C++ 正确使用 std::cout.precision() - 不打印尾随零
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Correct use of std::cout.precision() - not printing trailing zeros
提问by mahmood
I see many questions about the precision number for floating point numbers but specifically I want to know why this code
我看到很多关于浮点数精度数的问题，但特别是我想知道为什么这个代码
#include <iostream>
#include <stdlib.h>
int main()
{
int a = 5;
int b = 10;
std::cout.precision(4);
std::cout << (float)a/(float)b << "\n";
return 0;
}
shows 0.5? I expect to see 0.5000.
Is it because of the original integer data types?
显示0.5? 我期待看到0.5000。是因为原始整数数据类型吗？
回答by Nemanja Boric
#include <iostream>
#include <stdlib.h>
#include <iomanip>
int main()
{
int a = 5;
int b = 10;
std::cout << std::fixed;
std::cout << std::setprecision(4);
std::cout << (float)a/(float)b << "\n";
return 0;
}
You need to pass std::fixedmanipulator to coutin order to show trailing zeroes.
您需要将std::fixed操纵器传递给以cout显示尾随零。
回答by stardust
std::cout.precision(4);tells the maximum number of digits to usenot the minimum.
that means, for example, if you use
std::cout.precision(4);告诉使用的最大位数而不是最小位数。这意味着，例如，如果您使用
precision 4 on 1.23456 you get 1.235
precision 5 on 1.23456 you get 1.2346
If you want to get ndigits at all times you would have to use std::fixed.
如果您想n始终获得数字，则必须使用std::fixed.
回答by Nikos C.
The behavior is correct. The argument specifies the maximum meaningful amount of digits to use. It is not a minimum. If only zeroes follow, they are not printed because zeroes at the end of a decimal part are not meaningful. If you want the zeroes printed, then you need to set appropriate flags:
行为是正确的。该参数指定要使用的最大有意义的数字量。这不是最低限度。如果后面只有零，则不会打印它们，因为小数部分末尾的零没有意义。如果要打印零，则需要设置适当的标志：
std::cout.setf(std::ios::fixed, std::ios::floatfield);
This sets the notation to "fixed", which prints all digits.
这将符号设置为“固定”，打印所有数字。
