[已解决]大端字节顺序和栈存储数字的顺序很像

大可爱 · 发表于 2018-10-15 16:58:40

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本帖最后由大可爱于 2018-10-15 17:11 编辑

大端字节顺序和栈在把二进制转换为十进制的时候把数字当作字符来处理的顺序很像
是不是可以这样的？

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10
typedef char ElemType;
typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;
void InitStack(sqStack *s)
{
s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if( !s->base )
{
exit(0);
}
s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}
void Push(sqStack *s, ElemType e)
{
if( s->top - s->base >= s->stackSize )
{
s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType));
if( !s->base )
{
exit(0);
}
}
*(s->top) = e;
s->top++;
}
void Pop(sqStack *s, ElemType *e)
{
if( s->top == s->base )
{
return;
}
*e = *--(s->top);
}
int StackLen(sqStack s)
{
return (s.top - s.base);
}
int main()
{
ElemType c;
sqStack s;
int len, i, sum = 0;
InitStack(&s);
printf("请输入二进制数，输入#符号表示结束！\n");
scanf("%c", &c);
while( c != '#' )
{
Push(&s, c);
scanf("%c", &c);
}
getchar(); // 把'\n'从缓冲区去掉
len = StackLen(s);
printf("栈的当前容量是: %d\n", len);
for( i=0; i < len; i++ )
{
Pop(&s, &c);
sum = sum + (c-48) * pow(2, i);
}
printf("转化为十进制数是: %d\n", sum);
return 0;
}