问题描述
#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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