我想将 int 转换为 char 数组，其中 int 的 0-7 位是第一个 char 8-15 第二个等

Question

So here is the code but for some reason intToChar function returns "abcd" and some gibberish at the end. I was trying to change my newly created array to 0 first but then I'm getting access violation errors.

int charToInt(char x[])
{
    int z = 0;
    for (int i = 0; i < sizeof(int) / sizeof(char) && x[i]; i++)
    {
        z <<= sizeof(char) * 8;
        z |= x[i];
    }
    return z;
}
char* intToChar(int x)
{
    char* z = new char[5];
    for (int i = (sizeof(z) / sizeof(z[0]) - 1); i >= 0; i--)
    {
        z[i] = x;
        x >>= sizeof(char) * 8;
    }
    return z;
}

int main()
{
    char xd[5] = "abcd";
    cout << xd;
    cout << endl;
    int x = charToInt(xd);
    cout << x;
    cout << endl;
    char* y = intToChar(x);
    cout << y;
    cout << endl;
    return 0;
}

Answer 1

As noted in the comments, your use of sizeof(z) is sizeof(a_pointer) -- not what you want. Instead, given that sizeof(char) is always 1 , you can simply do:

#include <iostream>

int charToInt (char *x)
{
    int z = 0;
    
    for (size_t i = 0; i < sizeof(int) && x[i]; i++)
    {
        z <<= 8;
        z |= x[i];
    }
    
    return z;
}

char *intToChar (int x)
{
    char *z = new char[5];
    
    for (int i = (int)sizeof (int) - 1; i >= 0; i--)
    {
        z[i] = x;
        x >>= 8;
    }
    
    return z;
}

int main()
{
    char xd[5] = "abcd";
    std::cout << xd << '\n';
    
    int x = charToInt (xd);
    std::cout << x << '\n';
    
    char *y = intToChar(x);
    std::cout << y << '\n';
}

( note: the loop limits have been adjusted to reflect that fact you are utilizing only 4-bytes in your conversion either way)

Example Use/Output

$ ./bin/chr2int2char
abcd
1633837924
abcd

Alternatively, you can write intToChar() as follows:

char *intToChar (int x)
{
    char *z = new char[5];
    int i  = sizeof (int);
    z[i] = 0;               /* affirmatively nul-terminate z */
    
    while (i--)
    {
        z[i] = x;
        x >>= 8;
    }
    
    return z;
}

Also, if you were calling from a function other than main() , you would want to ensure you delete[] y; to avoid a memory leak, eg

    char *y = intToChar(x);
    std::cout << y << '\n';
    
    delete[] y;

Let me know if you have further questions.

Answer 2

You have couple of problems in the code you have write:

1. You do: "(sizeof(z) / sizeof(z[0]) - 1)" but size of 'z' is the size of pointer which is 8 bytes in most PCs today.

2. You haven't set the null terminator in the string, in the last byte of the char*

z[4] = '\0';

3. You forgot to delete the dynamic allocation of the char*, you should delete it in the main function after printing it.

If you will fix all of that, it will work as expected, but here you have a simple code to perform what you meant:

int charToInt(char x[])
{
    int z = 0;
    for (int i = 0; i < 4; i++)
    {
      z <<=  8;        
      z |= x[i];
    }
    return z;
}


char* intToChar(int x)
{  
  char* z = new char[5];
  for(int i = 0; i < 4; ++i) {
    z[i] = x >> (4 - i - 1) * 8;
  }
  z[4] = '\0';
  
  return z; 
}


int main()
{
  char xd[5] = "abcd";
  cout << xd << endl;
  
  int x = charToInt(xd);
  cout << x <<endl;
  
  char* y = intToChar(x);
  cout << y << endl;
  delete[] y;
  return 0;
}