c - 如何验证用户输入的数字实际上是 c 中的有效 unsigned int

Question

我正在用 c 语言编写一个程序，该程序在命令行上接收来自用户的参数，但该参数必须是有效的无符号整数。例如，如果用户输入值 -1，那么我将不得不打印一个错误代码。或者如果用户输入任何大于 4294967295 的值，那么我也会打印一个错误代码。

我不确定如何检查他们的输入是否在正确的范围内(0 和 4294967295)。

Answer 1

to verify that user input number is in fact a valid unsigned int in c

strtoul() 是正确的函数。它在接受 '-' 时会带来一些问题。让我们假设前导空格没问题，然后手动检查字符串的前端。

// 0: invalid, 1: valid
int unsigned_int_valid(const char *s) {

  while (isspace((unsigned char) *s) s++;
  // Fail on '-' as any negative number is out of range.
  if (*s == '-') return 0;  // Could add code to allow "-0"
  if (*s == '+') s++;
  if (!isdigit((unsigned char) *s) return 0;
  // Code knowns string begins with a digit

  errno = 0;     // Clear this global value as code tests it later.
  char *endptr;  // Pointer where parsing stopped.
  unsigned long ul = strtoul(s, &endptr, 0);

  // Usually this test is needed to see if _any_ conversion happened,
  // but code knowns the first character is a digit.
  #if 0
  if (s == endptr) return 0;
  #endif

  // Could march down the end of the string allowing trailing white-space
  while (isspace((unsigned char) *endptr) endptr++;

  // Extra text after the digits?
  if (*endptr) return 0;

  // Overflow?  strtoul sets `errno = ERANGE` when "outside the range" of unsigned long
  if (errno) return 0;

  #if ULONG_MAX > UINT_MAX
  // Does 'ul` value exceeds `unsigned` range?
  if (ul > UINT_MAX) return 0;
  #endif

  return 1;
}