atoi函数实现

最新推荐文章于 2025-03-12 10:22:29 发布

murongjianke001

最新推荐文章于 2025-03-12 10:22:29 发布

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参考代码：

int atoi(const char *str)
{
	if(str == NULL || *str == '\0') {
		printf("invalid input\n");
		return 0;
	}

	int sign = 1;
	int sum = 0;
	if(*str == ' ') ++str;
	if(*str == '+') ++str;
	else if(*str == '-') {
		sign = -1;
		++str;
	}

	while(*str >= '0' && *str <= '9')
	{
		sum = sum * 10 + *str - '0';
		++str;
	}


	return sum * sign;
}

测试用例：str = "+"， str = “+aaa”；输出为0，为有效输入，其实应该是无效输入

剑指offer代码：

#include<stdio.h>
#include<stdlib.h>

enum status {kvalid = 0, kinvalid};
int flag = kinvalid;

int atoi(const char *str)
{
	if(str == NULL || *str == '\0') {
		flag = kinvalid;
		return 0;
	}

	int sign = 1;
	int sum = 0;
	if(*str == ' ') ++str;
	if(*str == '+') ++str;
	else if(*str == '-') {
		sign = -1;
		++str;
	}

	if(*str != '\0')
	{
    	while(*str != '\0')
    	{
    		if(*str >= '0' && *str <= '9')
    		{
    			sum = sum * 10 + *str - '0';
				if((sign * sum > 0x7FFFFFFF)
				    || (sign * sum < 0x80000000))
				{
					sum = 0;
					break;
				}
    			++str;
            } else {
    			sum = 0;
    			break;
    		}
    	}
    	if(*str == '\0') flag = kvalid;
	}

	return sum * sign;
}

测试用例：str = “123a” 输出为0，为无效输入，按照atoi函数的本意，输出为123