ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<s_socket_pipe_listen>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  xor    %edi,%edi
  call   <DEBUGF@plt>
  lea    D19(%rip),%rsi
  xor    %edi,%edi
  call   <DEBUGF@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type: string
name: read_data
arguments: data_tag: D19
type: string

tool: D0: enter

tool: D19: leave

assistant: ```c
int s_socket_pipe_listen(int s, int backlog) {
  DEBUGF(0, "enter");
  DEBUGF(0, "leave");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x30(%rbp)
  movl   $0x14,-0x2c(%rbp)
  movl   $0x1e,-0x28(%rbp)
  movl   $0x28,-0x24(%rbp)
  movl   $0x32,-0x20(%rbp)
  movl   $0x3c,-0x1c(%rbp)
  movl   $0x46,-0x18(%rbp)
  movl   $0x50,-0x14(%rbp)
  movl   $0x5a,-0x10(%rbp)
  movl   $0x64,-0xc(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rdx
  lea    D37(%rip),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L27
L0:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L27:
  cmpl   $0x9,-0x34(%rbp)
  jle    L0
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D9
type_infered: string
name: retrieve_data
arguments: label: D37
type_infered: string
name: retrieve_data
arguments: label: D5
type_infered: string
name: retrieve_data
arguments: label: D38
type_infered: string

tool: D9: Value

tool: D37: Index

tool: D5: %s%9s


tool: D38: %3d%13d


assistant: ```c
int main() {
  int n[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
  int i;
  printf("%s%9s\n", "Index", "Value");
  for (i = 0; i < 10; i++) {
    printf("%3d%13d\n", i, n[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0xa,(%rsp)
  movl   $0x14,0x4(%rsp)
  movl   $0x1e,0x8(%rsp)
  movl   $0x28,0xc(%rsp)
  movl   $0x32,0x10(%rsp)
  movl   $0x3c,0x14(%rsp)
  movl   $0x46,0x18(%rsp)
  movl   $0x50,0x1c(%rsp)
  movl   $0x5a,0x20(%rsp)
  movl   $0x64,0x24(%rsp)
  lea    D25(%rip),%rdx
  lea    D29(%rip),%rsi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D35(%rip),%rbp
L5:
  mov    (%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  jne    L5
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D25
type_guessed: string
name: parse_data
arguments: label: D29
type_guessed: string
name: parse_data
arguments: label: D31
type_guessed: string
name: parse_data
arguments: label: D35
type_guessed: string

tool: D25: Value

tool: D29: Index

tool: D31: %s%9s


tool: D35: %3d%13d


assistant: ```c
int main() {
  int n[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
  int i;
  printf("%s%9s\n", "Index", "Value");
  for (i = 0; i < 10; i++) {
    printf("%3d%13d\n", i, n[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D11(%rip),%rdx
  lea    D15(%rip),%rsi
  push   %rbx
  lea    D23(%rip),%rdi
  xor    %ebx,%ebx
  lea    D27(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x14,0x4(%rsp)
  movabs $0x280000001e,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x3c00000032,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x5000000046,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x640000005a,%rax
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edx
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L3:
  mov    (%rsp,%rbx,4),%edx
L23:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xa,%rbx
  jne    L3
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D11
data_type: string
name: unpack_data
arguments: tag: D15
data_type: string
name: unpack_data
arguments: tag: D23
data_type: string
name: unpack_data
arguments: tag: D27
data_type: string

tool: D11: Value

tool: D15: Index

tool: D23: %s%9s


tool: D27: %3d%13d


assistant: ```c
int main() {
  int n[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
  int i;
  printf("%s%9s\n", "Index", "Value");
  for (i = 0; i < 10; i++) {
    printf("%3d%13d\n", i, n[i]);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D54(%rip),%rdx
  lea    D50(%rip),%rsi
  push   %rbp
  lea    D3(%rip),%rdi
  lea    D56(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x30,%rsp
  movdqa D44(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    %rsp,%r12
  movaps %xmm0,(%rsp)
  movdqa D31(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  call   <printf@plt>
  nopl   0x0(%rax)
L5:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xa,%rbx
  jne    L5
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D54
type: string
name: unpack_data
arguments: tag: D50
type: string
name: unpack_data
arguments: tag: D3
type: string
name: unpack_data
arguments: tag: D56
type: string
name: unpack_data
arguments: tag: D44
type: byte[16]
name: unpack_data
arguments: tag: D6
type: byte[8]
name: unpack_data
arguments: tag: D31
type: byte[16]

tool: D54: Value

tool: D50: Index

tool: D3: %s%9s


tool: D56: %3d%13d


tool: {"D44": [0x0a,0x00,0x00,0x00,0x14,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x28,0x00,0x00,0x00]}

tool: {"D6": [0x5a,0x00,0x00,0x00,0x64,0x00,0x00,0x00]}

tool: {"D31": [0x32,0x00,0x00,0x00,0x3c,0x00,0x00,0x00,0x46,0x00,0x00,0x00,0x50,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int n[10] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100};
  int i;
  printf("%s%9s\n", "Index", "Value");
  for (i = 0; i < 10; i++) {
    printf("%3d%13d\n", i, n[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<string_cat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x20(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x28(%rbp)
  jmp    L39
L27:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
  addl   $0x1,-0x28(%rbp)
L39:
  mov    -0x28(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L27
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L35
L15:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
  addl   $0x1,-0x24(%rbp)
L35:
  mov    -0x24(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L15
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *string_cat(char *s1, char *s2) {
  int size1 = strlen(s1);
  int size2 = strlen(s2);
  char *res = (char *)malloc(sizeof(char) * size1 + size2 + 1);
  char *pt_r = res;
  char *pt_s = s1;
  for (int i = 0; i < size1; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  pt_s = s2;
  for (int i = 0; i < size2; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  *pt_r = '\0';
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_cat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  lea    0x1(%rbx,%rax,1),%edi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L24
  lea    -0x1(%rbx),%edx
  lea    0x1(%rax,%rdx,1),%rdi
  mov    %rax,%rdx
  mov    %r13,%rsi
  sub    %rax,%rsi
L1:
  movzbl (%rdx,%rsi,1),%ecx
  mov    %cl,(%rdx)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L1
  mov    %ebx,%ecx
  add    %rax,%rcx
L32:
  test   %r12d,%r12d
  jle    L16
  mov    %r12d,%r12d
  mov    $0x0,%edx
L27:
  movzbl 0x0(%rbp,%rdx,1),%esi
  mov    %sil,(%rcx,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%r12
  jne    L27
  add    %r12,%rcx
L16:
  movb   $0x0,(%rcx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  mov    %rax,%rcx
  jmp    L32
```

assistant: ```c
char *string_cat(char *s1, char *s2) {
  int size1 = strlen(s1);
  int size2 = strlen(s2);
  char *res = (char *)malloc(sizeof(char) * size1 + size2 + 1);
  char *pt_r = res;
  char *pt_s = s1;
  for (int i = 0; i < size1; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  pt_s = s2;
  for (int i = 0; i < size2; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  *pt_r = '\0';
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<string_cat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  lea    0x1(%r13,%rax,1),%edi
  mov    %eax,%r12d
  call   <malloc@plt>
  mov    %rax,%r8
  test   %r13d,%r13d
  jle    L17
  lea    -0x1(%r13),%edi
  sub    %rax,%rbx
  lea    0x1(%rax,%rdi,1),%rsi
  mov    %rbx,%rcx
  nopl   0x0(%rax,%rax,1)
L30:
  movzbl (%rax,%rcx,1),%edx
  add    $0x1,%rax
  mov    %dl,-0x1(%rax)
  cmp    %rsi,%rax
  jne    L30
  lea    0x1(%r8,%rdi,1),%rcx
L18:
  test   %r12d,%r12d
  jle    L34
  lea    -0x1(%r12),%esi
  xor    %eax,%eax
  lea    0x1(%rsi),%rdi
  nopl   (%rax)
L25:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdx,%rsi
  jne    L25
  add    %rdi,%rcx
L34:
  movb   $0x0,(%rcx)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  mov    %rax,%rcx
  jmp    L18
```

assistant: ```c
char *string_cat(char *s1, char *s2) {
  int size1 = strlen(s1);
  int size2 = strlen(s2);
  char *res = (char *)malloc(sizeof(char) * size1 + size2 + 1);
  char *pt_r = res;
  char *pt_s = s1;
  for (int i = 0; i < size1; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  pt_s = s2;
  for (int i = 0; i < size2; i++) {
    *pt_r = *pt_s;
    pt_r++;
    pt_s++;
  }
  *pt_r = '\0';
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x2,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_infered: string
name: parse_data
arguments: data_tag: D18
type_infered: string
name: parse_data
arguments: data_tag: D27
type_infered: string

tool: D8: sizeof short int %d


tool: D18: sizeof float %d


tool: D27: sizeof double %d


assistant: ```c
int main() {
  printf("sizeof short int %d\n", sizeof(short int));
  printf("sizeof float %d\n", sizeof(float));
  printf("sizeof double %d\n", sizeof(double));
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type: string
name: retrieve_data
arguments: data_label: D5
type: string
name: retrieve_data
arguments: data_label: D11
type: string

tool: D2: sizeof short int %d


tool: D5: sizeof float %d


tool: D11: sizeof double %d


assistant: ```c
int main() {
  printf("sizeof short int %d\n", sizeof(short int));
  printf("sizeof float %d\n", sizeof(float));
  printf("sizeof double %d\n", sizeof(double));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string
name: extract_symbolic_data
arguments: data_tag: D18
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: string

tool: D7: sizeof short int %d


tool: D18: sizeof float %d


tool: D1: sizeof double %d


assistant: ```c
int main() {
  printf("sizeof short int %d\n", sizeof(short int));
  printf("sizeof float %d\n", sizeof(float));
  printf("sizeof double %d\n", sizeof(double));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x4,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D14
type_infered: string
name: parse_data
arguments: data_tag: D16
type_infered: string
name: parse_data
arguments: data_tag: D20
type_infered: string

tool: D14: sizeof short int %d


tool: D16: sizeof float %d


tool: D20: sizeof double %d


assistant: ```c
int main() {
  printf("sizeof short int %d\n", sizeof(short int));
  printf("sizeof float %d\n", sizeof(float));
  printf("sizeof double %d\n", sizeof(double));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<yesno>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x98(%rbp)
  jmp    L42
L16:
  cmpl   $0x0,-0x98(%rbp)
  jle    L46
  cmpl   $0x7e,-0x98(%rbp)
  jle    L68
L46:
  cmpl   $0x0,-0x98(%rbp)
  jne    L42
  mov    -0x94(%rbp),%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L42
L68:
  mov    -0x98(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x98(%rbp)
  mov    -0x94(%rbp),%edx
  cltq
  mov    %dl,-0x90(%rbp,%rax,1)
L42:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x94(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x94(%rbp)
  je     L51
  cmpl   $0xa,-0x94(%rbp)
  jne    L16
L51:
  mov    -0x98(%rbp),%eax
  cltq
  movb   $0x0,-0x90(%rbp,%rax,1)
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <rpmatch@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L73
  call   <__stack_chk_fail@plt>
L73:
  leave
  ret
```

assistant: ```c
int yesno(void) {
  char buf[128];
  int len = 0;
  int c;
  while ((c = getchar()) != EOF && c != '\n')
    if ((len > 0 && len < 127) || (len == 0 && !isspace(c)))
      buf[len++] = c;
  buf[len] = '\0';
  return rpmatch(buf) == 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<yesno>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  jmp    L34
L47:
  movslq %ebp,%rax
  mov    %bl,(%rsp,%rax,1)
  lea    0x1(%rbp),%ebp
L34:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  mov    D7(%rip),%rax
  cmp    %ebx,(%rax)
  je     L19
  cmp    $0xa,%ebx
  je     L19
  lea    -0x1(%rbp),%eax
  cmp    $0x7d,%eax
  jbe    L47
  test   %ebp,%ebp
  jne    L34
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L34
  jmp    L47
L19:
  movslq %ebp,%rbp
  movb   $0x0,(%rsp,%rbp,1)
  mov    %rsp,%rdi
  call   <rpmatch@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L60
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  ret
L60:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yesno(void) {
  char buf[128];
  int len = 0;
  int c;
  while ((c = getchar()) != EOF && c != '\n')
    if ((len > 0 && len < 127) || (len == 0 && !isspace(c)))
      buf[len++] = c;
  buf[len] = '\0';
  return rpmatch(buf) == 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<yesno>:
  endbr64
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x90,%rsp
  mov    D5(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
L52:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    %eax,(%r12)
  je     L63
L5:
  cmp    $0xa,%eax
  je     L63
  lea    -0x1(%rbp),%eax
  cmp    $0x7d,%eax
  jbe    L13
  test   %ebp,%ebp
  jne    L52
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L52
L13:
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %bl,(%rsp,%rax,1)
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    %eax,(%r12)
  jne    L5
L63:
  movslq %ebp,%rbp
  mov    %rsp,%rdi
  movb   $0x0,(%rsp,%rbp,1)
  call   <rpmatch@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L44
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yesno(void) {
  char buf[128];
  int len = 0;
  int c;
  while ((c = getchar()) != EOF && c != '\n')
    if ((len > 0 && len < 127) || (len == 0 && !isspace(c)))
      buf[len++] = c;
  buf[len] = '\0';
  return rpmatch(buf) == 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<yesno>:
  endbr64
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x90,%rsp
  mov    D3(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
L2:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    %eax,(%r12)
  je     L7
L1:
  cmp    $0xa,%eax
  je     L7
  lea    -0x1(%rbp),%eax
  cmp    $0x7d,%eax
  jbe    L56
  test   %ebp,%ebp
  jne    L2
  mov    %ebx,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L2
L56:
  movslq %ebp,%rax
  add    $0x1,%ebp
  mov    %bl,(%rsp,%rax,1)
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    %eax,(%r12)
  jne    L1
L7:
  movslq %ebp,%rbp
  mov    %rsp,%rdi
  movb   $0x0,(%rsp,%rbp,1)
  call   <rpmatch@plt>
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    0x88(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int yesno(void) {
  char buf[128];
  int len = 0;
  int c;
  while ((c = getchar()) != EOF && c != '\n')
    if ((len > 0 && len < 127) || (len == 0 && !isspace(c)))
      buf[len++] = c;
  buf[len] = '\0';
  return rpmatch(buf) == 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jg     L20
  mov    $0x0,%eax
  call   <uso@plt>
L20:
  movl   $0x1,-0x4(%rbp)
  jmp    L23
L0:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <SacaObjetos@plt>
  addl   $0x1,-0x4(%rbp)
L23:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L0
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
datatype: string

tool: D6: %i Procesando:	%s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  if (argc < 2)
    uso();
  for (i = 1; i < argc; i++) {
    printf("%i Procesando:\t%s\n", i, *(argv + i));
    SacaObjetos(*(argv + i));
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %edi,%r13d
  mov    $0x1,%ebx
  lea    D5(%rip),%r12
  cmp    $0x1,%edi
  jle    L20
L28:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x0(%rbp,%rbx,8),%rdi
  call   <SacaObjetos@plt>
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L28
L15:
  mov    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  mov    $0x0,%eax
  call   <uso@plt>
  jmp    L15
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
datatype: string

tool: D5: %i Procesando:	%s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  if (argc < 2)
    uso();
  for (i = 1; i < argc; i++) {
    printf("%i Procesando:\t%s\n", i, *(argv + i));
    SacaObjetos(*(argv + i));
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L11
  mov    %rsi,%rbp
  movslq %edi,%r13
  mov    $0x1,%ebx
  lea    D4(%rip),%r12
  nopl   (%rax)
L2:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x0(%rbp,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <SacaObjetos@plt>
  cmp    %r13,%rbx
  jne    L2
L7:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  xor    %eax,%eax
  call   <uso@plt>
  jmp    L7
```

tool_calls: name: parse_data
arguments: label: D4
type: string

tool: D4: %i Procesando:	%s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  if (argc < 2)
    uso();
  for (i = 1; i < argc; i++) {
    printf("%i Procesando:\t%s\n", i, *(argv + i));
    SacaObjetos(*(argv + i));
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L2
  mov    %rsi,%rbp
  movslq %edi,%r13
  mov    $0x1,%ebx
  lea    D1(%rip),%r12
  nopl   (%rax)
L3:
  mov    0x0(%rbp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x0(%rbp,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <SacaObjetos@plt>
  cmp    %rbx,%r13
  jne    L3
L27:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  xor    %eax,%eax
  call   <uso@plt>
  jmp    L27
```

tool_calls: name: parse_data
arguments: label: D1
type: string

tool: D1: %i Procesando:	%s


assistant: ```c
int main(int argc, char **argv) {
  int i;
  if (argc < 2)
    uso();
  for (i = 1; i < argc; i++) {
    printf("%i Procesando:\t%s\n", i, *(argv + i));
    SacaObjetos(*(argv + i));
  }
  return 1;
}
```