为什么＆QUOT;对于（I = 0.1; I = 1.0;！I + = 0.1）＆QUOT;在I = 1.0不破？ [英] Why "for( i = 0.1 ; i != 1.0 ; i += 0.1)" doesn't break at i = 1.0?

问题描述

我今天考试了C和有人问我类似的问题：

  

什么是错的这个计划：

 为（X = 0.1; X = 1.0;！X + = 0.1）
    的printf（％F \\ N，X）;
 

我不解决这个问题，因为我不得不写东西，我打上 .1 为错误。但是，当我回到家里，我运行这个程序，原来，当 X 等于 1.0 ，并停留在一个无限循环：

  $猫exam.c
＃包括LT＆;＆stdio.h中GT;INT主（INT ARGC，字符** argv的）
{
    浮X;    为（X = .1; X = 1.0;！X + = 0.1）
        的printf（％F \\ N，X）;    返回0;
}
$ GCC exam.c -o考试
$ ./exam
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.900000
1.000000＆LT;？ - 
1.100000
1.200000
1.300000
1.400000
1.500000
....
 

可能有人请解释为什么发生这种情况。

解决方案

这是功课典型的问题。

存在的问题是，0.1不能精确存储在一个浮更好检查＆下; = 1.0

然而这仅仅只适用范围非常有限像邪神说。我完全错过了这个问题。原因这是更好地使用 INT 后来划分其值

I had an exam today in C and I was asked a question similar to:

What is wrong with this program:

for( x = .1 ; x != 1.0 ; x += .1)
    printf("%f\n", x);

I couldn't solve it and since I had to write something I marked .1 as an error. But, when I went back home, I run this program, It turned out that it doesn't break when x equals to 1.0 and stuck in an infinite loop:

$ cat exam.c
#include <stdio.h>

int main(int argc, char **argv)
{
    float x;

    for(x = .1 ; x != 1.0 ; x += .1)
        printf("%f\n", x);

    return 0;
}
$ gcc exam.c -o exam
$ ./exam
0.100000
0.200000
0.300000
0.400000
0.500000
0.600000
0.700000
0.800000
0.900000
1.000000 <- ?
1.100000
1.200000
1.300000
1.400000
1.500000
....

Could someone please explain why this is happening.

解决方案

This is a typical question for homework.

The Problem is that the 0.1 cannot be exactly stored in a float better check for <= 1.0

However this just works only for a very limited range like Cthulhu said. I missed that problem completely. Of cause it is better to use int and divide its value later.

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