使用Linux shell脚本:

#include <stdio.h>
int main()
{
    int number = 30;
    char command[25];
    snprintf(command, 25, "echo $((%d %c 3)) ", number, 47);
    system( command );
    return 0;
}

请看我的另一个答案。

你可以使用(依赖于平台)内联程序集，例如，对于x86:(也适用于负数)

#include <stdio.h>

int main() {
  int dividend = -42, divisor = 5, quotient, remainder;

  __asm__ ( "cdq; idivl %%ebx;"
          : "=a" (quotient), "=d" (remainder)
          : "a"  (dividend), "b"  (divisor)
          : );

  printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
  return 0;
}

Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.

public static int DivideBy3(int a) {
    bool negative = a < 0;
    if (negative) a = Negate(a);
    int result;
    int sub = 3 << 29;
    int threes = 1 << 29;
    result = 0;
    while (threes > 0) {
        if (a >= sub) {
            a = Add(a, Negate(sub));
            result = Add(result, threes);
        }
        sub >>= 1;
        threes >>= 1;
    }
    if (negative) result = Negate(result);
    return result;
}
public static int Negate(int a) {
    return Add(~a, 1);
}
public static int Add(int a, int b) {
    int x = 0;
    x = a ^ b;
    while ((a & b) != 0) {
        b = (a & b) << 1;
        a = x;
        x = a ^ b;
    }
    return x;
}

这是c#，因为这是我手边的东西，但与c的区别应该很小。

要将一个32位数字除以3，可以将其乘以0x55555556，然后取64位结果的前32位。

现在剩下要做的就是使用位运算和移位来实现乘法…

(注意:查看下面的编辑2以获得更好的版本!)

这并不像听起来那么棘手，因为你说“没有使用[..+[…]运营商”。如果你想禁止同时使用+字符，请参见下面。

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    for (unsigned i = 0; i < by; i++)
      cmp++; // that's not the + operator!
    floor = r;
    r++; // neither is this.
  }
  return floor;
}

然后用div_by(100,3)将100除以3。

编辑:你可以继续并替换++操作符:

unsigned inc(unsigned x) {
  for (unsigned mask = 1; mask; mask <<= 1) {
    if (mask & x)
      x &= ~mask;
    else
      return x & mask;
  }
  return 0; // overflow (note that both x and mask are 0 here)
}

编辑2:稍快的版本，不使用任何包含+、-、*、/、%字符的操作符。

unsigned add(char const zero[], unsigned const x, unsigned const y) {
  // this exploits that &foo[bar] == foo+bar if foo is of type char*
  return (int)(uintptr_t)(&((&zero[x])[y]));
}

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    cmp = add(0,cmp,by);
    floor = r;
    r = add(0,r,1);
  }
  return floor;
}

我们使用add函数的第一个参数，因为不使用*字符就不能表示指针的类型，除非在函数形参列表中，其中的语法类型[]与类型* const相同。

FWIW，你可以很容易地实现一个乘法函数使用类似的技巧使用0x55555556技巧提出的AndreyT:

int mul(int const x, int const y) {
  return sizeof(struct {
    char const ignore[y];
  }[x]);
}

这真的很简单。

if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;

(当然，为了简洁起见，我省略了一些程序。)如果程序员厌倦了把这些都输入，我相信他或她可以编写一个单独的程序来为他生成这些。我碰巧认识一个能大大简化他工作的人。