[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);
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