[英]Print a variable number of digits of a double using printf
Anyone knows if it is possible to use printf to print a VARIABLE number of digits? 任何人都知道是否可以使用printf打印VARIABLE位数?
The following line of code prints exactly 2: 以下代码行正好打印2:
printf("%.2lf", x);
but let's say I have a variable: 但是我要说我有一个变量:
int precision = 2;
Is there a way to use it in printf to specify the number of digits? 有没有办法在printf中使用它来指定位数?
Otherwise I will have to write a 'switch' or 'if' structure. 否则我将不得不写一个'switch'或'if'结构。
Thanks 谢谢
It is possible: 有可能的:
#include <stdio.h>
int main() {
int precision = 3;
float b = 6.412355;
printf("%.*lf\n",precision,b);
return 0;
}
Yes, you can do that easily - 是的,你可以轻松地做到这一点 -
int precision = 2;
printf("%.*lf", precision, x);
Yes: printf("%*d", width, num)
: see here: http://linux.die.net/man/3/printf 是: printf("%*d", width, num)
:见: http : //linux.die.net/man/3/printf
If you are using C++ you could use std::cout
in combiation with ios_base::precision()
: http://www.cplusplus.com/reference/ios/ios_base/precision/ 如果您使用的是C ++,可以在与ios_base::precision()
组合中使用std::cout
: http : //www.cplusplus.com/reference/ios/ios_base/precision/
If you use C++, you can use setprecision : 如果使用C ++,则可以使用setprecision:
#include <iostream>
#include <iomanip> // std::setprecision
int main () {
int precision = 2;
double f =3.14159;
std::cout << std::setprecision(precision) << f << '\n';
++precision;
std::cout << std::setprecision(precision) << f << '\n';
return 0;
}
3.1
3.14
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