可以将上述代码综合一下吗
时间: 2024-09-28 21:06:10 浏览: 24
当然可以,以下是整合了上述步骤的完整代码片段,包括栈操作函数和输入验证部分:
```c
#include <stdio.h>
#include <stdlib.h>
// 定义栈操作函数
void push(SqStack *stack, int value) {
if (stack->top >= MAXSIZE - 1) {
printf("Stack overflow! Please increase the MAXSIZE.\n");
return;
}
stack->base[++stack->top] = value;
}
int pop(SqStack *stack) {
if (stack->top == -1) {
printf("Stack underflow!\n");
return -1;
}
return stack->base[stack->top--];
}
// 转换函数
char* convertToBase(int x, int m) {
SqStack stack = { .base = malloc(MAXSIZE * sizeof(int)), .top = -1 };
if (m < 2 || m > 16) {
free(stack.base);
printf("Invalid base! Only support 2, 8, or 16.\n");
return NULL;
}
while (x > 0) {
int remainder = x % m;
if (m == 2)
push(&stack, remainder);
else if (m == 16)
push(&stack, remainder + '0'); // 十六进制转字符
x /= m;
}
char *result = malloc((stack.top + 1) * sizeof(char)); // 包含空字符'\0'
for (int i = 0; i <= stack.top; ++i)
result[i] = stack.base[i];
// 结果字符串不需要包含前导零
while (result[0] == '0' && result[1] != '\0')
result++;
free(stack.base);
return result;
}
int main() {
int x, m;
scanf("%d %d", &x, &m);
char *converted = convertToBase(x, m);
if (converted) {
printf("%s\n", converted);
free(converted);
}
return 0;
}
```
这段代码首先检查基础是否有效,然后通过循环不断获取x除以m的余数,并压入栈中。最后,将栈顶的内容合并成字符串返回。在main函数中读取用户输入并调用转换函数。
阅读全文