利用栈实现十进制转八进制C语言
时间: 2024-02-03 08:03:35 浏览: 23
好的,下面是十进制转八进制的栈实现过程的 C 语言代码:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE 100
typedef struct {
int top;
int data[MAX_SIZE];
} Stack;
void push(Stack *s, int num) {
if (s->top == MAX_SIZE - 1) {
printf("Stack is full\n");
exit(1);
}
s->data[++s->top] = num;
}
int pop(Stack *s) {
if (s->top == -1) {
printf("Stack is empty\n");
exit(1);
}
return s->data[s->top--];
}
int is_empty(Stack *s) {
return s->top == -1;
}
char *decimal_to_octal(int decimal_num) {
Stack s;
s.top = -1;
while (decimal_num > 0) {
int remainder = decimal_num % 8;
push(&s, remainder);
decimal_num /= 8;
}
char *octal_num = (char *)malloc(sizeof(char) * (s.top + 2));
int index = 0;
while (!is_empty(&s)) {
octal_num[index++] = pop(&s) + '0';
}
octal_num[index] = '\0';
return octal_num;
}
int main() {
int decimal_num = 123;
char *octal_num = decimal_to_octal(decimal_num);
printf("%d in decimal is %s in octal\n", decimal_num, octal_num);
free(octal_num);
return 0;
}
```
你可以将需要转换的十进制数赋值给 `decimal_num` 变量。程序将输出转换后的八进制数。注意,在使用完动态分配的字符串之后,需要使用 `free` 函数释放内存空间。