c - 两个二进制数字串相加 - C

Question

我知道互联网上有很多关于这个问题的信息，特别是 Stack - 但我仍然陷入困境。

给定两个二进制数字字符串 - 我们必须将它们的二进制和作为字符串返回。

实际上，我想我和这个人有同样的问题:Padding and adding binary strings in C

但我做了一些补充和改动，如下所示:

    // This function validates the two strings are really representing two binary number, and we keep the size of each
int isNum(const char* num, int* sizeOfNum) {
  *sizeOfNum = 0;
  while (*num != '\0') {
    if (*num != '0' || *num != '1') {
      return 0;
    }
    *sizeOfNum += 1;
  }
  return 1;
}

char* sumOfBinString(const char* num1, const char* num2) {

    int* sizeOfNum1 = (int*)(malloc(sizeof(int));
    int* sizeOfNum2 = (int*)(malloc(sizeof(int));

    if (isNum(num1, sizeOfNum1) && isNum(num2, sizeOfNum2)) {

        if (*sizeOfNum1 > *sizeOfNum2) {
          int size = (*sizeOfNum1);
        }
        else {
          int size = (*sizeOfNum2);
        }
    }

        //This 'size' is actually the size of the longer binary number string, and plus one for the null-terminator
    char *num1Copy = (char *)malloc((size+1) * sizeof(char));
    char *num2Copy = (char *)malloc((size+1) * sizeof(char));

    int i;
    for(i = size; i >= 0; i--)
    {
      num1Copy[i] = '0';
      num2Copy[i] = '0';
    }

    for (i = *sizeOfNum1; i >= 0; i--)
      if (i == 0 && num1Copy[i]=='1') //Two's complement
        num1Copy[i] = '1';
      else
        num1Copy[size - i] = num1[*sizeOfNum1-i];

    for (i = *sizeOfNum2; i >= 0; i--)
      if (i == 0 && num2Copy[i]=='1') //Two's complement
        num2Copy[i] == '1';
      else
        num2Copy[size - i] = num2[*sizeOfNum2-i];
}

但我不知道如何从这里继续，因为在某些情况下进位可能会溢出，我需要“添加”这个数字作为 MSB。

有人可以帮助我从这里继续吗？

Answer 1

首先:

int* sizeOfNum1 = (int*)(malloc(sizeof(int));
int* sizeOfNum2 = (int*)(malloc(sizeof(int));

if (isNum(num1, sizeOfNum1) && isNum(num2, sizeOfNum2)) {

    if (*sizeOfNum1 > *sizeOfNum2) {
      int size = (*sizeOfNum1);
    }
    else {
      int size = (*sizeOfNum2);
    }
}

您不需要使用大小 bsc 的指针，这很容易忘记释放内存或陷入其他问题。您可以在 int 上执行此操作:

    int sizeOfNum1=0,sizeOfNum2=0;

    if (isNum(num1, &sizeOfNum1) && isNum(num2, &sizeOfNum2)) { //if u add & isNum can modify sizeofNum1 by itself (You give address of variable to function)

        if (sizeOfNum1 > sizeOfNum2) {
          int size = sizeOfNum1;
        }
        else {
          int size = sizeOfNum2;
        }
    }

对于加法背后的数学，就像在学校一样:

    111--- //carry line
      1100
   + 10100
    ------
    100000

在循环中，每个 sizeOfNum 有 2 个迭代器，当它达到 0 时，您只在输出中添加一个长度更长的迭代器(ofc 请记住，您需要始终检查进位，例如在 1100 迭代器将结束进位时)仍然是 1，所以你需要注意)这里有一些指南:

int i,j;    
i=(size==sizeOfNum1)?size:sizeOfNum2;
j=((i==sizeOfNum1)&&(i==size))?sizeOfNum2:sizeOfNum1;
for(;i>=0;i--,j--){
   //here you are adding
}