hexa到dec转换 [英] hexa to dec conversion

问题描述

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

void main()
{
  int dechex=0;
  int hexdec;
  printf("Please enter your input in Hexadecimal form: ");

  hexdec = getchar();

  while(hexdec != '\n')
  {
    if('0' <= hexdec && hexdec <= '9')
    {
      dechex = dechex * 16;
      dechex = dechex + (hexdec - '0');
      break;
    }
    else if('A' <= hexdec && hexdec <= 'F')
    {
      dechex = dechex * 16;
      dechex = dechex + (hexdec - 'A')+10;
      break;
    }
    else if('a' <= hexdec && hexdec <= 'f')
    {
      dechex = dechex * 16;
      dechex = dechex + (hexdec - 'a')+10;
      break;
    }
    else
    {
      dechex=0;
    }

    hexdec = getchar();
  }
  
  printf("\nYour input number in decimal form is %d",dechex);
}

请帮帮我们我们错了o / p ....

please help me guys am getting wrong o/p ....

thanku but output is giving
input-5 ;output-5
input-fff;output-15
input-aa;output-10
why this happens

请帮助大家

please help guys

推荐答案

您的 dechex 变量未初始化。因此它在执行程序时可能包含任何起始值。



因此将其初始化为零：

Your dechex variable is not initialised. So it may contain any start value when executing your program.

So initialise it with zero:

void main()
{
  int dechex = 0;
  // ...
}

使用stdlib的 strtol [ ^ ]功能...

1.获取你的十六进制字符串

2.在十六进制字符串的begginnig添加'0x'

3.调用strtol

Use stdlib's strtol[^] function...
1. Get your hex-string
2. Add '0x' at the begginnig of the hex-string
3. Call strtol

只是我们解决了我的问题通过删除代码中的中断自我.... lolll

just f guys i solved it my self by removing breaks in code....lolll

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