C++ x 的对数基数 n
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/18874255/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Log base n of x
提问by Amirhossein Mahdinejad
I have a little problem. Who knows how we can calculate the log base n with Shift_L or Shift_R?
我有一个小问题。谁知道我们如何使用 Shift_L 或 Shift_R 计算对数基数 n？
for example: for n=2 we had this solution:
例如：对于 n=2 我们有这个解决方案：
int log(int n){
int res = 0;
while((n>>=1))
res++;
return res;
}
回答by Carsten
You don't seem to want the logarithm for a base b, but the largest integer nso that n <= log_b(x). If that's the case, the following function should serve your needs:
您似乎不需要 base 的对数b，而是最大的整数，n以便n <= log_b(x). 如果是这种情况，以下功能应该可以满足您的需求：
int intlog(double base, double x) {
return (int)(log(x) / log(base));
}
回答by Teh Suu
well this is rather a math problem instead of an actuall programming problem, if i understand your problem correctly:
好吧，如果我正确理解您的问题，这更像是一个数学问题而不是实际的编程问题：
log_2 (x) = log_a (x) / log_a (2)where a can be any base.
log_2 (x) = log_a (x) / log_a (2)其中 a 可以是任何基数。
Therefore you could use the math.h's function log(double)
因此你可以使用math.h's 函数log(double)
double res = log(x)/log(2);
double res = log(x)/log(2);
