简体   繁体   English

检查命令行输入是否为有效整数(C)

[英]Checking if command line input is valid integer (C)

I'm trying to verify that my program's user enters a valid integer in the command line. 我正在尝试验证程序的用户是否在命令行中输入了有效的整数。 I've run into a problem: It rejects ALL input. 我遇到了一个问题:它拒绝所有输入。

Here's what I have 这就是我所拥有的

// Make sure input is a valid int
char *ptr = NULL;
long int input = strtol(argv[i+1], &ptr, 10);
if(ptr == NULL){
    userMinInt = input;
    minIntSet = true;    
}
else
    fprintf(stderr, "You must enter a valid integer for <min-int>. Using default value of %ld\n", minInt);

Code is checking if the end-pointer is NULL . 代码正在检查端点指针是否为NULL Instead, code should check for: 相反,代码应检查:

1) Does the end-pointer point to the null character '\\0' ? 1)端点指针是否指向空字符'\\0'
2) Does the end-pointer differ from the start? 2)终点指针是否与起点不同?

Additional checks below: 以下是其他检查:

char *start = argv[i+1];  // maybe should be argv[i]
char *ptr;
// set errno to 0 for subsequent check
errno = 0;
long int input = strtol(start, &ptr, 10);
if (ptr == start) {
  fprintf(stderr, "No conversion done.\n");
}
else if (*ptr != 0) {
  fprintf(stderr, "Extra data after the number.\n");
}
else if (errno) {
  fprintf(stderr, "Number outside long range.\n");
}
else if (input < INT_MIN || input > INT_MAX) {
  fprintf(stderr, "Number %ld outside int range.\n", input);
}
else {
  printf("Number is %d.\n", (int) input);
}

Thanks to peter, I've solved the problem. 感谢彼得,我解决了这个问题。 Here is the new working code: 这是新的工作代码:

// Make sure input is a valid int
char *ptr = NULL;
long int input = strtol(argv[i+1], &ptr, 10);
if(ptr != NULL && *ptr == (char)0){
    userMinInt = input;
    minIntSet = true;
}
else
    fprintf(stderr, "You must enter a valid integer for <min-int>. Using default value of %ld\n", minInt);

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM