我试过这个代码来找出前n个自然数的阶乘之和。但它没有显示出预期的结果.. [英] I tried this code to find out the sum of factorial of first n natural numbers . But it is not showing the desired result..

问题描述

#include<stdio.h>
#include<math.h>
int main()
{
	int n,i,j,sum=0,f=1;
	printf("enter the value of no upto which you want to print the sum");
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	 {
	 	
	 	for(j=i;j>=1;j--)
	 	{
		 
	 	f=f*j;
	 	sum=sum+f;
	    }
	 	
	 }
	 printf("sum is %d",sum);
	 return 0;
}

已添加代码块 - OriginalGriff [/ edit]



我尝试过：

[edit]Code block added - OriginalGriff[/edit]

What I have tried:

#include<stdio.h>
#include<math.h>
int main()
{
	int n,i,j,sum=0,f=1;
	printf("enter the value of no upto which you want to print the sum");
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	 {
	 	
	 	for(j=i;j>=1;j--)
	 	{
		 
	 	f=f*j;
	 	sum=sum+f;
	    }
	 	
	 }
	 printf("sum is %d",sum);
	 return 0;
}

推荐答案

编译并不意味着你的代码是对的！ ：笑：

将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。



所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。



首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。

所以，你可以查看代码和很明显，它在某处：

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

int Double(int value)
   {
   return value * value;
   }

一旦你知道可能出现的问题，就开始使用teh调试器找出原因。在您的行上设置一个断点：

Once you have an idea what might be going wrong, start using teh debugger to find out why. Put a breakpoint on your line:

for(i=1;i<=n;i++)

并运行你的应用程序。在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。

如果没有，为什么不呢？它有何不同？



这是一项非常值得开发的技能，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！



是的，我可能会告诉你问题是什么 - 但这并不难做到，并且你会同时学到一些非常值得的东西！

and run your app. Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?

This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

Yes, I could probably tell you what "the problem" is - but it's not difficult to do this yourself, and you will learn something really worthwhile at the same time!

Quote：

但它没有显示想要的结果..

But it is not showing the desired result..

这不是提供信息。

因为它是一些HomeWork，我不会给你代码，只是建议，学习调试器，它是一个很棒的工具，它可以帮助你更快地学习。



拿一张纸和一支铅笔。手工解决小数字，1,2,3,4的问题。记下你期望变量如何进化，每次变化使用1行。

This is not informative.
As it is some HomeWork, I don't give you code, just advices, learn debugger, it is a great tool and it will helps you to learn faster.

Take a sheet of paper and a pencil. Solve the problem by hand for small numbers, 1, 2, 3, 4. Write down how you expect variables to evolve, use 1 line per change.

sum= 0
... f= 1!
sum= sum+ f
... f= 2!
sum= sum+ f
...

然后在调试器上启动程序，逐行执行并在变量变化时检查变量开始做一些意想不到的事情，你接近一个bug。

-----

有一个工具可以让你看到你的代码正在做什么，它的name是调试器。它也是一个很好的学习工具，因为它向你展示了现实，你可以看到哪种期望与现实相符。

当你不明白你的代码在做什么或为什么它做它做的时候，答案就是答案是调试器。

使用调试器查看代码正在执行的操作。只需设置断点并查看代码执行情况，调试器允许您逐行执行第1行并在执行时检查变量。



调试器 - 维基百科，免费的百科全书 [ ^ ]



掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]

使用Visual Studio 2010进行基本调试 - YouTube [ ^ ]

调试器在这里向您展示您的代码正在做什么，您的任务是与什么进行比较应该这样做。

调试器中没有魔法，它没有找到错误，它只是帮助你。当代码没有达到预期的效果时，你就接近了一个bug。

Then launch your program on debugger, execute line by line and inspect variables as they change, when variables start to do something unexpected, you are close to a bug.
-----
There is a tool that allow you to see what your code is doing, its name is debugger. It is also a great learning tool because it show you reality and you can see which expectation match reality.
When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.

Debugger - Wikipedia, the free encyclopedia[^]

Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

你有一个太多的循环。

试试

You have one too many loops.
Try

#include<stdio.h>
#include<math.h>

int main()
{
  int n , i, sum=0, f = 1;
  printf("enter the value of no upto which you want to print the sum ");
  scanf("%d",&n);
  for(i=1;i<=n;i++)
  {
    f=f*i;
    sum=sum+f;
  }
  printf("\nsum is %d\n",sum);
  return 0;
}

请注意，这样的程序可以在 n< 13 ：

• 13！对于 int 来说太大了（假设 32 位）。
• 21！对于 long long来说太大了（假设 64 位）。



Please note, such a program works while n<13:

• 13! is too large for a int (assuming 32 bit).
• 21! would be too large for a long long (assuming 64 bit).




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