ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<is_white>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  cmp    $0x20,%eax
  seta   %dl
  test   %dl,%dl
  jne    L6
  movabs $0x100003600,%rdx
  mov    %eax,%ecx
  shr    %cl,%rdx
  mov    %rdx,%rax
  and    $0x1,%eax
  test   %rax,%rax
  setne  %al
  test   %al,%al
  je     L6
  mov    $0x1,%eax
  jmp    L4
L6:
  nop
  mov    $0x0,%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int is_white(char c) {
  switch (c) {
  case ' ':
  case '\t':
  case '\f':
  case '\n':
  case '\r':
    return 1;
  default:
    break;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_white>:
  endbr64
  cmp    $0x20,%dil
  ja     L1
  movabs $0x100003600,%rax
  mov    %edi,%ecx
  shr    %cl,%rax
  and    $0x1,%eax
  ret
L1:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int is_white(char c) {
  switch (c) {
  case ' ':
  case '\t':
  case '\f':
  case '\n':
  case '\r':
    return 1;
  default:
    break;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_white>:
  endbr64
  cmp    $0x20,%dil
  ja     L7
  movabs $0x100003600,%rax
  mov    %edi,%ecx
  shr    %cl,%rax
  and    $0x1,%eax
  ret
  nopl   (%rax)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
int is_white(char c) {
  switch (c) {
  case ' ':
  case '\t':
  case '\f':
  case '\n':
  case '\r':
    return 1;
  default:
    break;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_white>:
  endbr64
  cmp    $0x20,%dil
  ja     L8
  movabs $0x100003600,%rax
  mov    %edi,%ecx
  shr    %cl,%rax
  and    $0x1,%eax
  ret
  nopl   (%rax)
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
int is_white(char c) {
  switch (c) {
  case ' ':
  case '\t':
  case '\f':
  case '\n':
  case '\r':
    return 1;
  default:
    break;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x49(%rbp)
  movl   $0x0,-0x48(%rbp)
  movl   $0x0,-0x44(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x44(%rbp)
  jmp    L2
L5:
  mov    -0x48(%rbp),%eax
  cltq
  movzbl -0x40(%rbp,%rax,1),%eax
  mov    %al,-0x49(%rbp)
  mov    -0x44(%rbp),%eax
  cltq
  movzbl -0x40(%rbp,%rax,1),%edx
  mov    -0x48(%rbp),%eax
  cltq
  mov    %dl,-0x40(%rbp,%rax,1)
  mov    -0x44(%rbp),%eax
  cltq
  movzbl -0x49(%rbp),%edx
  mov    %dl,-0x40(%rbp,%rax,1)
  addl   $0x1,-0x48(%rbp)
  subl   $0x1,-0x44(%rbp)
L2:
  mov    -0x48(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L5
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D26
data_type: string
name: retrieve_data
arguments: label: D13
data_type: string
name: retrieve_data
arguments: label: D25
data_type: string

tool: D26: 
 Input string: 

tool: D13: %s

tool: D25: 
 Reversed string is: %s 

assistant: ```c
int main() {
  char str[50], rts = 0;
  int k = 0, n = 0;
  printf("\n Input string: ");
  scanf("%s", str);
  n = strlen(str) - 1;
  while (k < n) {
    rts = str[k];
    str[k] = str[n];
    str[n] = rts;
    k++;
    n--;
  }
  printf("\n Reversed string is: %s ", str);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L6
  mov    %rsp,%rsi
  movslq %eax,%rdx
  add    %rbx,%rdx
L11:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  mov    %r8b,(%rsi)
  mov    %dil,-0x1(%rdx)
  mov    %ecx,%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %eax,%edi
  sub    %r8d,%edi
  cmp    %edi,%ecx
  jg     L11
L6:
  mov    %rsp,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
data_type: string
name: extract_symbolic_data
arguments: label: D9
data_type: string
name: extract_symbolic_data
arguments: label: D3
data_type: string

tool: D17: 
 Input string: 

tool: D9: %s

tool: D3: 
 Reversed string is: %s 

assistant: ```c
int main() {
  char str[50], rts = 0;
  int k = 0, n = 0;
  printf("\n Input string: ");
  scanf("%s", str);
  n = strlen(str) - 1;
  while (k < n) {
    rts = str[k];
    str[k] = str[n];
    str[n] = rts;
    k++;
    n--;
  }
  printf("\n Reversed string is: %s ", str);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L23
  movslq %eax,%rdx
  mov    %rbp,%rsi
  add    %rbp,%rdx
  nop
L25:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %r8b,-0x1(%rsi)
  mov    %dil,(%rdx)
  mov    %eax,%edi
  sub    %ecx,%edi
  sub    $0x1,%edi
  cmp    %edi,%ecx
  jg     L25
L23:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_infered: string
name: retrieve_data
arguments: data_tag: D18
type_infered: string
name: retrieve_data
arguments: data_tag: D13
type_infered: string

tool: D2: 
 Input string: 

tool: D18: %s

tool: D13: 
 Reversed string is: %s 

assistant: ```c
int main() {
  char str[50], rts = 0;
  int k = 0, n = 0;
  printf("\n Input string: ");
  scanf("%s", str);
  n = strlen(str) - 1;
  while (k < n) {
    rts = str[k];
    str[k] = str[n];
    str[n] = rts;
    k++;
    n--;
  }
  printf("\n Reversed string is: %s ", str);
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ecx
  test   %ecx,%ecx
  jle    L29
  movslq %eax,%rdx
  mov    %rbp,%rsi
  add    %rbp,%rdx
  nop
L11:
  movzbl (%rsi),%edi
  movzbl -0x1(%rdx),%r8d
  sub    $0x1,%ecx
  add    $0x1,%rsi
  sub    $0x1,%rdx
  mov    %r8b,-0x1(%rsi)
  mov    %dil,(%rdx)
  mov    %eax,%edi
  sub    %ecx,%edi
  sub    $0x1,%edi
  cmp    %edi,%ecx
  jg     L11
L29:
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
type_infered: string
name: extract_symbolic_data
arguments: tag: D2
type_infered: string
name: extract_symbolic_data
arguments: tag: D14
type_infered: string

tool: D4: 
 Input string: 

tool: D2: %s

tool: D14: 
 Reversed string is: %s 

assistant: ```c
int main() {
  char str[50], rts = 0;
  int k = 0, n = 0;
  printf("\n Input string: ");
  scanf("%s", str);
  n = strlen(str) - 1;
  while (k < n) {
    rts = str[k];
    str[k] = str[n];
    str[n] = rts;
    k++;
    n--;
  }
  printf("\n Reversed string is: %s ", str);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<print_sys_cfg_eapol_gwk_hsk_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D36
data_type: string
name: disassemble_data
arguments: data_tag: D38
data_type: string
name: disassemble_data
arguments: data_tag: D28
data_type: string
name: disassemble_data
arguments: data_tag: D33
data_type: string

tool: D36: 
Usage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]


tool: D38: 
If TIMEOUT and number of RETRIES are both provided, a 'set' is performed.

tool: D28: 
If no parameter is provided,a 'get' is performed.

tool: D33: 
TIMEOUT and number of RETRIES should be greater than or equal to zero


assistant: ```c
void print_sys_cfg_eapol_gwk_hsk_usage(void) {
  printf("\nUsage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]\n");
  printf("\nIf TIMEOUT and number of RETRIES are both provided, a 'set' is performed.");
  printf("\nIf no parameter is provided,a 'get' is performed.");
  printf("\nTIMEOUT and number of RETRIES should be greater than or equal to zero\n");
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_sys_cfg_eapol_gwk_hsk_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D22(%rip),%rdi
  call   <printf@plt>
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D28
data_type: string
name: extract_symbolic_data
arguments: data_tag: D22
data_type: string
name: extract_symbolic_data
arguments: data_tag: D17
data_type: string
name: extract_symbolic_data
arguments: data_tag: D29
data_type: string

tool: D28: 
Usage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]


tool: D22: 
If TIMEOUT and number of RETRIES are both provided, a 'set' is performed.

tool: D17: 
If no parameter is provided,a 'get' is performed.

tool: D29: 
TIMEOUT and number of RETRIES should be greater than or equal to zero


assistant: ```c
void print_sys_cfg_eapol_gwk_hsk_usage(void) {
  printf("\nUsage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]\n");
  printf("\nIf TIMEOUT and number of RETRIES are both provided, a 'set' is performed.");
  printf("\nIf no parameter is provided,a 'get' is performed.");
  printf("\nTIMEOUT and number of RETRIES should be greater than or equal to zero\n");
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<print_sys_cfg_eapol_gwk_hsk_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
type_infered: string
name: parse_data
arguments: data_label: D4
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D21
type_infered: string

tool: D8: 
Usage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]


tool: D4: 
If TIMEOUT and number of RETRIES are both provided, a 'set' is performed.

tool: D10: 
If no parameter is provided,a 'get' is performed.

tool: D21: 
TIMEOUT and number of RETRIES should be greater than or equal to zero


assistant: ```c
void print_sys_cfg_eapol_gwk_hsk_usage(void) {
  printf("\nUsage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]\n");
  printf("\nIf TIMEOUT and number of RETRIES are both provided, a 'set' is performed.");
  printf("\nIf no parameter is provided,a 'get' is performed.");
  printf("\nTIMEOUT and number of RETRIES should be greater than or equal to zero\n");
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_sys_cfg_eapol_gwk_hsk_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D0
datatype: string
name: parse_data
arguments: label: D19
datatype: string
name: parse_data
arguments: label: D4
datatype: string
name: parse_data
arguments: label: D30
datatype: string

tool: D0: 
Usage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]


tool: D19: 
If TIMEOUT and number of RETRIES are both provided, a 'set' is performed.

tool: D4: 
If no parameter is provided,a 'get' is performed.

tool: D30: 
TIMEOUT and number of RETRIES should be greater than or equal to zero


assistant: ```c
void print_sys_cfg_eapol_gwk_hsk_usage(void) {
  printf("\nUsage : sys_cfg_eapol_gwk_hsk [<TIMEOUT> <RETRIES>]\n");
  printf("\nIf TIMEOUT and number of RETRIES are both provided, a 'set' is performed.");
  printf("\nIf no parameter is provided,a 'get' is performed.");
  printf("\nTIMEOUT and number of RETRIES should be greater than or equal to zero\n");
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<string_nconcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L99
L28:
  addl   $0x1,-0x10(%rbp)
L99:
  cmpq   $0x0,-0x18(%rbp)
  je     L111
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L111:
  movl   $0x0,-0xc(%rbp)
  jmp    L31
L130:
  addl   $0x1,-0xc(%rbp)
L31:
  cmpq   $0x0,-0x20(%rbp)
  je     L114
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L130
L114:
  mov    -0x24(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  cmovbe %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L86
  mov    $0x0,%eax
  jmp    L16
L86:
  movl   $0x0,-0x10(%rbp)
  jmp    L59
L122:
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%ecx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
L59:
  cmpq   $0x0,-0x18(%rbp)
  je     L69
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L122
L69:
  movl   $0x0,-0xc(%rbp)
  jmp    L17
L1:
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  add    %ecx,%edx
  mov    %edx,%ecx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L17:
  cmpq   $0x0,-0x20(%rbp)
  je     L115
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jb     L1
L115:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
char *string_nconcat(char *s1, char *s2, unsigned int n) {
  unsigned int i, j;
  char *p;
  for (i = 0; s1 && s1[i]; ++i)
    ;
  for (j = 0; s2 && s2[j]; ++j)
    ;
  n = (n > j ? j : n);
  p = malloc(sizeof(char) * (i + n + 1));
  if (p == NULL)
    return (NULL);
  for (i = 0; s1 && s1[i]; ++i)
    p[i] = s1[i];
  for (j = 0; s2 && j < n; ++j)
    p[i + j] = s2[j];
  p[i + j] = '\0';
  return (p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_nconcat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x0,%ecx
  test   %rdi,%rdi
  jne    L114
  jmp    L3
L135:
  add    $0x1,%ecx
L114:
  mov    %ecx,%eax
  cmpb   $0x0,(%rbx,%rax,1)
  jne    L135
L3:
  mov    $0x0,%eax
  test   %rbp,%rbp
  jne    L99
  jmp    L42
L115:
  add    $0x1,%eax
L99:
  mov    %eax,%esi
  cmpb   $0x0,0x0(%rbp,%rsi,1)
  jne    L115
L42:
  cmp    %eax,%edx
  cmovbe %edx,%eax
  mov    %eax,%r12d
  lea    0x1(%rcx,%rax,1),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L139
  mov    $0x0,%ecx
  test   %rbx,%rbx
  jne    L75
  jmp    L32
L2:
  mov    %dl,(%rax,%rsi,1)
  add    $0x1,%ecx
L75:
  mov    %ecx,%esi
  movzbl (%rbx,%rsi,1),%edx
  test   %dl,%dl
  jne    L2
L32:
  test   %rbp,%rbp
  je     L96
  mov    $0x0,%edx
  test   %r12d,%r12d
  je     L96
L117:
  movzbl 0x0(%rbp,%rdx,1),%edi
  lea    (%rcx,%rdx,1),%esi
  mov    %dil,(%rax,%rsi,1)
  add    $0x1,%rdx
  cmp    %edx,%r12d
  ja     L117
  test   %r12d,%r12d
  mov    $0x1,%edx
  cmove  %edx,%r12d
L82:
  lea    (%rcx,%r12,1),%edx
  movb   $0x0,(%rax,%rdx,1)
L139:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L96:
  mov    $0x0,%r12d
  jmp    L82
```

assistant: ```c
char *string_nconcat(char *s1, char *s2, unsigned int n) {
  unsigned int i, j;
  char *p;
  for (i = 0; s1 && s1[i]; ++i)
    ;
  for (j = 0; s2 && s2[j]; ++j)
    ;
  n = (n > j ? j : n);
  p = malloc(sizeof(char) * (i + n + 1));
  if (p == NULL)
    return (NULL);
  for (i = 0; s1 && s1[i]; ++i)
    p[i] = s1[i];
  for (j = 0; s2 && j < n; ++j)
    p[i + j] = s2[j];
  p[i + j] = '\0';
  return (p);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<string_nconcat>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L53
  xor    %edi,%edi
  nopl   0x0(%rax)
L59:
  mov    %edi,%eax
  add    $0x1,%edi
  cmpb   $0x0,(%rbx,%rax,1)
  jne    L59
L121:
  xor    %eax,%eax
  xor    %r12d,%r12d
  test   %rbp,%rbp
  jne    L16
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L19:
  add    $0x1,%eax
L16:
  mov    %eax,%ecx
  cmpb   $0x0,0x0(%rbp,%rcx,1)
  jne    L19
  cmp    %eax,%edx
  cmovbe %edx,%eax
  mov    %eax,%r12d
  add    %eax,%edi
L5:
  call   <malloc@plt>
  test   %rax,%rax
  je     L61
  xor    %esi,%esi
  mov    %rax,%rdx
  test   %rbx,%rbx
  jne    L42
  jmp    L73
  nopl   0x0(%rax)
L9:
  mov    %cl,(%rdx)
  add    $0x1,%esi
L42:
  mov    %esi,%edx
  movzbl (%rbx,%rdx,1),%ecx
  add    %rax,%rdx
  test   %cl,%cl
  jne    L9
L73:
  test   %rbp,%rbp
  je     L34
  xor    %ecx,%ecx
  test   %r12d,%r12d
  je     L34
  xchg   %ax,%ax
L13:
  movzbl 0x0(%rbp,%rcx,1),%edi
  lea    (%rsi,%rcx,1),%edx
  add    $0x1,%rcx
  mov    %dil,(%rax,%rdx,1)
  cmp    %ecx,%r12d
  ja     L13
  test   %r12d,%r12d
  mov    $0x1,%edx
  cmove  %edx,%r12d
  lea    (%r12,%rsi,1),%edx
  add    %rax,%rdx
L34:
  movb   $0x0,(%rdx)
L61:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L53:
  mov    $0x1,%edi
  jmp    L121
```

assistant: ```c
char *string_nconcat(char *s1, char *s2, unsigned int n) {
  unsigned int i, j;
  char *p;
  for (i = 0; s1 && s1[i]; ++i)
    ;
  for (j = 0; s2 && s2[j]; ++j)
    ;
  n = (n > j ? j : n);
  p = malloc(sizeof(char) * (i + n + 1));
  if (p == NULL)
    return (NULL);
  for (i = 0; s1 && s1[i]; ++i)
    p[i] = s1[i];
  for (j = 0; s2 && j < n; ++j)
    p[i + j] = s2[j];
  p[i + j] = '\0';
  return (p);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_nconcat>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L84
  xor    %edi,%edi
  nopl   0x0(%rax)
L32:
  mov    %edi,%eax
  add    $0x1,%edi
  cmpb   $0x0,(%rbx,%rax,1)
  jne    L32
L137:
  test   %rbp,%rbp
  je     L3
  xor    %eax,%eax
  cmpb   $0x0,0x0(%rbp)
  je     L73
  nopl   0x0(%rax)
L14:
  lea    0x1(%rax),%ecx
  cmpb   $0x0,0x0(%rbp,%rcx,1)
  mov    %rcx,%rax
  jne    L14
L73:
  cmp    %eax,%edx
  cmovbe %edx,%eax
  mov    %eax,%r12d
  add    %eax,%edi
L16:
  call   <malloc@plt>
  test   %rax,%rax
  je     L116
  mov    %rax,%rdx
  xor    %esi,%esi
  test   %rbx,%rbx
  je     L91
  movzbl (%rbx),%ecx
  xor    %esi,%esi
  test   %cl,%cl
  je     L91
  nopl   0x0(%rax)
L92:
  mov    %cl,(%rdx)
  lea    0x1(%rsi),%edx
  movzbl (%rbx,%rdx,1),%ecx
  mov    %rdx,%rsi
  add    %rax,%rdx
  test   %cl,%cl
  jne    L92
L91:
  test   %rbp,%rbp
  je     L124
  xor    %ecx,%ecx
  test   %r12d,%r12d
  je     L124
  nopw   0x0(%rax,%rax,1)
L34:
  movzbl 0x0(%rbp,%rcx,1),%edi
  lea    (%rsi,%rcx,1),%edx
  add    $0x1,%rcx
  mov    %dil,(%rax,%rdx,1)
  cmp    %ecx,%r12d
  ja     L34
  test   %r12d,%r12d
  mov    $0x1,%edx
  cmove  %edx,%r12d
  lea    (%r12,%rsi,1),%edx
  add    %rax,%rdx
L124:
  movb   $0x0,(%rdx)
L116:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  xor    %r12d,%r12d
  jmp    L16
L84:
  mov    $0x1,%edi
  jmp    L137
```

assistant: ```c
char *string_nconcat(char *s1, char *s2, unsigned int n) {
  unsigned int i, j;
  char *p;
  for (i = 0; s1 && s1[i]; ++i)
    ;
  for (j = 0; s2 && s2[j]; ++j)
    ;
  n = (n > j ? j : n);
  p = malloc(sizeof(char) * (i + n + 1));
  if (p == NULL)
    return (NULL);
  for (i = 0; s1 && s1[i]; ++i)
    p[i] = s1[i];
  for (j = 0; s2 && j < n; ++j)
    p[i + j] = s2[j];
  p[i + j] = '\0';
  return (p);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <rdtsc@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rdtsc@plt>
  sub    -0x8(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: Time is %d


assistant: ```c
int main() {
  unsigned long long cycles = rdtsc();
  cycles = rdtsc() - cycles;
  printf("Time is %d\n", (unsigned)cycles);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <rdtsc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <rdtsc@plt>
  mov    %eax,%esi
  sub    %ebx,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
type_infered: string

tool: D3: Time is %d


assistant: ```c
int main() {
  unsigned long long cycles = rdtsc();
  cycles = rdtsc() - cycles;
  printf("Time is %d\n", (unsigned)cycles);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <rdtsc@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <rdtsc@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  sub    %ebx,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
data_type: string

tool: D5: Time is %d


assistant: ```c
int main() {
  unsigned long long cycles = rdtsc();
  cycles = rdtsc() - cycles;
  printf("Time is %d\n", (unsigned)cycles);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <rdtsc@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <rdtsc@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  sub    %ebx,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D2
type: string

tool: D2: Time is %d


assistant: ```c
int main() {
  unsigned long long cycles = rdtsc();
  cycles = rdtsc() - cycles;
  printf("Time is %d\n", (unsigned)cycles);
  return 0;
}
```