用C转换十进制数为二进制 [英] Converting decimal number to binary in C

问题描述

我想一个十进制数转换为二进制，但我不知何故最终得到随机ASCII符号作为输出。下面是我的程序：

 的#include＆LT;＆stdio.h中GT;诠释的main（）
{
    INT数= 0;
    INT I = 50;
    烧焦二进制[10];
    的printf（请输入一个数字：）;
    scanf函数（％d个，＆安培;号码）;
     而（数字！= 0）{
         二元[I] =编号％2;
         一世 - ;
         数=数/ 2;
    }
    的printf（\\ N％的，二进制）;    返回0;
}
 

解决方案

有您的code多的问题（没有特定的顺序，但一个我发现他们）：

• 您正在初始化字符与 INT 在行二进制[我] =编号％2 ;你应适当变换整数一个字符，像这样的：

   二元[I] ='0'+（编号％2）;
   




• 您正在写的printf（\\ n％S，二进制）; ，其中二进制是一个的char [10] ，因此不一定空值终止。你应该多一个字符添加到二进制并初始化它 \\ 0 ：

   二进制字符[11];
  二进制[10] ='\\ 0';
   




• 您错误地初始化您的数组：您开始使用 I = 50 和二进制[10] 。你应该添加一个常数，并使用它两次：

   的#define MAX_SIZE 32
  INT I = MAX_SIZE  -  1;
  焦炭二进制[MAX_SIZE + 1];
  二进制[MAX_SIZE] ='\\ 0';
   

  请注意的是避免了您的字符串结束后写的，而 +1 1 你忘了， C>这基本上是最后一个问题




• 您正在阅读的 INT 并试图获得其二进制重新presentation;但（真）的二进制重新$ P $负值的psentation通常不是你所期望的。 preFER读取 unsigned int类型：

   无符号整型数;
  scanf函数（％U，＆安培;号码）;
   




• 您可以取代号％2 与数字功放＆; 1 和数/ = 2 与若干＆GT;＆GT; = 1 ;这是更重presentative你在号上运行什么（二）操作。此外，这实际上应该正确的负整数，你的版本应该不会（未经测试，反正你平时没有理由希望得到一个负整数的二进制重新presentation）的工作：

   二进制[我 - ] ='0'+（数字放大器和1）;
  数字＆GT;＆GT; = 1;
   




• 您停止后您的号码是零，但是这意味着你没有必要填写二进制阵列;这意味着你可以在二进制（VC ++的调试模式这应该IIRC显示为是字符）。你应该添加，你的循环后，一些像这样的，从而填补了数组：

   而（I＆GT; = 0）
      二进制[我 - ] ='0';
   




希望帮助！

I'm trying to convert a decimal number to binary but I somehow end up getting 'random' ASCII symbols as an output. Here is my program:

#include <stdio.h>

int main()
{
    int number = 0;
    int i = 50;
    char binary[10];
    printf("Enter a number: ");
    scanf("%d", &number);
     while(number!=0){
         binary[i] = number % 2;
         i--;
         number = number / 2;
    }
    printf("\n%s", binary);

    return 0;
}

解决方案

There are multiple problems with your code (no specific order but the one I found them):

• You are initializing a char with an int at line binary[i] = number % 2; you should properly transform the integer to a character, like this:

  binary[i] = '0' + (number % 2);
  

• You are writing printf("\n%s", binary); where binary is a char[10], thus not necessarily null-terminated. You should add one more character to binary and initialize it to \0:

  char binary[11];
  binary[10] = '\0';
  

• You are wrongly initializing your array: you start with i = 50 and binary[10]. You should add a constant and use it twice:

  #define MAX_SIZE 32
  int i = MAX_SIZE - 1;
  char binary[MAX_SIZE + 1];
  binary[MAX_SIZE] = '\0';
  

  Note the -1 you forgot, that avoids that you write after the end of the string, and the +1 that is essentially the last problem

• You are reading an int and trying to get its binary representation; but the (true) binary representation of a negative value is usually not what you would expect. Prefer reading an unsigned int:

  unsigned int number;
  scanf("%u", &number);
  

• You could replace number % 2 with number & 1 and number /= 2 with number >>= 1; which is more representative of exactly what (binary) operations you are running on number. Besides, this should actually work correctly on negative integers, where your version should not (untested, anyway you usually have no reason of wanting to get the binary representation of a negative integer):

  binary[i--] = '0' + (number & 1);
  number >>= 1;
  

• You stop after your number is zero, but that means you did not necessarily fill the binary array; this implies you may have uninitialized characters at the beginning of binary (on VC++ in debug mode this should IIRC be displayed as ÿ characters). You should add, after your loop, something such as this so as to fill the array:

  while (i >= 0)
      binary[i--] = '0';
  

Hope that helps!

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