C++ 如何在不将数字转换为字符串/字符数组的情况下获取数字的数字？
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How to get the digits of a number without converting it to a string/ char array?
提问by chustar
How do I get what the digits of a number are in C++ without converting it to strings or character arrays?
如何在不将其转换为字符串或字符数组的情况下获取 C++ 中数字的数字？
回答by Vinay Sajip
The following prints the digits in order of ascending significance (i.e. units, then tens, etc.):
以下按重要性升序（即单位，然后是十位等）打印数字：
do {
int digit = n % 10;
putchar('0' + digit);
n /= 10;
} while (n > 0);
回答by tunnuz
What about floor(log(number))+1?
怎么样floor(log(number))+1？
With ndigits and using base byou can express any number up to pow(b,n)-1. So to get the number of digits of a number xin base byou can use the inverse function of exponentiation: base-b logarithm. To deal with non-integer results you can use the floor()+1trick.
使用n位数字并使用b基数，您可以表示最多pow(b,n)-1. 因此，要获得以b为底数的数字x的位数，您可以使用求幂的反函数：以 b 为底数的对数。要处理非整数结果，您可以使用该技巧。floor()+1
PS: This works for integers, not for numbers with decimals (in that case you should know what's the precision of the type you are using).
PS：这适用于整数，不适用于带小数的数字（在这种情况下，您应该知道您使用的类型的精度是多少）。
回答by Martin York
Since everybody is chiming in without knowing the question.
Here is my attempt at futility:
因为每个人都在不知道问题的情况下插话。
这是我徒劳的尝试：
#include <iostream>
template<int D> int getDigit(int val) {return getDigit<D-1>(val/10);}
template<> int getDigit<1>(int val) {return val % 10;}
int main()
{
std::cout << getDigit<5>(1234567) << "\n";
}
回答by P Shved
I have seen many answers, but they allforgot to use do {...} while()loop, which is actually the canonical way to solve this problem and handle 0properly.
看了很多答案，但是都忘记用do {...} while()loop了，这其实是解决这个问题的规范方法，处理0得当。
My solution is based on thisone by Naveen.
int n = 0;
std::cin>>n;
std::deque<int> digits;
n = abs(n);
do {
digits.push_front( n % 10);
n /= 10;
} while (n>0);
回答by Naveen
You want to some thing like this?
你想要这样的东西吗？
int n = 0;
std::cin>>n;
std::deque<int> digits;
if(n == 0)
{
digits.push_front(0);
return 0;
}
n = abs(n);
while(n > 0)
{
digits.push_front( n % 10);
n = n /10;
}
return 0;
回答by Steve Rowe
Something like this:
像这样的东西：
int* GetDigits(int num, int * array, int len) {
for (int i = 0; i < len && num != 0; i++) {
array[i] = num % 10;
num /= 10;
}
}
The mod 10's will get you the digits. The div 10s will advance the number.
mod 10's 将为您提供数字。div 10s 将推进数字。
回答by Ivan Gelov
Integer version is trivial:
整数版本是微不足道的：
int fiGetDigit(const int n, const int k)
{//Get K-th Digit from a Number (zero-based index)
switch(k)
{
case 0:return n%10;
case 1:return n/10%10;
case 2:return n/100%10;
case 3:return n/1000%10;
case 4:return n/10000%10;
case 5:return n/100000%10;
case 6:return n/1000000%10;
case 7:return n/10000000%10;
case 8:return n/100000000%10;
case 9:return n/1000000000%10;
}
return 0;
}
回答by user unknown
simple recursion:
简单递归：
#include <iostream>
// 0-based index pos
int getDigit (const long number, int pos)
{
return (pos == 0) ? number % 10 : getDigit (number/10, --pos);
}
int main (void) {
std::cout << getDigit (1234567, 4) << "\n";
}
回答by Mac
Those solutions are all recursive or iterative. Might a more direct approach be a little more efficient?
这些解决方案都是递归的或迭代的。更直接的方法会更有效吗？
Left-to-right:
左到右：
int getDigit(int from, int index)
{
return (from / (int)pow(10, floor(log10(from)) - index)) % 10;
}
Right-to-left:
右到左：
int getDigit(int from, int index)
{
return (from / pow(10, index)) % 10;
}
回答by Deep Panchal
A simple solution would be to use the log 10 of a number. It returns the total digits of the number - 1. It could be fixed by using converting the number to an int.
一个简单的解决方案是使用数字的对数 10。它返回数字的总位数 - 1。可以通过将数字转换为 int 来修复它。
int(log10(number)) + 1
