解析整数而不在C中附加字符 [英] parse integer without appending char in C

问题描述

我想解析一个整数，但我下面的代码也接受像3b"这样的字符串，它以数字开头但附加了字符.我如何拒绝这样的字符串?

I want to parse an integer but my following code also accepts Strings like "3b" which start as a number but have appended chars. How do I reject such Strings?

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

int main(int argc, char *argv[])
{
    int n;
    if(argc==2 && sscanf(argv[1], "%d", &n)==1 && n>0){
        f(n);
        return 0;
    }
    else{
        exit(EXIT_FAILURE);
    }
}

推荐答案

对于你的问题，你可以使用 #include 中的 strtol() 函数代码>库.

For your problem, you can use strtol() function from the #include <stdlib.h> library.

如何使用 strtol(教程要点中的示例代码)

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

int main(){
   char str[30] = "2030300 This is test";
   char *ptr;
   long ret;

   ret = strtol(str, &ptr, 10);
   printf("The number(unsigned long integer) is %ld\n", ret);
   printf("String part is |%s|", ptr);

   return(0);
}

在strtol 中，它扫描str，将单词存储在一个指针中，然后是被转换的数字的基数.如果基数在 2 到 36 之间，则用作数字的基数.但我建议将 0 放在 10 所在的位置，这样它就会自动选择正确的数字.其余的存储在 ret 中.

Inside the strtol, it scans str, stores the words in a pointer, then the base of the number being converted. If the base is between 2 and 36, it is used as the radix of the number. But I recommend putting zero where the 10 is so it will automatically pick the right number. The rest is stored in ret.

这篇关于解析整数而不在C中附加字符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！