[lldb] Issue with structure string

[DhruvSrivastavaX]

As part of some of our general testing, identified the following issue:

Test case:

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

int main() {

  enum empcats {management, research, clerical, sales};
  struct
  {
    char name[30];
    float salary;
    enum empcats category;
  } employee;

  strcpy (employee.name, "Benjamin Franklin");
  employee.salary = 118.50;
  employee.category = research;

  printf ("Name = %s\n", employee.name);
  printf ("Salary = %6.2f \n", employee.salary);
  printf ("Category = %d\n", employee.category);

  if (employee.category == clerical)
    printf ("Employee category is clerical \n");
  else
  {
    printf ("Employee category is not clerical. \n");
    if (employee.category == research)
    {
      printf ("PASSED! \n");
    }
  }

  return (0);

}

Complied with clang:
clang test2.c -o test2 -g

And then this error is seen with the character array access:

(lldb) p employee
((unnamed struct))  (name = "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001", salary = 118.5, category = research)
(lldb) p employee.name
(char[30]) "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001"
(lldb) fr v --show-types employee
((unnamed struct)) employee = {
  (char[30]) name = "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001"
  (float) salary = 118.5
  (empcats) category = research
}

I see this issue with Mac, Linux as well as the Demo branch in AIX.
Please check.

[llvmbot]

@llvm/issue-subscribers-lldb

Author: Dhruv Srivastava (DhruvSrivastavaX)

As part of some of our general testing, identified the following issue:

Test case:

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

int main() {

  enum empcats {management, research, clerical, sales};
  struct
  {
    char name[30];
    float salary;
    enum empcats category;
  } employee;

  strcpy (employee.name, "Benjamin Franklin");
  employee.salary = 118.50;
  employee.category = research;

  printf ("Name = %s\n", employee.name);
  printf ("Salary = %6.2f \n", employee.salary);
  printf ("Category = %d\n", employee.category);

  if (employee.category == clerical)
    printf ("Employee category is clerical \n");
  else
  {
    printf ("Employee category is not clerical. \n");
    if (employee.category == research)
    {
      printf ("PASSED! \n");
    }
  }

  return (0);

}

Complied with clang:
clang test2.c -o test2 -g

And then this error is seen with the character array access:

(lldb) p employee
((unnamed struct))  (name = "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001", salary = 118.5, category = research)
(lldb) p employee.name
(char[30]) "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001"
(lldb) fr v --show-types employee
((unnamed struct)) employee = {
  (char[30]) name = "Benjamin Franklin\0\xdfo\U00000001\0\0\0<<\U0000001c\x91\U00000001"
  (float) salary = 118.5
  (empcats) category = research
}

I see this issue with Mac, Linux as well as our Demo branch in AIX.
Please check.

[bulbazord]

I think LLDB's current behavior is expected and fine. Allow me to elaborate.

From LLDB's perspective, a char[30] is not a const char *. C-style strings are expected to end in a null terminator, and you seem to be implying lldb should stop showing you characters in the buffer when it sees the null terminator after "Benjamin Franklin". You can create a const char * from a char[30], but they are not the same thing. You can assign your own semantics to a char[30], for example, maybe it could be a const char **? Something like: "foo\0bar\0baz\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0". LLDB would want to show you the entire buffer here instead of stopping at the first c-style string it finds in the buffer.

[jimingham]

I agree with Alex. lldb has no way of knowing how you meant to use that char[30], so it should not assume you meant a null-terminated C-string since assuming that would lose information. There are many other possible uses of this data structure in which eliding everything after the first \0 would be wrong.