ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<trimwhitespace>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  jmp    L60
L64:
  add    $0x1,%r12
L60:
  movzbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L64
  cmpb   $0x0,(%r12)
  jne    L40
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L40:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  cmp    %rbx,%r12
  jb     L32
  jmp    L4
  xchg   %ax,%ax
L49:
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  je     L4
L32:
  movzbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L49
L4:
  movb   $0x0,0x1(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *trimwhitespace(char *str) {
  char *end;
  while (isspace((unsigned char)*str))
    str++;
  if (*str == 0)
    return str;
  end = str + strlen(str) - 1;
  while (end > str && isspace((unsigned char)*end))
    end--;
  *(end + 1) = 0;
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<trimwhitespace>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  jmp    L54
L16:
  add    $0x1,%r12
L54:
  movzbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L16
  cmpb   $0x0,(%r12)
  jne    L4
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   (%rax)
L4:
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  cmp    %rbx,%r12
  jb     L48
  jmp    L40
  xchg   %ax,%ax
L64:
  sub    $0x1,%rbx
  cmp    %rbx,%r12
  je     L40
L48:
  movzbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L64
L40:
  movb   $0x0,0x1(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *trimwhitespace(char *str) {
  char *end;
  while (isspace((unsigned char)*str))
    str++;
  if (*str == 0)
    return str;
  end = str + strlen(str) - 1;
  while (end > str && isspace((unsigned char)*end))
    end--;
  *(end + 1) = 0;
  return str;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D47(%rip),%xmm0
  mov    $0x64,%esi
  lea    D8(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D58(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x78,%esi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D67(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x61,%esi
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D47
type: f64
name: unpack_data
arguments: data_label: D8
type: string
name: unpack_data
arguments: data_label: D58
type: string
name: unpack_data
arguments: data_label: D27
type: string
name: unpack_data
arguments: data_label: D30
type: string
name: unpack_data
arguments: data_label: D67
type: string
name: unpack_data
arguments: data_label: D23
type: string

tool: D47: 1.2

tool: D8: You have %d bugs at the imaginary rate of %f.


tool: D58: The entire universe has %ld bugs.


tool: D27: You are expected to have %f bugs.


tool: D30: That is only a %e portion of the universe.


tool: D67: Which means you should care %d%%.


tool: D23: Value is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int bugs = 100;
  double bug_rate = 1.2;
  printf("You have %d bugs at the imaginary rate of %f.\n", bugs, bug_rate);
  unsigned long universe_of_defects = 1L * 1024L * 1024L * 1024L * 1024L * 1024L * 1024L;
  printf("The entire universe has %ld bugs.\n", universe_of_defects);
  double expected_bugs = bugs * bug_rate;
  printf("You are expected to have %f bugs.\n", expected_bugs);
  double part_of_universe = expected_bugs / universe_of_defects;
  printf("That is only a %e portion of the universe.\n", part_of_universe);
  char nul_byte = '\0';
  int care_percentage = bugs * nul_byte;
  printf("Which means you should care %d%%.\n", care_percentage);
  printf("Value is %d\n", 'a');
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  mov    $0x1,%eax
  movsd  D47(%rip),%xmm0
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x78,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %esi,%esi
  lea    D44(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %esi,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x61,%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D47
datatype: f64
name: unpack_data
arguments: label: D8
datatype: string
name: unpack_data
arguments: label: D27
datatype: string
name: unpack_data
arguments: label: D11
datatype: string
name: unpack_data
arguments: label: D44
datatype: string
name: unpack_data
arguments: label: D0
datatype: string
name: unpack_data
arguments: label: D30
datatype: string

tool: D47: 1.2

tool: D8: You have %d bugs at the imaginary rate of %f.


tool: D27: The entire universe has %ld bugs.


tool: D11: You are expected to have %f bugs.


tool: D44: That is only a %e portion of the universe.


tool: D0: Which means you should care %d%%.


tool: D30: Value is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int bugs = 100;
  double bug_rate = 1.2;
  printf("You have %d bugs at the imaginary rate of %f.\n", bugs, bug_rate);
  unsigned long universe_of_defects = 1L * 1024L * 1024L * 1024L * 1024L * 1024L * 1024L;
  printf("The entire universe has %ld bugs.\n", universe_of_defects);
  double expected_bugs = bugs * bug_rate;
  printf("You are expected to have %f bugs.\n", expected_bugs);
  double part_of_universe = expected_bugs / universe_of_defects;
  printf("That is only a %e portion of the universe.\n", part_of_universe);
  char nul_byte = '\0';
  int care_percentage = bugs * nul_byte;
  printf("Which means you should care %d%%.\n", care_percentage);
  printf("Value is %d\n", 'a');
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%esi
  mov    $0x1,%eax
  movsd  D61(%rip),%xmm0
  lea    D39(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x78,%esi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %esi,%esi
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %esi,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x61,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D61
data_type: f64
name: unpack_data
arguments: label: D39
data_type: string
name: unpack_data
arguments: label: D19
data_type: string
name: unpack_data
arguments: label: D56
data_type: string
name: unpack_data
arguments: label: D64
data_type: string
name: unpack_data
arguments: label: D15
data_type: string
name: unpack_data
arguments: label: D9
data_type: string

tool: D61: 1.2

tool: D39: You have %d bugs at the imaginary rate of %f.


tool: D19: The entire universe has %ld bugs.


tool: D56: You are expected to have %f bugs.


tool: D64: That is only a %e portion of the universe.


tool: D15: Which means you should care %d%%.


tool: D9: Value is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int bugs = 100;
  double bug_rate = 1.2;
  printf("You have %d bugs at the imaginary rate of %f.\n", bugs, bug_rate);
  unsigned long universe_of_defects = 1L * 1024L * 1024L * 1024L * 1024L * 1024L * 1024L;
  printf("The entire universe has %ld bugs.\n", universe_of_defects);
  double expected_bugs = bugs * bug_rate;
  printf("You are expected to have %f bugs.\n", expected_bugs);
  double part_of_universe = expected_bugs / universe_of_defects;
  printf("That is only a %e portion of the universe.\n", part_of_universe);
  char nul_byte = '\0';
  int care_percentage = bugs * nul_byte;
  printf("Which means you should care %d%%.\n", care_percentage);
  printf("Value is %d\n", 'a');
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sprint_hinted_integer>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x128,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  movzbl (%rdx),%r13d
  cmpb   $0x2d,0x1(%rdx)
  je     L46
  movb   $0x25,0x6(%rsp)
  movb   $0x6c,0x7(%rsp)
  mov    %r13b,0x8(%rsp)
  movb   $0x0,0x9(%rsp)
  lea    0x6(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %rbp,%rdx
  call   <sprintf@plt>
L4:
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L94
  add    $0x128,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L46:
  lea    0x2(%rdx),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  movb   $0x25,0x6(%rsp)
  movb   $0x6c,0x7(%rsp)
  mov    %r13b,0x8(%rsp)
  movb   $0x0,0x9(%rsp)
  lea    0x6(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  mov    %rbp,%rdx
  call   <sprintf@plt>
  test   %r12d,%r12d
  je     L4
  lea    0x10(%rsp),%rdi
  call   <strlen@plt>
  cmp    %eax,%r12d
  jg     L10
  lea    0x1(%rax),%edx
  movslq %edx,%rdx
  movb   $0x0,0x10(%rsp,%rdx,1)
  movslq %eax,%rcx
  lea    0x10(%rsp,%rcx,1),%rdx
  lea    0xf(%rsp),%rsi
  lea    -0x1(%r12),%edi
  sub    %rdi,%rsi
  add    %rcx,%rsi
L41:
  movzbl -0x1(%rdx),%ecx
  mov    %cl,(%rdx)
  sub    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L41
  sub    %r12d,%eax
  cltq
  movb   $0x2e,0x10(%rsp,%rax,1)
  jmp    L4
L10:
  lea    0x1(%r12),%edx
  movslq %edx,%rdx
  movb   $0x0,0x10(%rsp,%rdx,1)
  cltq
  lea    -0x1(%r12),%edx
  mov    %rax,%rdi
  sub    %rdx,%rdi
  mov    %rax,%rdx
  mov    $0x30,%esi
  lea    0xf(%rsp),%r8
  movslq %r12d,%r12
  sub    %rax,%r12
  lea    0x10(%rsp,%r12,1),%rcx
  jmp    L62
L92:
  mov    %al,(%rcx,%rdx,1)
  lea    -0x1(%rdx),%rax
  cmp    %rdx,%rdi
  je     L20
  mov    %rax,%rdx
L62:
  mov    %esi,%eax
  test   %edx,%edx
  jle    L92
  movzbl (%r8,%rdx,1),%eax
  jmp    L92
L20:
  movb   $0x2e,0x10(%rsp)
  jmp    L4
L94:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sprint_hinted_integer(char *buf, long val, const char *hint, const char *units) {
  char code;
  int shift, len;
  char tmp[256];
  char fmt[10];
  code = hint[0];
  if (hint[1] == '-') {
    shift = atoi(hint + 2);
  } else
    shift = 0;
  fmt[0] = '%';
  fmt[1] = 'l';
  fmt[2] = code;
  fmt[3] = 0;
  sprintf(tmp, fmt, val);
  if (shift != 0) {
    len = strlen(tmp);
    if (shift <= len) {
      tmp[len + 1] = 0;
      while (shift--) {
        tmp[len] = tmp[len - 1];
        len--;
      }
      tmp[len] = '.';
    } else {
      tmp[shift + 1] = 0;
      while (shift) {
        if (len-- > 0)
          tmp[shift] = tmp[len];
        else
          tmp[shift] = '0';
        shift--;
      }
      tmp[0] = '.';
    }
  }
  strcpy(buf, tmp);
}
```
user: What is the c source code of the assembly code below?

```asm
<sprint_hinted_integer>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x120,%rsp
  movzbl (%rdx),%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  cmpb   $0x2d,0x1(%rdx)
  je     L24
  lea    0x10(%rsp),%r14
  mov    $0x6c25,%eax
  lea    0x6(%rsp),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  mov    %ax,0x6(%rsp)
  mov    %bl,0x8(%rsp)
  movb   $0x0,0x9(%rsp)
  call   <sprintf@plt>
L85:
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x120,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L24:
  lea    0x2(%rdx),%rdi
  lea    0x10(%rsp),%r14
  call   <atoi@plt>
  mov    $0x6c25,%edx
  mov    %r14,%rdi
  mov    %bl,0x8(%rsp)
  mov    %dx,0x6(%rsp)
  movslq %eax,%r12
  lea    0x6(%rsp),%rsi
  mov    %r13,%rdx
  movb   $0x0,0x9(%rsp)
  call   <sprintf@plt>
  test   %r12d,%r12d
  je     L85
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  cmp    %eax,%r12d
  jle    L79
  lea    0x1(%r12),%eax
  lea    -0x1(%r12),%edx
  cltq
  lea    0xf(%rsp),%rsi
  movb   $0x0,0x10(%rsp,%rax,1)
  movslq %ebx,%rax
  mov    %rax,%rcx
  sub    %rax,%r12
  sub    %rdx,%rcx
  add    %r14,%r12
  jmp    L84
  nop
L7:
  mov    %rdx,%rax
L84:
  mov    $0x30,%edx
  test   %eax,%eax
  jle    L88
  movzbl (%rsi,%rax,1),%edx
L88:
  mov    %dl,(%r12,%rax,1)
  lea    -0x1(%rax),%rdx
  cmp    %rax,%rcx
  jne    L7
  movb   $0x2e,0x10(%rsp)
  jmp    L85
  nopw   0x0(%rax,%rax,1)
L79:
  lea    0x1(%rax),%eax
  movslq %ebx,%rdi
  sub    %r12d,%ebx
  cltq
  movslq %ebx,%rbx
  movb   $0x0,0x10(%rsp,%rax,1)
  lea    -0x1(%r12),%eax
  sub    %rax,%rdi
  lea    0x1(%rax),%rdx
  lea    -0x1(%r14,%rdi,1),%rsi
  add    %r14,%rdi
  call   <memmove@plt>
  movb   $0x2e,0x10(%rsp,%rbx,1)
  jmp    L85
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sprint_hinted_integer(char *buf, long val, const char *hint, const char *units) {
  char code;
  int shift, len;
  char tmp[256];
  char fmt[10];
  code = hint[0];
  if (hint[1] == '-') {
    shift = atoi(hint + 2);
  } else
    shift = 0;
  fmt[0] = '%';
  fmt[1] = 'l';
  fmt[2] = code;
  fmt[3] = 0;
  sprintf(tmp, fmt, val);
  if (shift != 0) {
    len = strlen(tmp);
    if (shift <= len) {
      tmp[len + 1] = 0;
      while (shift--) {
        tmp[len] = tmp[len - 1];
        len--;
      }
      tmp[len] = '.';
    } else {
      tmp[shift + 1] = 0;
      while (shift) {
        if (len-- > 0)
          tmp[shift] = tmp[len];
        else
          tmp[shift] = '0';
        shift--;
      }
      tmp[0] = '.';
    }
  }
  strcpy(buf, tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sprint_hinted_integer>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x120,%rsp
  movzbl (%rdx),%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  cmpb   $0x2d,0x1(%rdx)
  je     L65
  lea    0x10(%rsp),%r14
  mov    $0x6c25,%eax
  lea    0x6(%rsp),%rsi
  mov    %r13,%rdx
  mov    %r14,%rdi
  mov    %ax,0x6(%rsp)
  mov    %bl,0x8(%rsp)
  movb   $0x0,0x9(%rsp)
  call   <sprintf@plt>
L32:
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x120,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L65:
  lea    0x2(%rdx),%rdi
  lea    0x10(%rsp),%r14
  call   <atoi@plt>
  mov    $0x6c25,%edx
  mov    %r14,%rdi
  mov    %bl,0x8(%rsp)
  mov    %dx,0x6(%rsp)
  movslq %eax,%r12
  lea    0x6(%rsp),%rsi
  mov    %r13,%rdx
  movb   $0x0,0x9(%rsp)
  call   <sprintf@plt>
  test   %r12d,%r12d
  je     L32
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  cmp    %eax,%r12d
  jle    L95
  lea    0x1(%r12),%eax
  lea    -0x1(%r12),%edx
  cltq
  lea    0xf(%rsp),%rsi
  movb   $0x0,0x10(%rsp,%rax,1)
  movslq %ebx,%rax
  mov    %rax,%rcx
  sub    %rax,%r12
  sub    %rdx,%rcx
  add    %r14,%r12
  jmp    L74
  nop
L29:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  lea    -0x1(%rax),%rdx
  cmp    %rcx,%rax
  je     L38
L92:
  mov    %rdx,%rax
L74:
  test   %eax,%eax
  jg     L29
  movb   $0x30,(%r12,%rax,1)
  lea    -0x1(%rax),%rdx
  cmp    %rcx,%rax
  jne    L92
L38:
  movb   $0x2e,0x10(%rsp)
  jmp    L32
L95:
  lea    0x1(%rax),%eax
  movslq %ebx,%rdi
  sub    %r12d,%ebx
  cltq
  movslq %ebx,%rbx
  movb   $0x0,0x10(%rsp,%rax,1)
  lea    -0x1(%r12),%eax
  sub    %rax,%rdi
  lea    0x1(%rax),%rdx
  lea    -0x1(%r14,%rdi,1),%rsi
  add    %r14,%rdi
  call   <memmove@plt>
  movb   $0x2e,0x10(%rsp,%rbx,1)
  jmp    L32
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sprint_hinted_integer(char *buf, long val, const char *hint, const char *units) {
  char code;
  int shift, len;
  char tmp[256];
  char fmt[10];
  code = hint[0];
  if (hint[1] == '-') {
    shift = atoi(hint + 2);
  } else
    shift = 0;
  fmt[0] = '%';
  fmt[1] = 'l';
  fmt[2] = code;
  fmt[3] = 0;
  sprintf(tmp, fmt, val);
  if (shift != 0) {
    len = strlen(tmp);
    if (shift <= len) {
      tmp[len + 1] = 0;
      while (shift--) {
        tmp[len] = tmp[len - 1];
        len--;
      }
      tmp[len] = '.';
    } else {
      tmp[shift + 1] = 0;
      while (shift) {
        if (len-- > 0)
          tmp[shift] = tmp[len];
        else
          tmp[shift] = '0';
        shift--;
      }
      tmp[0] = '.';
    }
  }
  strcpy(buf, tmp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<funkcja_pi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  fldz
  fstpt  -0x20(%rbp)
  fld1
  fstpt  -0x10(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L19
L15:
  fldt   -0x10(%rbp)
  fldt   0xed9(%rip)        # 2000 <_fini+0xe9c>
  fmulp  %st,%st(1)
  fildl  -0x28(%rbp)
  fdivrp %st,%st(1)
  fldt   -0x20(%rbp)
  faddp  %st,%st(1)
  fstpt  -0x20(%rbp)
  fldt   -0x10(%rbp)
  fchs
  fstpt  -0x50(%rbp)
  mov    -0x50(%rbp),%rax
  mov    -0x48(%rbp),%edx
  mov    %rax,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  addl   $0x1,-0x24(%rbp)
  addl   $0x2,-0x28(%rbp)
L19:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L15
  fldt   -0x20(%rbp)
  pop    %rbp
  ret
```

assistant: ```c
long double funkcja_pi(int p) {
  int i, j;
  long double pi, znak;
  pi = 0;
  znak = 1;
  for (i = 1, j = 0; j < p; j++, i += 2) {
    pi += (znak * 4) / i;
    znak *= (-1);
  }
  return pi;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<funkcja_pi>:
  endbr64
  test   %edi,%edi
  jle    L29
  lea    0x1(%rdi,%rdi,1),%edx
  fld1
  fldz
  mov    $0x1,%eax
  flds   0xeec(%rip)        # 2000 <_fini+0xec4>
  jmp    L39
L10:
  fxch   %st(2)
L39:
  fld    %st(2)
  fmul   %st(1),%st
  mov    %eax,-0xc(%rsp)
  fildl  -0xc(%rsp)
  fdivrp %st,%st(1)
  faddp  %st,%st(2)
  fxch   %st(2)
  fchs
  add    $0x2,%eax
  cmp    %edx,%eax
  jne    L10
  fstp   %st(0)
  fstp   %st(1)
L2:
  ret
L29:
  fldz
  jmp    L2
```

assistant: ```c
long double funkcja_pi(int p) {
  int i, j;
  long double pi, znak;
  pi = 0;
  znak = 1;
  for (i = 1, j = 0; j < p; j++, i += 2) {
    pi += (znak * 4) / i;
    znak *= (-1);
  }
  return pi;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<funkcja_pi>:
  endbr64
  test   %edi,%edi
  jle    L1
  fld1
  lea    0x1(%rdi,%rdi,1),%edx
  mov    $0x1,%eax
  fldz
  flds   0xee5(%rip)        # 2000 <_fini+0xeb4>
  jmp    L17
  nopl   (%rax)
L20:
  fxch   %st(2)
L17:
  fld    %st(2)
  mov    %eax,-0xc(%rsp)
  add    $0x2,%eax
  fmul   %st(1),%st
  fildl  -0xc(%rsp)
  fdivrp %st,%st(1)
  faddp  %st,%st(2)
  fxch   %st(2)
  fchs
  cmp    %eax,%edx
  jne    L20
  fstp   %st(0)
  fstp   %st(1)
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  fldz
  ret
```

assistant: ```c
long double funkcja_pi(int p) {
  int i, j;
  long double pi, znak;
  pi = 0;
  znak = 1;
  for (i = 1, j = 0; j < p; j++, i += 2) {
    pi += (znak * 4) / i;
    znak *= (-1);
  }
  return pi;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<funkcja_pi>:
  endbr64
  test   %edi,%edi
  jle    L22
  fld1
  lea    0x1(%rdi,%rdi,1),%edx
  mov    $0x1,%eax
  fldz
  flds   0xee5(%rip)        # 2000 <_fini+0xeb4>
  jmp    L11
  nopl   (%rax)
L15:
  fxch   %st(2)
L11:
  fld    %st(2)
  mov    %eax,-0xc(%rsp)
  add    $0x2,%eax
  fmul   %st(1),%st
  fildl  -0xc(%rsp)
  fdivrp %st,%st(1)
  faddp  %st,%st(2)
  fxch   %st(2)
  fchs
  cmp    %edx,%eax
  jne    L15
  fstp   %st(0)
  fstp   %st(1)
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  fldz
  ret
```

assistant: ```c
long double funkcja_pi(int p) {
  int i, j;
  long double pi, znak;
  pi = 0;
  znak = 1;
  for (i = 1, j = 0; j < p; j++, i += 2) {
    pi += (znak * 4) / i;
    znak *= (-1);
  }
  return pi;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pandigital>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movl   $0x1,-0x30(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L83
L11:
  mov    -0x48(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    -0x30(%rbp,%rdx,4),%eax
  test   %eax,%eax
  je     L53
  mov    $0x0,%eax
  jmp    L1
L53:
  addl   $0x1,-0x38(%rbp)
  mov    -0x48(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    -0x30(%rbp,%rdx,4),%eax
  add    $0x1,%eax
  mov    %eax,-0x30(%rbp,%rdx,4)
  mov    -0x48(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x48(%rbp)
L83:
  cmpq   $0x0,-0x48(%rbp)
  jg     L11
  movl   $0x1,-0x34(%rbp)
  jmp    L13
L38:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  test   %eax,%eax
  jne    L14
  mov    $0x0,%eax
  jmp    L1
L14:
  addl   $0x1,-0x34(%rbp)
L13:
  mov    -0x34(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jle    L38
  mov    $0x1,%eax
L1:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
bool pandigital(long n) {
  int digits[10] = {1, 0, 0, 0, 0, 0, 0, 0, 0};
  int n_digits = 0;
  for (; n > 0; n /= 10) {
    if (digits[n % 10]) {
      return false;
    }
    n_digits++;
    digits[n % 10]++;
  }
  for (int i = 1; i <= n_digits; i++) {
    if (!digits[i]) {
      return false;
    }
  }
  return true;
}
```