为什么我的电源运算符 (^) 不起作用?
#include <stdio.h>
void main(void)
{
int a;
int result;
int sum = 0;
printf("Enter a number: ");
scanf("%d", &a);
for( int i = 1; i <= 4; i++ )
{
result = a ^ i;
sum += result;
}
printf("%d
", sum);
}
为什么 ^
不能作为幂运算符工作?
Why is ^
not working as the power operator?
推荐答案
首先,C/C++ 中的 ^
运算符是按位异或.与权力无关.
Well, first off, the ^
operator in C/C++ is the bit-wise XOR. It has nothing to do with powers.
现在,关于您使用 pow()
函数的问题,一些谷歌搜索表明将参数之一转换为双重帮助:
Now, regarding your problem with using the pow()
function, some googling shows that casting one of the arguments to double helps:
result = (int) pow((double) a,i);
请注意,我还将结果转换为 int
,因为所有 pow()
重载都返回双精度,而不是 int
.我没有可用的 MS 编译器,所以我无法检查上面的代码.
Note that I also cast the result to int
as all pow()
overloads return double, not int
. I don't have a MS compiler available so I couldn't check the code above, though.
从 C99 开始,还有 float
和 long double
函数分别称为 powf
和 powl
,如果有帮助的话.
Since C99, there are also float
and long double
functions called powf
and powl
respectively, if that is of any help.
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