C ++中小数点的位数限制 [英] Digit limitation from decimal point in C++

问题描述

我是C ++的新手。我有一个双精度变量 double a = 0.1239857 ，我想将变量 a 限制为小数点后两位数字。因此 a 将为 0.12 。我知道C ++具有返回大于或小于 a 的最大或最小整数的函数，例如ceil或floor。

I'm new to C++. I have a double variable double a=0.1239857 and I want to limit variable a from decimal point two digits. So a will be 0.12. I know C++ have functions that return largest or smallest integer that is greater or lower than a like ceil or floor.

是否存在一个函数实现浮点变量的数字限制？或如何更改 a 变量的精度？

Is there a function that implements digit limitation of floating-point variable? Or How can I change precision of the a variable?

推荐答案

是您实际上是尝试舍入数字，还是只是更改其显示精度？

Are you actually trying to round the number, or just change its displayed precision?

对于前者（将多余的数字截断）：

For the former (truncating the extra digits):

double scale = 0.01;  // i.e. round to nearest one-hundreth
value = (int)(value / scale) * scale;

或（根据jheriko的回答，四舍五入）

or (rounding up/down as appropriate, per jheriko's answer)

double scale = 0.01;  // i.e. round to nearest one-hundreth
value = floor(value / scale + 0.5) * scale;

对于后者：

cout << setprecision(2) << value;

其中 setprecision（）的参数为小数点后显示的最大位数。

where the parameter to setprecision() is the maximum number of digits to show after the decimal point.

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