创建一个atoi函数 [英] Creating an atoi function

问题描述

我正在尝试创建自己的atoi函数.通过以下操作，我得到的返回值为0.无论我在函数中更改数字变量如何，都将获得返回值.关于修改代码有什么建议吗?

I'm attempting to create my own atoi function. With the following I'm getting a return value of 0. Whatever I change the number variable within the function is what I get as a return value. Any suggestions on modifying the code?

//my atoi function
int atoi_me(char *numstring)
{
    int number = 0;
    while((*numstring >= '0') && (*numstring <= '9'))
    {
        number = (number * 10) + (*numstring - '0');
        numstring++;
    }

    return number;
}

int main()
{
    char *number[MAXSIZE];
    int num;

    printf("Please enter a number:\n");
    scanf("%c", &number);
    num = atoi_me(*number);
    printf("%d", num);
    return 0;
}

推荐答案

1. 您要声明一个char *数组，即一个字符串数组，而不是单个字符串.您可能想要:

1. You're declaring an array of char *, that is, an array of strings, rather than a single string. You probably want:

char number[MAXSIZE];

您的scanf格式字符串错误.如果要读取字符串，则应使用%s. %c仅读取一个字符.

Your scanf format string is wrong. If you want to read a string, you should use %s. %c reads only a single character.

您的scanf参数错误-传递number本身(或者，如果愿意，可以传递&number[0])，而不是&number.

Your scanf parameter is wrong - pass number itself (or &number[0] if you prefer), not &number.

您要传递给atoi_me的参数是错误的.用number(或等效的&number[0])而不是*number调用它.

The parameter you're passing to atoi_me is wrong. Call it with number (or equivalently &number[0]), not *number.

将所有内容放在一起，您应该有一个main例程，如下所示:

Putting all of that together, you should have a main routine something like this:

int main(void)
{
    char number[MAXSIZE];
    int num;
    printf("Please enter a number: ");
    scanf("%s", number);
    num = atoi_me(number);
    printf("%d\n", num);
    return 0;
} 

编者注:scanf行可能会导致缓冲区溢出.您最好使用类似 fgets(3) 这样的函数，防范此类问题.

Editorial notes: You have a potential buffer overflow with the scanf line. You'd be better off using a function like fgets(3) that makes it easy to protect against that kind of problem.

atoi(3) 传统上也支持负数(前导- )和可选的前导+表示正数，而您的实现无法处理.

atoi(3) also traditionally supports negative numbers (with a leading -) and an optional leading + for positive numbers, which your implementation doesn't handle.

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