二进制转化成十进制
各位大佬,为什么我的代码没有报错,但就是运行到输完二进制数按下#就直接结束了{:10_266:}{:10_266:}#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define STACK_INIT_SIZE 20
#define STACKINCREMENT10
#define MAXSIZE 10
typedef char ElemType;
typedef struct{//创建栈
ElemType *base;
ElemType *top;
int StackSize;
}sqStack;
void InitStack(sqStack *s)
{
s->base = (ElemType *)malloc(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;
printf("请输入二进制数,输入#表示结束:\n");
scanf("%c", &c);
while(c != '#')
{
Push(&s, c);
scanf("%c", &c);
}
getchar();
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);//将字符变成数字
}
printf("转化为十进制数是:%d\n", sum);
return 0;
} 本帖最后由 jackz007 于 2021-12-7 12:30 编辑
#include <stdio.h>
int main(void)
{
char c , s ;
unsigned long long d , i , n ;
for(n = 0 ; (c = getchar()) != '\n'; s = c , s = '\0') ;
if(n) {
for(i = 0 ; i < n && s >= '0' && s <= '1' ; i ++) ;
if(i == n) {
if(n < 65) {
for(d = i = 0 ; i < n ; i ++) d = d * 2 + s - '0' ; // 二进制化十进制,只要这一行代码就足够了
printf("%I64u\n" , d) ;
} else {
fprintf(stderr , "Error : data overflow.\n") ;
}
} else {
fprintf(stderr , "Error : invalid character in data string\n") ;
}
}
}
编译、运行实况:
D:\0002.Exercise\C>g++ -o x x.c
D:\0002.Exercise\C>x
111
7
D:\0002.Exercise\C>x
100000000
256
D:\0002.Exercise\C>x
1111111111111111111111111111111111111111111111111111111111111111
18446744073709551615
D:\0002.Exercise\C>
页:
[1]