<sizeof(int)<<" "<<sizeof(char)<<" "<<sizeof(long long)<<" "<<sizeof(uint64_t)<<endl;
输出：
4 4 1 8 8
所以，看起来我的 long 和 int 是大小均为 32 位。进一步研究表明 intmax_t 类型也是 64 位。实际上每个 PDT 的上限都是 64 位，因此如果我需要表示 128 位整数，C++ 是否有一个内置类（类似于 Java 中的 BigInteger ）？
The power function (written in c++)...
long power (long a, long b){
long result=1l;
for (int i = 0;i<b;i++){
result*=a;
}
return result;
}
Now I do some output testing...
cout<<power(2l,2l)<<endl;
cout<<power(2l,4l)<<endl;
cout<<power(2l,31l)<<endl;
cout<<power(2l,32l)<<endl;
cout<<power(2l,61l)<<endl;
Output:
4
16
-2147483648
0
0
Well there seems to be some problem with the long falling back to a 32 bit size (instead of staying as a 64 bit). I'm wondering why this doesn't work, yet if I use the long long type, everything works fine.
Some extra info:
I'm using C++ and the compiler MinGW
I am running a 64-bit OS (Windows 7)
UPDATE:
You guys are awesome! Never thought that this type of thing would be going on.
I just checked some arbitrary PDTs using sizeof and this is what I found...
cout<<sizeof(long)<<" "<<sizeof(int)<<" "<<sizeof(char)<<" "<<sizeof(long long)<<" "<<sizeof(uint64_t)<<endl;
Output:
4 4 1 8 8
So, it looks like my long and int are both 32 bit in size. Some more playing around shows that the intmax_t type is also 64 bit. Practically every single PDT is capped at 64 bits, so if I ever needed to represent a 128-bit integer, does c++ have a built in class for that (something similar to BigInteger in Java)?
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显然，在您的环境中，
long类型是 32 位。为了解决类似的问题，我建议您使用像 uint64_t 这样的类型，而不是依赖于本机类型具有特定大小的假设。
编辑
回答你的第二个问题（c++有一个用于128位整数的内置类吗？）：不，没有。或者更确切地说，它没有强制要求这样做。但是，如果实现可以提供这样的功能，您就可以使用诸如 uint128_t 之类的东西。但就我个人而言，我还没有看到任何系统可以做到这一点。不过，有像 GMP 这样的第三方库提供了该功能。
Apparently, the type
longis 32 bits in your environment.To get around similar problems, I would suggest that you use types like
uint64_tinstead of relying in the assumption that a native type has a specific size.EDIT
To answer your second question (does c++ have a built in class for 128-bit integers?): No, it does not. Or rather, it does not mandate one. However, if an implementation would provide one you would be able to use something like
uint128_t. Personally, I haven't seen any system that does this, though. There are third party libraries like GMP that provide that functionality, though.A.尝试运行：
看看它是32位的。
B.我猜是你的项目定义有问题。即使您在 64 位机器上工作，它也可以编译 32 位程序，因此您可以在 32 位和 64 位机器上使用它。
long 始终是指针的大小...
C.
uint64_t是最佳实践。A. trying running:
To see that it's a 32 bit.
B. I guess it is a problem with the definition of your project. even if you work on a 64 bit machine it can compile a 32 bit program so you can use it both on 32 and 64 but machines.
and long is always the size of a pointer...
C.
uint64_tis the best practice.那么 long 类型需要 32 位，因此最大正值是 4294967296-1。但是你的函数计算值5842587018385982521381124421=21^21。
Well long type takes 32 bits, so max positive value is 4294967296-1. But your function calculates value 5842587018385982521381124421=21^21.
在 C++ 和 C 中，
int的大小取决于体系结构，这都是事实，但问题是 32 位有符号int的大小介于-2^31 和 (2^31-1)，不是 2^31。您确实溢出了 32 位数字。您应该使用unsigned int来代替。介于 0 和 (2^32-1) 之间。In C++, as well as in C, size of
intis architecture-dependent, and that's all true, but that catch is that 32-bit signedint's go between -2^31 and (2^31-1), not 2^31. You are indeed overflowing a 32-bit number. You should use anunsigned intinstead. That goes between 0 and (2^32-1).