isdigit的签名

int isdigit(int c);
atoi的签名
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,当我传递有效数字时,输出结果与预期不符:
$ ./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。

最佳答案

当您引用argv[1]时,它引用包含值123的字符数组。 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;
}

关于c - C:检查命令行参数是否为整数?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/29248585/

10-10 07:58