Converting a char* into an int - c++

Question

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;
}
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.
Any explanations or solutions? Thanks in advance.

Answer 1

If you don't want to use C-style pointers, perhaps you can try this.
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 2

Sorry for my primitive syntax. Have no really good and full explanation, wait for some, but this works.
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 3

You could do it like this:
//to a char*
char *P2 = static_cast<char *>((void*)Tst);
//from a char *
int *S1 = static_cast<int *>((void *)P2);
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);
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];
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);
and then to receive:
char *Buff = new char[sizeof(dataBlock)];
recv(socket, Buff, sizeof(dataBlock),0);
dataBlock * Data = static_cast<dataBlock*>((void*)Buff);