C:检查命令行参数是否为整数？

Question

签名 isdigit

int isdigit(int c);

签名 atoi

int atoi(const char *nptr);

我只是想检查传递的命令行参数是否是整数.这是C代码:

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

int main(int argc, char *argv[])
{
    if (argc == 1)
        return -1;

    printf ("Hai, you have executed the program : %s\n", argv[0]);
    if (isdigit(atoi(argv[1])))
        printf ("%s is a number\n", argv[1]);
    else
        printf ("%s is not a number\n", argv[1]);
    return 0;
}

但是当我传递一个有效数字时,输出并不像预期的那样:

$ ./a.out 123
Hai, you have executed the program : ./a.out
123 is not a number
$ ./a.out add
Hai, you have executed the program : ./a.out
add is not a number

我无法弄清楚错误.

Answer 1

引用时argv[1],它引用包含值的字符数组123.isdigit函数是为单个字符输入定义的.

因此,要处理这种情况,最好定义一个函数,如下所示:

bool isNumber(char number[])
{
    int i = 0;

    //checking for negative numbers
    if (number[0] == '-')
        i = 1;
    for (; number[i] != 0; i++)
    {
        //if (number[i] > '9' || number[i] < '0')
        if (!isdigit(number[i]))
            return false;
    }
    return true;
}

Answer 2

if (isdigit(atoi(argv[1]))) 

将会:

if (isdigit(atoi("123")))

这将是:

if (isdigit(123))

这将是:

if ( 0 )

因为123代表ASCII字符'{'.