Ç - 比较花车if语句 [英] C - Comparing floats with if statements
问题描述
我目前正在写一个贪心算法,但比较花车的时候我已经在一个问题绊倒了。
I am currently writing a greedy algorithm, but I have stumbled upon a problem when comparing floats.
我会用code是这样的:
I would use code like this:
float f = 0;
if (f == 0) {
// code
}
我已经在我的工作程序在一个单独的程序测试这一点,它工作得很好,但不是。
I have tested this on a seperate program and it worked fine, but not on the program I am working on.
下面是从我自己的程序中提取。
Here is an extract from my own program.
float chf2 = fmod(chf, 0.1);
float ch3 = chf - chf2;
if (chf2 == 0) {
/* Divide user's number by 0.1 */
float ch3 = chf / 0.1;
/* Round the number */
int ch4 = round(ch3);
/* Print the amount of coins and end */
printf("%d\n", ch4 + coin2);
return 0;
}
奇怪的是,这似乎与检查一个previous if语句来上班的时候0.25从用户的输入FMOD。
Oddly, this seems to work with a previous if statement that checks when a fmod of 0.25 from the user's input.
有没有检查,如果一个浮动等于另一个浮动一个更好的办法?
Is there a better way of checking if a float is equal to another float?
推荐答案
您code正常工作。您的期望可能需要调整一点点,但是。
Your code works correctly. Your expectations might need a little bit of adjusting, however.
的 FMOD
函数总是返回精确的结果---当你写 C = FMOD(A,B)
, C
是一个数字,这样ç+ K * b
(无限precision计算)恰好等于 A
对于某个整数 K
。所以,你的code实际上是声音--- 如果(C == 0)
将触发什么时候 A
正是A b 的倍数。
The fmod
function always returns an exact result---when you write c = fmod(a, b)
, c
is a number such that c + k*b
(evaluated in infinite precision) exactly equals a
for some integer k
. So your code is actually sound---if (c == 0)
will trigger exactly when a
is exactly a multiple of b
.
您把 B
是双击
0.1
,这实际上是分数36028797018963968分之3602879701896397
,而不是分数 1/10
。所以,如果我计算 FMOD(1,0.1)
,我应该会得到分数(36028797018963968%3602879701896397)/ 36028797018963968
,这是 3602879701896395 36028797018963968
。这是非常略小于双击
名为 0.1
。
You took b
to be the double
0.1
, which is actually the fraction 3602879701896397/36028797018963968
, not the fraction 1/10
. So if I compute fmod(1, 0.1)
, I should expect to get the fraction (36028797018963968 % 3602879701896397) / 36028797018963968
, which is 3602879701896395 36028797018963968
. That's very slightly smaller than the double
called 0.1
.
这也解释了为什么当您使用code符合您的期望 0.25
而不是 0.1
; 0.25
给你确切的1/4。
This also explains why your code meets your expectations when you use 0.25
instead of 0.1
; 0.25
gives you 1/4 exactly.
您可以阅读这浮点指南了解浮动的重新presentation - 点号。
You might read this floating-point guide to learn about the representation of floating-point numbers.
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