C问题：最经济的整数检验是2的幂 [英] C Question: Most Economical Test For Integer Is a Power of 2
问题描述
'C'中最经济的测试是什么？整数是2的幂？
例如，更好的东西：
void is_2_pow（int arg）
{
return（（x == 1）||（x == 2）||（x == 4）||（x == 8）||（x == 16）/ *和
等等* /）;
}
What is the most economical test in ''C'' for "integer is a power of 2"?
For example, something better than:
void is_2_pow(int arg)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
推荐答案
" David T. Ashley" < dt*@e3ft.com写在留言中
新闻：B4 ***************************** *@giganews.com ...
"David T. Ashley" <dt*@e3ft.comwrote in message
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C中最经济的测试整数是2的幂是什么？
例如，更好的东西：
void is_2_pow（int arg）
{
return（（x == 1）||（x == 2）||（x == 4）||（x == 8）||（x == 16）/ *和
等等* /）;
}
What is the most economical test in ''C'' for "integer is a power of 2"?
For example, something better than:
void is_2_pow(int arg)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
让我们再试一次：
void is_2_pow（int x）
{
return（（x == 1）||（x == 2）||（x == 4） ||（x == 8）||（x == 16）/ *和
等等* /）;
}
但我想每个人都会知道我的意思。
Let''s try that again:
void is_2_pow(int x)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
but I think everyone would have known what I meant.
David T. Ashley写道：
David T. Ashley wrote:
" David T. Ashley" < dt*@e3ft.com写在留言中
新闻：B4 ***************************** *@giganews.com ...
"David T. Ashley" <dt*@e3ft.comwrote in message
news:B4******************************@giganews.com ...
C中最经济的测试整数是2的幂是什么？
例如，更好的东西：
void is_2_pow（int arg）
{
return（（x == 1）||（x == 2）||（x == 4）||（x == 8）||（x == 16）/ *和
等等* /）;
}
What is the most economical test in ''C'' for "integer is a power of 2"?
For example, something better than:
void is_2_pow(int arg)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
让我们再试一次：
void is_2_pow（int x）
{
return（（x == 1）||（x == 2）||（x == 4） ||（x == 8）||（x == 16）/ *和
等等* /）;
}
Let''s try that again:
void is_2_pow(int x)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
你在void函数中返回一个值。应该是：int
is_2_pow（int x）
可能不是最经济的方法，但我会这样做：
int isPowerof2（int x）{
unsigned int i;
int count = 0;
int n = 1;
for（i = INT_MAX; i; i>> = 1）{
if（x& n）
count + +;
n<< = 1;
}
if（count 1）
count = 0 ;
返回计数;
}
You''re returning a value in a void function. Should be: int
is_2_pow(int x)
Probably not the most economical method, but I''d do:
int isPowerof2(int x) {
unsigned int i;
int count = 0;
int n = 1;
for (i = INT_MAX; i; i >>= 1) {
if (x & n)
count ++;
n <<= 1;
}
if (count 1)
count = 0;
return count;
}
David T. Ashley写道：
David T. Ashley wrote:
>
''C'中最经济的测试是什么？整数是2的幂？
例如，比以下更好的东西：
void is_2_pow（int arg）
{
return（（x == 1）||（x == 2）||（x == 4）||（x == 8）||（x == 16）/ *和
依此类推* /）;
}
>
What is the most economical test in ''C'' for "integer is a power of 2"?
For example, something better than:
void is_2_pow(int arg)
{
return((x == 1) || (x == 2) || (x == 4) || (x == 8) || (x == 16) /* and
so on */ );
}
int n_is_Power_of_two（长期无符号n）
{
return（n& n - 1）== 0&& n！= 0;
}
-
pete
int n_is_Power_of_two(long unsigned n)
{
return (n & n - 1) == 0 && n != 0;
}
--
pete
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