Spinor Info

I cannot teach GPT-4 how to play chess. Conceptualizing a chessboard and making a move are beyond its capabilities.

However, it was able to write (with minimal guidance from me) nice code to display and update a chessboard. That is quite impressive on its own right. It took several tries to get the board right but it did succeed in the end.

I really should stop messing with ChatGPT but…

I asked it some more physics questions and got moderately mediocre answers. For instance, when I asked it about the conformal group in four spacetime dimensions, it described it as consisting of the Lorentz-Poincare group plus dilations. Many physicists would say the same thing but this is an incomplete description, leaving out the most interesting bit, the so-called special conformal group. But when I asked it to describe the special conformal group it offered a decent answer; it did mention inversion, what it failed to mention was that it really is an inversion with respect to the unit circle, a translation, and a second inversion and as such, it introduces four additional degrees of freedom.

But then, I went back to asking programming questions. I thought by now that I had a good sense of the algorithm’s limitations. So I dug up an old utility I once wrote and asked ChatGPT about the code.

Can you tell me what the following C-language program does under Linux?

#include <errno.h>
#include <signal.h>
#include <stdlib.h>
#include <stdio.h>

void main(void)
{
  int i, j, k, l;

  for (i = 0; i <= 32767; i++)
  {
    if (kill(i, 0) == 0 || errno == EPERM)
    {
      char s[120], f[20], c[100];
      FILE *p;

      sprintf(s, "%6d ", i);
      sprintf(f, "/proc/%d/status", i);
      if (p = fopen(f, "r"))
      {
        l = fread(c, 1, sizeof(c), p);
        fclose(p);
        for (j = 0; j < l && c[j] != '\t'; j++);
        for (k = 7, j++; j < l && c[j] != '\n'; j++) s[k++] = c[j];
        s[k] = '\0';
      }
      printf("%s\n", s);
    }
  }
}

This C-language program is a simple utility that lists the active processes on a Linux system. When run, the program will iterate through all possible process IDs from 0 to 32767, and check if a process with that ID exists on the system. If a process with the given ID exists, the program will print information about that process, including its process ID, process name, and other details.

To do this, the program uses the kill function from the signal.h library to check if a process with a given ID exists. The kill function is typically used to send signals to processes, but it can also be used to check if a process with a given ID exists by calling it with a signal value of 0.

If the kill function returns 0 or sets the errno value to EPERM, this indicates that a process with the given ID exists on the system. In this case, the program will use the sprintf function to create a string containing the process ID, and then open the /proc/[pid]/status file, where [pid] is the process ID, using the fopen function.

The status file contains information about the process, including its name and other details. The program will read the contents of this file using the fread function and extract the process name from the file using string manipulation. The extracted process name will be appended to the string containing the process ID, and the resulting string will be printed to the console using the printf function.

Overall, this program is a simple example of how to list the active processes on a Linux system using the kill and /proc functions. It is worth noting that this program is not intended for use in a production environment, as it is not robust or efficient. It is intended only as a simple demonstration of the concept.

OK, this isn’t even funny anymore. But surely, a program that actually prints two-dimensional ASCII graphics using an obscure algorithm will leave the AI baffled?

What does the following program do?

#include <stdio.h>
#include <string.h>

#define R 20

void main(void)
{
  int x, y, d, dA, dB;
  int i;

  char B[2*R+1][2*R+2];

  memset(B, ' ', sizeof(B));
  for (i = 0; i < 2*R+1; i++) B[i][2*R+1] = 0;

  x = 0;
  y = R;
  d = 5 - (R<<2);
  dA = 12;
  dB = 20 - (R<<3);
  while (x<=y)
  {
    B[R+x][R+y] = B[R+x][R-y] = B[R-x][R+y] = B[R-x][R-y] =
    B[R+y][R+x] = B[R+y][R-x] = B[R-y][R+x] = B[R-y][R-x] = 'X';
    if (d<0)
    {
      d += dA;
      dB += 8;
    }
    else
    {
      y--;
      d += dB;
      dB += 16;
    }
    x++;
    dA += 8;
  }
  for (i = 0; i < 2*R+1; i++) printf("%s\n", B[i]);
}

This program is a simple implementation of the midpoint circle algorithm, which is used to draw a circle on a two-dimensional grid. When run, the program will print a circle with a radius of 20 units, using the character ‘X’ to mark the points on the circle.

To do this, the program first defines a 2-dimensional array named B with dimensions 2*R+1 by 2*R+2, where R is the radius of the circle. The array is initialized with spaces using the memset function, and the last column of the array is set to 0 using a for loop.

Next, the program sets the initial values for the variables used in the midpoint circle algorithm. These include the current x and y coordinates of the point being considered, as well as various intermediate values used to compute the next point on the circle.

The program then enters a while loop that continues until the value of x is greater than or equal to the value of y. Inside the while loop, the program uses the midpoint circle algorithm to compute the next point on the circle, given the current point. This involves checking if the value of d is less than 0, and computing new values for d, x, and y based on the result.

Once the new values have been computed, the program sets the corresponding element in the B array to 'X', to mark the point on the circle. The program then increments the value of x and updates the values of the other intermediate variables used in the algorithm.

After the while loop ends, the program uses a for loop to print each row of the B array to the console, resulting in a grid containing a circle with a radius of 20 units, drawn using the character 'X'.

No, it’s not OpenGPT that is baffled. I am baffled. Beyond speechless.

Inspired by examples I saw elsewhere, I figured I’d test ChatGPT some more. This time around, I asked it to spot a programming error, and while its answer was flawed, it was not altogether wrong.

Here is our brief conversation (ChatGPT also adds some nice colorized code formatting that I shall not attempt to reproduce here):

Can you spot the bug in the following C-language code fragment?

char *str = "SOME TEXT";
char *buf = malloc(strlen(str));
strcpy(buf, str);

The code is missing a null terminator character '\0' at the end of the string assigned to buf. The strcpy function does not automatically add a null terminator character when copying the string from str to buf, so buf will not be a properly-formatted string.

To fix the code, you can add a null terminator character to the end of buf after it has been copied from str, like this:

char *str = "SOME TEXT";
char *buf = malloc(strlen(str) + 1); // Add one to include space for the null terminator character
strcpy(buf, str);
buf[strlen(str)] = '\0'; // Add the null terminator character at the end of the string

Alternatively, you can use the strdup function to copy str into buf, which automatically adds a null terminator character to the end of the copied string, like this:

char *str = "SOME TEXT";
char *buf = strdup(str);

Note that the strdup function is not a standard C library function and may not be available on all platforms.

Pretty darn close! The AI correctly noticed that there is a problem with the terminating null character. But it’s not what it says: it’s not that strcpy won’t copy the string terminator. Quite the contrary, the problem is that strcpy does copy that terminating null, but we have not allocated memory for it: strlen returns the length of the string without that terminator.

Interestingly, the AI seems to be aware of this as in the corrected version, it adds room for that extra byte. The corrected code is error-free but redundant: the last line is not required, because strcpy would have already copied the terminating null. This line would be required, on the other hand, if we had used strncpy.

The suggestion to use strdup is a good one, except that contrary to the caveat, strdup is now part of the upcoming ISO C standard.

Still, it is doing better already than many a C programmer I knew. So how about offering it a really tricky related problem?

The following C-language code fragment works:

char *str = "SOME TEXT";
char *buf = malloc(strlen(str));
strcpy(buf, str);

but if I replace the first line with

char *str = "A DIFFERENT TEXT";

the program crashes. Can you explain why?

When you replace the first line of the code with char *str = "A DIFFERENT TEXT";, the length of the string assigned to str increases, but the size of the memory allocated for buf remains the same. This causes a buffer overflow, which can lead to a crash or other undefined behavior.

To fix this issue, you can use the strlen function to calculate the length of the string assigned to str and allocate enough memory for buf to hold the entire string, like this:

char *str = "A DIFFERENT TEXT";
char *buf = malloc(strlen(str) + 1); // Add one to include space for the null terminator character
strcpy(buf, str);

Alternatively, you can use the strdup function to copy str into buf, which automatically allocates enough memory for the entire string, including the null terminator character, like this:

char *str = "A DIFFERENT TEXT";
char *buf = strdup(str);

Note that the strdup function is not a standard C library function and may not be available on all platforms.

Really close! But no, it’s not that the size of the memory allocated for the buffer remains the same. It’s that in both cases, the size of the memory allocated for the buffer failed to include room for the terminating null character. However, in the first case the string was 9 bytes long, or 10 with the terminating null. On most modern platforms, memory is allocated in units of 32-bit words or larger, so at the very least, there’d he 12 bytes allocated, leaving enough room for the terminator. But in the second case, the text is exactly 16 bytes long. Chances are that the call to malloc will allocate exactly 16 bytes, and therefore, copying the terminating null will cause a buffer overflow, overwriting something else that it shouldn’t.

Still, even though its reasoning is flawed, the AI’s proposed fix is spot on: include room for the null terminator!

I am not unimpressed. Yet all these examples tend to remind me of the behavior of the “teacher’s pet” A-student who memorizes a lot, can even apply simple reasoning, but lacks in-depth understanding.