笔记

整形间的进制转换

  1. 使用输入输出格式说明符：

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    int a = 07442313637;
    int b = 1015650207;
    int c = 0x3c89979f;

    printf("方法一：\n");
    // 16进制 运行结果：3c89979f 3C89979F
    printf("%x %X\n", a, b);
    // 8进制 运行结果：7442313637 7442313637
    printf("%o %o\n", b, c);
    // 10进制 运行结果：1015650207 1015650207
    printf("%d %d\n", a, c);

  2. 使用栈进行转换：

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    #include <stdio.h>
    #include <malloc.h>

    typedef struct
    {
       	char data[10];
    	int top;
       	int bottom;
    } stack;

    stack *StackCreate()
    {
       	stack *p = (stack *)malloc(sizeof(stack));
       	if (p == NULL)
       	{
       	    return 0;
       	}
       	p->bottom = p->top = 0;
       	return p;
    }
    void StackInput(stack *p, char str)
    {
       	p->data[p->top] = str;
       	p->top++;
    }
    char StackOutput(stack *p, char str)
    {
       	if (p->top != p->bottom)
       	{
       	    str = p->data[p->top - 1];
       	    p->top--;
       	    return str;
       	}
    }
    void StackPrint(stack *p)
    {
       	while (p->top != p->bottom)
       	{
       	    printf("%c", p->data[p->top - 1]);
       	    p->top--;
       	}
    }

    int main()
    {
       	int b = 1015650207;

       	printf("方法二：\n");
       	// 16进制，修改base和index实现其他进制
       	int base = 16;
       	char index[] = "0123456789abcdef";
       	int last = 0;
       	stack *p;
       	p = StackCreate();
       	while (b > 0)
       	{
       	    last = b % base;
       	    StackInput(p, index[last]);
       	    b = (b - last) / base;
       	}
       	StackPrint(p);
       	return 0;
    }

数值字符串转换为整形

  1. 通过遍历字符数组来匹配对应字符，并返回相应下标。需要注意的是循环遍历的时候要保证少一次循环，以确保最后一位数只进行加操作。代码如下： 
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    // 字符串转整形 运行结果：1015650207
    printf("字符串转整形：\n");
    char index[] = "0123456789abcdef";
    int base = 16;
    char d[] = "3c89979f";
    int i = 0;
    int j = 0;
    int tj = 0;
    long result = 0;
    while (d[i] != '\0')
    {
        j = 0;
        while (index[j] != '\0')
        {
            if (index[j] == d[i])
            {
                break;
            }
            j++;
        }
        result = (result + tj) * base;
        tj = j;
        i++;
    }
    result += tj;
    printf("%d", result);