temp1=temp1>>1

temp1=temp1>>1是C++中的位运算，意思是将temp1向右移动1位，即将temp1的二进制数表示向右移动一位。这个操作会将temp1的二进制数表示的每一位向右移动一位，并在左边补0。这个操作相当于将temp1除以2取整，因为二进制中向右移动一位相当于除以2取整。这个操作可以用来对二进制数进行快速的除以2操作。

相关问题

用python语言翻译如下代码://线性移位寄存器生成密钥 int IFSR(int K) { int feature = 143; int temp1, temp2; temp1 = feature & K; while (temp1) { temp2 = (temp2 + temp1 & 1) % 2; temp1 = temp1 >> 1; } K = ((K << 1) & 255) + temp2; return K; } //生成16轮密钥 void getKeys(int round) { for (int i = 0; i < round; i++) { int temp = 0; for (int j = 0; j < 8; j++) { K = IFSR(K); temp += K & 1; temp = temp << 1; } temp = temp >> 1; Keys[i] = temp; } } //轮函数 int F(int RE, int Ki) { Ki = Ki >> 4; int temp = Ki ^ RE; return temp; } //明文加密 void encrypy(int round) { int m, temp1, temp_j; int le, re; m = strlen(str_plaincode); for (int i = 0; i < m; i++) { int temp1; if (i == 0) temp1 = IV; else temp1 = ((IV << j) ^ temp_j) & 255; le = temp1 >> 4; re = temp1 & 15; for (int j = 0; j < round; j++) { int temp2 = le; le = re; re = (temp2 ^ F(re, Keys[j])); } temp_j = (re << 4) + le; temp_j = temp_j >> (8 - j); str_ciphertext[i] = str_plaincode[i] ^ temp_j; } } //解密

以下是Python代码的翻译：

# 线性移位寄存器生成密钥
def IFSR(K):
    feature = 143
    temp1, temp2 = 0, 0
    temp1 = feature & K
    while temp1:
        temp2 = (temp2 + temp1 & 1) % 2
        temp1 >>= 1
    K = ((K << 1) & 255) + temp2
    return K

# 生成16轮密钥
def getKeys(round):
    global Keys
    Keys = [0] * round
    for i in range(round):
        temp = 0
        for j in range(8):
            global K
            K = IFSR(K)
            temp += K & 1
            temp <<= 1
        temp >>= 1
        Keys[i] = temp

# 轮函数
def F(RE, Ki):
    Ki = Ki >> 4
    temp = Ki ^ RE
    return temp

# 明文加密
def encrypy(round):
    global str_ciphertext
    m = len(str_plaincode)
    for i in range(m):
        temp1 = 0
        if i == 0:
            temp1 = IV
        else:
            temp1 = ((IV << j) ^ temp_j) & 255
        le = temp1 >> 4
        re = temp1 & 15
        for j in range(round):
            temp2 = le
            le = re
            re = (temp2 ^ F(re, Keys[j]))
        temp_j = (re << 4) + le
        temp_j = temp_j >> (8 - j)
        str_ciphertext.append(str_plaincode[i] ^ temp_j)

# 解密
def decode():
    global test
    m = len(str_ciphertext)
    for i in range(m):
        temp1 = 0
        if i == 0:
            temp1 = IV
        else:
            temp1 = ((IV << j) ^ temp_j) & 255
        le = temp1 >> 4
        re = temp1 & 15
        for j in range(round):
            temp2 = le
            le = re
            re = (temp2 ^ F(re, Keys[round-j-1]))
        temp_j = (re << 4) + le
        temp_j = temp_j >> (8 - j)
        test += chr(str_ciphertext[i] ^ temp_j)

struct node* insert(struct node* head, int newData, int targetData) { struct node *temp, *p, *q; p = NULL; // 创建一个节点p p = (struct node*) malloc (sizeof(struct node)); p ->data = newData; p->next = NULL; // 查找存储targetData的节点 temp = head ; while (temp != NULL) { if (temp ->data == targetData){ return head;} temp = temp -> next;} temp = head; while (temp != NULL) { printf("%d ", temp->data); temp = temp->next; } printf("\n"); return head; } // 如果找到存储targetData的节点 if (temp != NULL) { // 将节点p插入到链表存储targetData节点temp的后面 p -> data = newData; p -> next = temp -> next; temp -> next = p; } return head; }

在insert函数中，查找存储targetData的节点的逻辑存在问题。如果找到该节点，则直接返回原链表的头指针head，没有进行插入操作。如果没有找到该节点，则会遍历整个链表并输出链表的值。正确的逻辑应该是，如果找到存储targetData的节点，则将新节点插入到该节点的后面；如果没有找到该节点，则返回原链表的头指针head，不进行插入操作。

以下是修正后的代码：

struct node* insert(struct node* head, int newData, int targetData) {
    struct node *temp = head, *p;
    p = (struct node*)malloc(sizeof(struct node));
    p->data = newData;
    p->next = NULL;

    while (temp != NULL) {
        if (temp->data == targetData) { // 找到存储targetData的节点
            p->next = temp->next; // 将节点p插入到存储targetData节点temp的后面
            temp->next = p;
            return head;
        }
        temp = temp->next;
    }

    // 没有找到存储targetData的节点
    return head;
}