首先检查numBunies == 0，以下是正确的代码：

int bunnyEars(int numBunnies) {

//int a = 2 * bunnyEars(bunnyEars(numBunnies-1));

if (numBunnies == 0)
{
    return 0;
}
else if (numBunnies % 2 == 0) {
    printf("%d", 3 + bunnyEars(numBunnies - 1));
    return 3 + bunnyEars(numBunnies - 1);    
}
else {
    printf("%d", 2 + bunnyEars(numBunnies - 1));
    return 2 + bunnyEars(numBunnies - 1);
}

return 0;
}

int main()
{
    bunnyEars(4);
    return 0;
}

首先，进行一次小型代码评审：

#include <stdio.h>

int bunnyEars(int numBunnies){

    //int a = 2 * bunnyEars(bunnyEars(numBunnies-1));

    if(numBunnies%2 == 0){
        // NOTE: 0 % 2 = 0
        // Always check your base case(s) first
        printf("%d", 3+bunnyEars(numBunnies-1));
        return 3+bunnyEars(numBunnies-1);}
        // NOTE: Poor formatting. Wrong math per the requirements
        // you stated. 
    else if (numBunnies==0){
        return 0;
    }
    else{
        printf("%d", 2+bunnyEars(numBunnies-1));
        return 2+bunnyEars(numBunnies-1);
    }

return 0;
}

int main()
{
    bunnyEars(4);  // This is where you should be printing the result
    return 0;
}

你需要把你的基本情况放在第一位。你的代码遇到了堆栈溢出，因为你的第一个条件对0为真，这意味着你永远不会停止递归。
你还需要对偶数的兔子做正确的计算。你说要求是K % 3，其中K是兔子在行中的位置。你的代码只是加3。例如：兔子#2应该有两只耳朵，兔子#4应该有一只耳朵，兔子#6应该没有耳朵。

#include <stdio.h>

int bunnyEars(int numBunnies) {
  if (numBunnies == 0) {
    return 0;
  }

  if (numBunnies & 1) {
    return 2 + bunnyEars(numBunnies - 1);
  } else {
    return (numBunnies % 3) + bunnyEars(numBunnies - 1);
  }
}

int main() {
  printf("%d\n", bunnyEars(4));
  return 0;
}

输出：

7