Question

//想法是，如何将用户限制为这些特定符号？ num_1 / 2 iput中的p.s为：10有效asd10无效

do
    {
        //Input
        cout << "Enter first the arithmatic operation (+, -, *, /, %) and then both operands: "; cin >> operation; cin >> num_1; cin>> num_2;
        if (cin.fail()) {
            cin.clear();
            cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
            continue;
        }

    } while (   (operation != '+' && operation != '-' && operation != '*' && operation != '/' && operation != '%' )
            && ((num_1 <= '0' || num_1 <= '9') || (num_1 >= 'A' && num_1 <= 'F'))
            && ((num_2 >= '0' && num_2 <= '9') || (num_2 >= 'A' && num_2 <= 'F'))   );

Answer 1

标准库使您几乎无法控制这些东西。

这是个老歌，但是当我需要花哨的输入时，readline仍然是我的首选：

#include <cstdio>
#include <cstdlib>
#include <readline/readline.h>

int snuff_key(int, int) { return 0; }

bool is_key_allowed(char c) {
    return (c >= '0' && c <= '9')
        || (c >= 'A' && c <= 'F')
        || (c == '*')
        || (c == '/')
        || (c == '+')
        || (c == '-');
}

int main() {
  char* buf;

  rl_initialize();

  for (unsigned char c = 0; c < 128; c++) {
      if (!(is_key_allowed(c))) {
          rl_bind_key(c, snuff_key);
      }
  }

  if ((buf = readline("")) != nullptr) {
    printf("you typed: '%s'\n", buf);
    free(buf);   
  }

  return 0;
}

有没有办法限制字符/字符串？

1 个答案: