将 char* 转换为 int [英] Converting a char* into an int

问题描述

我目前正在尝试将整数转换为 char* 以便通过套接字发送它.在接收方法中，我在逻辑上尝试再次将 char* 视为整数，但我似乎遗漏了一些东西，因为我无法正确处理.

I'm currently trying to convert an integer into a char* in order to send it over a socket. In the receiving method I logically try to treat the char* as an integer again, but I seem to be missing something because I can't get it right.

#include <iostream>

using namespace std;

int main(int argc, char** argv) {

    int num1 = 42;          // Works
    int num2 = 100;         // Works
    int num3 = 126;         // Works
    int num4 = 517;         // Doesn't seem to work for int > 127

    char p1[sizeof(int)];
    *p1 = num1;
    char p2[sizeof(int)];
    *p2 = num2;
    char p3[sizeof(int)];
    *p3 = num3;
    char p4[sizeof(int)];
    *p4 = num4;

    void* pA = p4;
    void* pB = &num4;

    int result1 = static_cast<int>(*p1);
    int result2 = static_cast<int>(*p2);
    int result3 = static_cast<int>(*p3);
    int result4 = static_cast<int>(*p4);

    int resultV1 = *static_cast<int*>(pA);
    int resultV2 = *reinterpret_cast<int*>(p3);
    unsigned int resultV3 = static_cast<int>(*p4);
    int resultV4 = *static_cast<int*>(pB);              // Works, but I need to convert a char* into an int, not a void* into an int

    cout << "R1:        " << result1 << endl;
    cout << "Expected:  " << num1 << endl << endl;
    cout << "R2:        " << result2 << endl;
    cout << "Expected:  " << num2 << endl << endl;
    cout << "R3:        " << result3 << endl;
    cout << "Expected:  " << num3 << endl << endl;
    cout << "R4:        " << result4 << endl;
    cout << "Expected:  " << num4 << endl << endl;
    cout << "RV1:       " << resultV1 << endl;
    cout << "Expected:  " << num4 << endl << endl;
    cout << "RV2:       " << resultV2 << endl;
    cout << "Expected:  " << num4 << endl << endl;
    cout << "RV3:       " << resultV3 << endl;
    cout << "Expected:  " << num4 << endl << endl;
    cout << "RV4:       " << resultV4 << endl;
    cout << "Expected:  " << num4 << endl << endl;

    getchar();
    return 0;
}

我现在完全不知道如何解决这个问题.我尝试了其他几种方法，但它们似乎都无法正常工作.必须先将整数转换为 char*，因为 WinSock-API 中的 recv() 方法将其读取的字节存储在缓冲区字符数组中.

I have absolutely no idea how to solve this problem at this point. I tried several other ways, but none of them seemed to work correctly yet. It is essential that the integer is converted into a char* first as the recv() method in the WinSock-API stores its read bytes in a buffer char-array.

有什么解释或解决办法吗?提前致谢.

Any explanations or solutions? Thanks in advance.

推荐答案

你可以这样做:

//to a char*
char *P2 = static_cast<char *>((void*)Tst);
//from a char *
int *S1 = static_cast<int *>((void *)P2);

但是，如果每次发送时发送相同数量的整数，您可能需要考虑发送数据块，如下所示:

However, if you are sending the same number of ints each time you send, you may want to consider sending blocks of data instead, like so:

int Data[4] ={42,100,126,517};
char *C1 = static_cast<char*>((void*)Data);
send(socket,C1, sizeof(int[4]),NULL);

并接收:字符 * Buff = 新字符 [16];recv(socket, buff, sizeof(int[4]), 0);int Data = static_cast((void)buff);int R1 = 数据[0];int R2 = 数据[1];int R3 = 数据[2]；int R4 = 数据[3];

and to receive: char * Buff = new char[16]; recv(socket, buff, sizeof(int[4]), 0); int Data = static_cast((void)buff); int R1 = Data[0]; int R2 = Data[1]; int R3 = Data[2]; int R4 = Data[3];

或者如果您要发送混合数据，但每次传输的格式都相同，请使用结构体，如下所示:

or if you are sending mixed data, but the format is the same for each transmission, use structs, like so:

struct dataBlock
{
char ID[20];
int R1;
int R2;
int R3;
int R4;
};
...
dataBlock Data = {"test\0", 57,100,127,156};
char *Buff = static_cast<char*>((void*)&Data);
send(socket, Buff, sizeof(dataBlock), 0);

然后接收:

char *Buff = new char[sizeof(dataBlock)];
recv(socket, Buff, sizeof(dataBlock),0);
dataBlock * Data = static_cast<dataBlock*>((void*)Buff);

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