Ç分配，平均整数 [英] c assignment, averaging integers

问题描述

我工作这么昨天约5小时，得到了code。使用帮助，从这个网站的工作，但我觉得我做这是一个骗子的方式方法，我用scanf函数命令。不管怎样，我想这是否解决了正确的方法。多谢你们！哦code编译但被吐出平均是错误的。我想从概念上理解我在做什么错误以及完成这项任务。

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
双get_number（INT NUM）;
主（）{
  双N1，N2，N3;
  双平均水平;  的printf（\\ nCompute的平均3整数的\\ n）;
  的printf（-------------------------------- \\ n）;
  N1 = get_number（1）;
  N2 = get_number（2）;
  N3 = get_number（3）;
  平均=（N1 + N2 + N3）/ 3;
  的printf（平均为％0.2F \\ n，平均）;
}双get_number（INT NUM）{
  双值= 0;
  INT℃;
  的printf（请输入编号％我：NUM）;
  而（（C =的getchar（））！='\\ n'）{
    如果（（℃下'0'）||（c取代;'9'））{
      的printf（不正确输入的字符为数字 - ％I \\ N，C）;
      出口（-1）;
    }
    其他{
      C =价值;
    }
  }
  返回（值）;
}
 

解决方案

您 get_number 函数总是返回0，因为你不指定任何的值变量。我猜你想要的：

  =值值* 10 +（C  - '0'）;
 

而不是

的

  C =价值;
 

要澄清：

您正在阅读由数字一个数字的数字（与的getchar 你读单个字符）。所以我们可以说你读了数字 1 2 3 ：

当你读 1 ，您值=值* 10 +（'1' - '0'）含义值= 0 + 1 = 1 。

当你读 2 ，你做的一样，让值= 1×10 +（'2' - '0'）意值= 12 。

同为 3 。

你所需要了解的是，乘以10的东西在的东西年底增加了0。添加一个数字的东西，以0结尾，替换0由该数字。

您还必须明白，你正在处理的字符，这实际上是整数。字符 0 重新$ P $通过假设 X 的ASCII code psented。 1 重新由psented $ P $ X + 1 如果你要打印的价值等。 0'+'3'它不会是 3 如你所期望。这就是为什么我们减去 0 （这是 X ）的字符，以获得实际的数字。

希望扫清一些事情了。

I worked on this yesterday about 5 hours and got the code to work using help from this site, but I think the way I did it was a cheater way, I used a scanf command. Anyways I want to fix this the correct way. Thanks guys! Oh the code compiles but the average that gets spit out is wrong. I would like to conceptually understand what I am doing wrong as well as finish the assignment.

#include <stdio.h> 
#include <stdlib.h>
double get_number(int num);


main () {
  double n1,n2,n3;
  double average;

  printf("\nCompute the average of 3 integers\n");
  printf("--------------------------------\n");
  n1 = get_number(1);
  n2 = get_number(2);
  n3 = get_number(3);
  average = (n1 + n2 + n3)/3;
  printf("The average is %0.2f\n",average);
}

double get_number(int num) { 
  double value = 0;
  int c;
  printf("Please input number %i: ", num);
  while ((c = getchar()) != '\n') { 
    if ( (c<'0') || (c>'9') ) { 
      printf("Incorrect character entered as a number - %i\n",c);
      exit(-1);
    }
    else {
      c =value;
    }
  }
  return(value);
}

解决方案

Your get_number function always returns 0, because you never assign anything to the value variable. I'm guessing you want:

value = value*10 + (c - '0');

Instead of

c = value;

To clarify:

You are reading a number digit by digit (with getchar you read a single character). So let's say you read the digits 1 2 3:

When you read 1, you do value = value*10 + ('1' - '0') meaning value = 0 + 1 = 1.

When you read 2, you do the same and get value = 1*10 + ('2' - '0') meaning value = 12.

Same for 3.

What you need to understand is that multiplying something by 10 adds a 0 at the end of that something. Adding a digit to something that ends with a zero replaces that 0 by that digit.

You must also understand that you are dealing with characters, which are actually integers. The character '0' is represented by an ASCII code of let's say x. '1' is represented by x + 1 etc. If you were to print the value of '0'+'3' it would not be 3 as you'd expect. This is why we subtract '0' (which is x) from the characters, to get the actual digits.

Hope that clears some things up.

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