C语言 如何打印长长
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How to printf long long
提问by Carlos
I'm doing a program that aproximate PI and i'm trying to use long long, but it isn't working. Here is the code
我正在做一个近似 PI 的程序，我正在尝试使用 long long，但它不起作用。这是代码
#include<stdio.h>
#include<math.h>
typedef long long num;
main(){
num pi;
pi=0;
num e, n;
scanf("%d", &n);
for(e=0; 1;e++){
pi += ((pow((-1.0),e))/(2.0*e+1.0));
if(e%n==0)
printf("%15lld -> %1.16lld\n",e, 4*pi);
//printf("%lld\n",4*pi);
}
}
回答by karadoc
%lldis the standard C99 way, but that doesn't work on the compiler that I'm using (mingw32-gcc v4.6.0). The way to do it on this compiler is: %I64d
%lld是标准的 C99 方式，但这不适用于我使用的编译器（mingw32-gcc v4.6.0）。在此编译器上执行此操作的方法是：%I64d
So try this:
所以试试这个：
if(e%n==0)printf("%15I64d -> %1.16I64d\n",e, 4*pi);
and
和
scanf("%I64d", &n);
The only way I know of for doing this in a completely portable way is to use the defines in <inttypes.h>.
我所知道的以完全可移植的方式执行此操作的唯一方法是使用<inttypes.h>.
In your case, it would look like this:
在你的情况下，它看起来像这样：
scanf("%"SCNd64"", &n);
//...
if(e%n==0)printf("%15"PRId64" -> %1.16"PRId64"\n",e, 4*pi);
It really is very ugly... but at least it is portable.
它真的很丑……但至少它是便携的。
回答by Jonathan Leffler
- Your
scanf()statement needs to use%lldtoo. - Your loop does not have a terminating condition.
There are far too many parentheses and far too few spaces in the expression
pi += pow(-1.0, e) / (2.0*e + 1.0);- You add one on the first iteration of the loop, and thereafter zero to the value of 'pi'; this does not change the value much.
- You should use an explicit return type of
intformain(). - On the whole, it is best to specify
int main(void)when it ignores its arguments, though that is less of a categorical statement than the rest. - I dislike the explicit licence granted in C99 to omit the return from the end of
main()and don't use it myself; I writereturn 0;to be explicit.
- 您的
scanf()语句也需要使用%lld。 - 您的循环没有终止条件。
表达式中有太多的括号和太少的空格
pi += pow(-1.0, e) / (2.0*e + 1.0);- 您在循环的第一次迭代中添加一个，然后将 'pi' 的值设为零；这不会改变太多的价值。
- 您应该使用显式返回类型
intformain()。 - 总的来说，最好指定
int main(void)它何时忽略其参数，尽管与其他语句相比，这不是一个明确的声明。 - 我不喜欢 C99 中授予的显式许可，以省略末尾的返回
main()并且不自己使用它；我写return 0;得很明确。
I think the whole algorithm is dubious when written using long long; the data type probably should be more like long double(with %Lffor the scanf()format, and maybe %19.16Lffor the printf()formats.
我认为整个算法在使用long long;编写时是可疑的；数据类型可能应该更像long double（%Lf用于scanf()格式，也可能%19.16Lf用于printf()格式。
回答by Atul Kumar
// acos(0.0) will return value of pi/2, inverse of cos(0) is pi/2
double pi = 2 * acos(0.0);
int n; // upto 6 digit
scanf("%d",&n); //precision with which you want the value of pi
printf("%.*lf\n",n,pi); // * will get replaced by n which is the required precision
回答by Vinicius Kamakura
First of all, %d is for a int
首先， %d 是一个 int
So %1.16lldmakes no sense, because %d is an integer
所以%1.16lld没有意义，因为 %d 是一个整数
That typedef you do, is also unnecessary, use the type straight ahead, makes a much more readable code.
你做的 typedef 也是不必要的，直接使用类型，使代码更具可读性。
What you want to use is the type double, for calculating pi
and then using %for %1.16f.
您要使用的是类型double，用于计算 pi 然后使用%for %1.16f。
