在 C++ 中将 int 设置为无穷大
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Setting an int to Infinity in C++
提问by Daniel Gratzer
I have an int athat needs to be equal to "infinity". This means that if
我有一个int a需要等于“无穷大”的。这意味着如果
int b = anyValue;
a>bis always true.
a>b永远是真的。
Is there any feature of C++ that could make this possible?
C++ 有什么特性可以使这成为可能吗？
采纳答案by Daniel Gratzer
Integers are finite, so sadly you can't have set it to a true infinity. However you can set it to the max value of an int, this would mean that it would be greater or equal to any other int, ie:
整数是有限的，所以遗憾的是你不能将它设置为真正的无穷大。但是，您可以将其设置为 int 的最大值，这意味着它将大于或等于任何其他 int，即：
a>=b
is always true.
永远是真的。
You would do this by
你会这样做
#include <limits>
//your code here
int a = std::numeric_limits<int>::max();
//go off and lead a happy and productive life
This will normally be equal to 2,147,483,647
这通常等于 2,147,483,647
If you really need a true "infinite" value, you would have to use a double or a float. Then you can simply do this
如果您真的需要一个真正的“无限”值，则必须使用双精度或浮点数。然后你可以简单地做到这一点
float a = std::numeric_limits<float>::infinity();
Additional explanations of numeric limits can be found here
可以在此处找到有关数字限制的其他说明
Happy Coding!
快乐编码！
Note: As WTP mentioned, if it is absolutely necessary to have an int that is "infinite" you would have to write a wrapper class for an int and overload the comparison operators, though this is probably not necessary for most projects.
注意：正如 WTP 所提到的，如果绝对有必要拥有一个“无限”的 int，则必须为 int 编写一个包装类并重载比较运算符，尽管这对于大多数项目来说可能不是必需的。
回答by Etienne de Martel
Integers are inherently finite. The closest you can get is by setting ato int's maximum value:
整数本质上是有限的。您可以获得的最接近的是通过设置a为int的最大值：
#include <limits>
// ...
int a = std::numeric_limits<int>::max();
Which would be 2^31 - 1(or 2 147 483 647) if intis 32 bits wide on your implementation.
这将是2^31 - 1（或2 147 483 647）如果int是32个位宽您的实现。
If you reallyneed infinity, use a floating point number type, like floator double. You can then get infinity with:
如果您确实需要无穷大，请使用浮点数类型，例如float或double。然后你可以得到无穷大：
double a = std::numeric_limits<double>::infinity();
回答by bdonlan
intis inherently finite; there's no value that satisfies your requirements.
int本质上是有限的；没有满足您要求的值。
If you're willing to change the type of b, though, you can do this with operator overrides:
但是，如果您愿意更改 的类型b，则可以使用运算符覆盖来执行此操作：
class infinitytype {};
template<typename T>
bool operator>(const T &, const infinitytype &) {
return false;
}
template<typename T>
bool operator<(const T &, const infinitytype &) {
return true;
}
bool operator<(const infinitytype &, const infinitytype &) {
return false;
}
bool operator>(const infinitytype &, const infinitytype &) {
return false;
}
// add operator==, operator!=, operator>=, operator<=...
int main() {
std::cout << ( INT_MAX < infinitytype() ); // true
}
回答by Wilmer E. Henao
This is a message for me in the future:
这是对我未来的信息：
Just use: (unsigned)!((int)0)
只需使用： (unsigned)!((int)0)
It creates the largest possible number in any machine by assigning all bits to 1s (ones) and then casts it to unsigned
它通过将所有位分配为 1（一个），然后将其强制转换为无符号数，从而在任何机器中创建最大可能的数字
Even better
甚至更好
#define INF (unsigned)!((int)0)
And then just use INF in your code
然后在你的代码中使用 INF
回答by nicodjimenez
You can also use INT_MAX:
您还可以使用 INT_MAX：
http://www.cplusplus.com/reference/climits/
http://www.cplusplus.com/reference/climits/
it's equivalent to using numeric_limits.
它相当于使用 numeric_limits。
回答by Shaun07776
int min and max values
int 最小值和最大值
Int -2,147,483,648 / 2,147,483,647 Int 64 -9,223,372,036,854,775,808 / 9,223,372,036,854,775,807
国际 -2,147,483,648 / 2,147,483,647 国际 64 -9,223,372,036,854,775,808 / 9,223,372,036,854,775,807
i guess you could set a to equal 9,223,372,036,854,775,807 but it would need to be an int64
我想你可以将 a 设置为等于 9,223,372,036,854,775,807 但它需要是一个 int64
if you always want a to be grater that b why do you need to check it? just set it to be true always
如果你总是希望 a 更磨碎 b 为什么你需要检查它？只需将其设置为始终为真
