利用栈实现十进制对二进制，八进制、十六进制的任意转换（N进制）

Myth丶恋晨 2022-03-25 08:52 268阅读 0赞

**1、程序代码如下（利用栈的动态分配实现）：**

#include<stdio.h>
    #include<stdlib.h>
    
    #define  ERROR   0
    #define  OK      1
    #define  INITSIZE 10
    #define  INCREMENT 2
    
    typedef int Status;
    typedef int Elemtype;
    typedef struct{
    	Elemtype *base;
    	Elemtype *top;
    	int StackSize;
    }SqStack;
    
    Status InitStack(SqStack *S)
    {
    	S->base=(Elemtype*)malloc(sizeof(Elemtype)*INITSIZE);
    	if(!S->base)
    	{
    		return ERROR;
    	}
    	S->top=S->base;
    	S->StackSize=INITSIZE;
    	return OK;
    }
    
    Status PushStack(SqStack *S,Elemtype e)
    {
    	if( S->top - S->base >= S->StackSize)
    	{
    		S->base=(Elemtype*)realloc(S->base,(S->StackSize+INCREMENT)*sizeof(Elemtype));
    		if(!S->base)
    		{
    			return ERROR;
    		}
    		S->top = S->base+S->StackSize;
    		S->StackSize+=INCREMENT;
    	}
    	*S->top=e;
    	S->top++;
    	return OK;
    }
    
    Status StackEmpty(SqStack *S)
    {
    	if(S->top == S->base)
    	{
    		return OK;
    	}
    	return ERROR;
    }
    
    Status PopStack(SqStack *S,Elemtype *e)
    {
    	if(S->top == S->base)
    	{
    		return ERROR;
    	}
    	*e=*--S->top;
    	return OK;
    }
    
    void Fun_10_2_8_16(int number,int cet)
    {
    	SqStack S;
    	Elemtype e;
    	InitStack(&S);
    	if(number==0) 
    	{
    		printf("转换后的%d进制数为:0",cet);
    		return ; 
    	}
    	while(number)
    	{
    		PushStack(&S,number%cet);
    		number=number/cet;
    	}
    	
    	printf("转换后的%d进制数为：",cet); 
    	while(!StackEmpty(&S))
    	{
    		PopStack(&S,&e);
    		switch(e)          //整型数转成字符输出 
    		{
    			case 10:
    			case 11:
    			case 12:
    			case 13:
    			case 14:
    			case 15: e+=55;break;
    			default :e+=48;
    		}
    		printf("%c",e);
    	}
    }
    
    int main()
    {
    	int number,cet;
    	printf("请输入要转换的十进制数：");
    	scanf("%d",&number);
    	printf("请输入要转换的数制：");
    	scanf("%d",&cet);
    	Fun_10_2_8_16(number,cet);
    	return 0;
    }

**2、运行截图：**  
![在这里插入图片描述][20190119204202775.png]  
![在这里插入图片描述][20190119204231497.png]  
![在这里插入图片描述][20190119204257558.png]

[20190119204202775.png]: /images/20220325/ddea9a9dbd1747a28ece7b31087a8201.png
[20190119204231497.png]: /images/20220325/fbc8e0cea9f04e008841014ce01afb4b.png
[20190119204257558.png]: /images/20220325/015b4e63fb91478e88a92bcec796c316.png