c++ - 负数的strtoul

Question

以下 C++ 程序调用 strtoul负数 1。因为在任何无符号类型中都不能表示负数，所以我原以为它会失败并返回 0

If no valid conversion could be performed, a zero value is returned.

而是返回一个大的正数

If the value read is out of the range of representable values by an unsigned long int, the function returns ULONG_MAX (defined in <climits>), and errno is set to ERANGE.

#include <cstdlib>
#include <iostream>

int main () {
  {char s[] = "-1";
    for (int b=0; b<17; ++b)
      std::cout << "strtoul (unsigned) of " << s
                << " with base arg " << b
                << ": " << strtoul(s,0,b) << std::endl;}
}

为什么 strtoul 不会失败并为负数返回 0？

Answer 1

对于 documentation，您最好使用 cppreference.com它似乎更准确:

if the minus sign was part of the input sequence, the numeric value calculated from the sequence of digits is negated as if by unary minus in the result type, which applies unsigned integer wraparound rules.

如前所述，可选的加号或减号是有效符号