如何提升力量 [英] How to raise to the power of

问题描述

我对C很陌生

我正在尝试将某些东西发挥某些作用

I am trying to put something to the power of something

例如5 ^ 3 = 125;

eg. 5^3=125;

但是当我编码时

#include <math.h>
...

printf("%d", 5^3);

我得到6.我在做根本上是错的事情吗?

I get 6. Am i doing something fundamentally wrong??

推荐答案

^是C语言中按位XOR运算符.

^ is the bitwise XOR operator in C.

5是二进制形式的101，3是011.所以101 XOR 011 = 110，它是十进制的6.

5 is 101 in binary representation and 3 is 011. So 101 XOR 011 = 110, which is 6 in decimal.

您想要的是pow(5,3)(在包含math.h之后即可使用).但是请注意，它作用于双精度数而不是整数，因此您的返回值将是一个浮点数，并且通常不精确.如果需要精确的整数幂运算，则要么必须使用其他库，要么实现自己的幂函数.这并不难，对于算法，请参见通过平方求幂.标准C不包含用于整数幂运算的函数.无论如何，对于整数幂运算，重要的是要注意溢出，对于较小的输入值可能已经发生

What you want is pow(5,3) (which is available after including math.h as you did). Note however that it acts on doubles not integer, so your return value will be a floating point and in general not exact. If you need exact integer exponentiation, then you either have to use an additional library or implement your own power-function. This is not that hard, for an algorithm see exponentiation by squaring. Standard C does not include a function for integer exponentiation. Anyway with integer exponentiation it is important to take care of overflows, which may happen already for small input values.

也要打印返回的双精度型，您要么将双精度型转换为整数printf("%d", (int)pow(5,3))，要么将其打印为浮点型printf("%f", pow(5,3))

Also to print the returned double you either have to cast the double to an int printf("%d", (int)pow(5,3)) or print it as floating point printf("%f", pow(5,3))

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