C语言 C - 如何划分浮点数？
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C - how to divide floats?
提问by Waypoint
I get input from command line as a int d. Now I am facing this problem:
我从命令行输入 int d。现在我面临这个问题：
float a,b;
int d;
float piece;
printf("Please enter the parts to divide the interval: ");
scanf("%d", &d);
a=0;
b=1;
piece=b-a/(float)d;
printf("%f\n",piece);
All I want is to printf some float number dependent on &d. e.g. when I write here 5, I would get 0.20000, for 6 - 0,166666 but I am still getting 1.000000 for all numbers, does anyone knows solution?
我想要的只是打印一些依赖于 &d 的浮点数。例如，当我在这里写 5 时，对于 6 - 0,166666，我会得到 0.20000，但对于所有数字，我仍然得到 1.000000，有人知道解决方案吗？
回答by Vamana
Division has precedence over subtraction, so you need to put the subtraction inside parentheses. You don't have to explicitly cast d to float; dividing a float by it will promote it to float.
除法优先于减法，因此您需要将减法放在括号内。您不必显式地将 d 强制转换为浮动；除以浮点数将促进它浮动。
piece = (b - a) / d;
回答by Sjoerd
Use parenthesis:
使用括号：
piece=(b-a)/(float)d;
回答by Stephen Canon
I believe you want:
我相信你想要：
piece = (b - a)/d;
I.e., the problem isn't division, but order of operations.
即，问题不是除法，而是运算顺序。
回答by Brian Driscoll
I think this line:
piece=b-a/(float)d;
我认为这一行：
piece=b-a/(float)d;
should be:
piece=(float)(b-a)/(float)d;
应该：
piece=(float)(b-a)/(float)d;
Just my 2 cents.
只有我的 2 美分。
EDIT
编辑
Since dis an int, perhaps try this instead:
由于d是一个整数，也许可以试试这个：
piece=(float)((b-a)/d);
piece=(float)((b-a)/d);
