Question

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，我逻辑上尝试将char *再次视为整数，但我似乎错过了一些东西，因为我无法正确处理它。

#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，但它们似乎都没有正常工作。必须首先将整数转换为char *，因为WinSock-API中的recv（）方法将其读取字节存储在缓冲区char数组中。

任何解释或解决方案？提前谢谢。

Answer 1

抱歉我的原始语法。没有真正好的和完整的解释，等待一些，但这是有效的。

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

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

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

Answer 2

如果你不想使用C风格的指针，也许你可以试试这个。

void int2CharArr(int num, char* p1){
    p1[0] = num & 0xFF;
    p1[1] = (num >> 8) & 0xFF;
    p1[2] = (num >> 16) & 0xFF;
    p1[3] = (num >> 24) & 0xFF;
}

int charArr2Int(char* p1){
    return (p1[3] << 24) + (p1[2] << 16) + (p1[1] << 8) + p1[0];
}

void test(){
    int a = 0x12345678;
    char q[sizeof(int)];
    int2CharArr(a, q);
    int b = charArr2Int(q);
    printf("%x , %x , %x , %x\n", *q, *(q+1), *(q+2), *(q+3));
    printf("%x\n", b);
}

Answer 3

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，如果您每次发送时都发送相同数量的整数，则可能需要考虑发送数据块，如下所示：

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

并收到： char * Buff = new char [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];

或者如果您要发送混合数据，但每次传输的格式相同，请使用结构，如下所示：

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);

将char *转换为int

3 个答案: