ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<print_help>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D20
datatype: string
name: extract_symbolic_data
arguments: data_label: D21
datatype: string
name: extract_symbolic_data
arguments: data_label: D27
datatype: string

tool: D20: Usage:  raw_extract --pal <palette-file>


tool: D21: 


tool: D27: The input file is the standard input, the output file the standard output.


assistant: ```c
void print_help() {
  printf("Usage:  raw_extract --pal <palette-file>\n");
  printf("\n");
  printf("The input file is the standard input, the output file the standard output.\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<print_help>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D0
datatype: string
name: parse_data
arguments: label: D7
datatype: string
name: parse_data
arguments: label: D18
datatype: string

tool: D0: Usage:  raw_extract --pal <palette-file>


tool: D7: 


tool: D18: The input file is the standard input, the output file the standard output.


assistant: ```c
void print_help() {
  printf("Usage:  raw_extract --pal <palette-file>\n");
  printf("\n");
  printf("The input file is the standard input, the output file the standard output.\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_help>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_infered: string
name: extract_symbolic_data
arguments: tag: D11
type_infered: string
name: extract_symbolic_data
arguments: tag: D20
type_infered: string

tool: D2: Usage:  raw_extract --pal <palette-file>


tool: D11: 


tool: D20: The input file is the standard input, the output file the standard output.


assistant: ```c
void print_help() {
  printf("Usage:  raw_extract --pal <palette-file>\n");
  printf("\n");
  printf("The input file is the standard input, the output file the standard output.\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_help>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D16
type_guessed: string
name: extract_symbolic_data
arguments: label: D11
type_guessed: string
name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D16: Usage:  raw_extract --pal <palette-file>


tool: D11: 


tool: D9: The input file is the standard input, the output file the standard output.


assistant: ```c
void print_help() {
  printf("Usage:  raw_extract --pal <palette-file>\n");
  printf("\n");
  printf("The input file is the standard input, the output file the standard output.\n");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<execle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    -0x20(%rbp),%rbx
  nop
L9:
  mov    %rbx,%rax
  lea    0x8(%rax),%rbx
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L9
  mov    (%rbx),%rdx
  lea    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <execve@plt>
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
execle(file, args) char *file;
char *args;
{
  register char **p;
  p = &args;
  while (*p++)
    ;
  return (execve(file, &args, *p));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<execle>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rax
L2:
  add    $0x8,%rax
  cmpq   $0x0,-0x8(%rax)
  jne    L2
  lea    0x8(%rsp),%rsi
  mov    (%rax),%rdx
  call   <execve@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
execle(file, args) char *file;
char *args;
{
  register char **p;
  p = &args;
  while (*p++)
    ;
  return (execve(file, &args, *p));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<execle>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  mov    %rsi,%rax
  nopl   (%rax)
L5:
  add    $0x8,%rax
  cmpq   $0x0,-0x8(%rax)
  jne    L5
  mov    (%rax),%rdx
  call   <execve@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
execle(file, args) char *file;
char *args;
{
  register char **p;
  p = &args;
  while (*p++)
    ;
  return (execve(file, &args, *p));
}
```
user: What is the c source code of the assembly code below?

```asm
<execle>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  mov    %rsi,%rax
  nopl   (%rax)
L2:
  add    $0x8,%rax
  cmpq   $0x0,-0x8(%rax)
  jne    L2
  mov    (%rax),%rdx
  call   <execve@plt>
  add    $0x18,%rsp
  ret
```

assistant: ```c
execle(file, args) char *file;
char *args;
{
  register char **p;
  p = &args;
  while (*p++)
    ;
  return (execve(file, &args, *p));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int rc = 0;
  rc = 0;
  return rc;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int rc = 0;
  rc = 0;
  return rc;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int rc = 0;
  rc = 0;
  return rc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int rc = 0;
  rc = 0;
  return rc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inverse_transform_msimplex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L8
L36:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  movq   %rax,%xmm0
  call   <exp@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x20(%rbp)
L8:
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x20(%rbp)
  jl     L36
  movsd  -0x18(%rbp),%xmm1
  movsd  D6(%rip),%xmm0
  addsd  %xmm0,%xmm1
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  D6(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x0,-0x1c(%rbp)
  jmp    L20
L33:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L20:
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L33
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
data_type: f64

tool: D6: 1

assistant: ```c
void inverse_transform_msimplex(double *x, const double *y, int K) {
  double sum = 0;
  int k = 0;
  for (; k < K - 1; k++) {
    x[k] = exp(y[k]);
    sum += x[k];
  }
  x[k] = 1.0 / (1.0 + sum);
  for (int i = 0; i < K - 1; i++) {
    x[i] /= x[k];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inverse_transform_msimplex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  cmp    $0x1,%edx
  jle    L18
  mov    %rsi,%r12
  mov    %edx,%r14d
  lea    -0x2(%rdx),%r13d
  mov    $0x0,%ebx
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
L27:
  movsd  (%r12,%rbx,8),%xmm0
  call   <exp@plt>
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  mov    %rbx,%rax
  add    $0x1,%rbx
  cmp    %rax,%r13
  jne    L27
  sub    $0x1,%r14d
  movslq %r14d,%r14
  lea    0x0(%rbp,%r14,8),%rdx
  movsd  D4(%rip),%xmm1
  movsd  0x8(%rsp),%xmm0
  addsd  %xmm1,%xmm0
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  mov    %rbp,%rax
  lea    0x8(%rbp,%r13,8),%rcx
L15:
  movsd  (%rax),%xmm0
  divsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L15
L26:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  mov    D4(%rip),%rax
  mov    %rax,(%rdi)
  jmp    L26
```

tool_calls: name: unpack_data
arguments: data_label: D4
datatype: f64

tool: D4: 1

assistant: ```c
void inverse_transform_msimplex(double *x, const double *y, int K) {
  double sum = 0;
  int k = 0;
  for (; k < K - 1; k++) {
    x[k] = exp(y[k]);
    sum += x[k];
  }
  x[k] = 1.0 / (1.0 + sum);
  for (int i = 0; i < K - 1; i++) {
    x[i] /= x[k];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inverse_transform_msimplex>:
  endbr64
  push   %r14
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  test   %r13d,%r13d
  jle    L3
  mov    %rsi,%r14
  lea    -0x2(%rdx),%r12d
  pxor   %xmm1,%xmm1
  xor    %ebp,%ebp
  nopl   (%rax)
L21:
  movsd  (%r14,%rbp,8),%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <exp@plt>
  movsd  0x8(%rsp),%xmm1
  mov    %rbp,%rax
  movsd  %xmm0,(%rbx,%rbp,8)
  add    $0x1,%rbp
  addsd  %xmm0,%xmm1
  cmp    %rax,%r12
  jne    L21
  movsd  D5(%rip),%xmm2
  movslq %r13d,%r13
  mov    %rbx,%rax
  lea    0x8(%rbx,%r12,8),%rdx
  lea    (%rbx,%r13,8),%rcx
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm2
  movsd  %xmm2,(%rcx)
  jmp    L5
  nopl   0x0(%rax)
L13:
  movsd  (%rcx),%xmm2
L5:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  divsd  %xmm2,%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %rax,%rdx
  jne    L13
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L3:
  mov    D5(%rip),%rax
  mov    %rax,(%rdi)
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
datatype: f64

tool: D5: 1

assistant: ```c
void inverse_transform_msimplex(double *x, const double *y, int K) {
  double sum = 0;
  int k = 0;
  for (; k < K - 1; k++) {
    x[k] = exp(y[k]);
    sum += x[k];
  }
  x[k] = 1.0 / (1.0 + sum);
  for (int i = 0; i < K - 1; i++) {
    x[i] /= x[k];
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inverse_transform_msimplex>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  test   %r13d,%r13d
  jle    L61
  lea    -0x2(%rdx),%ebp
  mov    %rsi,%r14
  mov    %edx,%r12d
  xor    %ebx,%ebx
  lea    0x1(%rbp),%rax
  mov    %ebp,0x14(%rsp)
  pxor   %xmm1,%xmm1
  mov    %rax,0x18(%rsp)
  xchg   %ax,%ax
L7:
  movsd  (%r14,%rbx,8),%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <exp@plt>
  movsd  0x8(%rsp),%xmm1
  mov    %rbx,%rax
  movsd  %xmm0,(%r15,%rbx,8)
  add    $0x1,%rbx
  addsd  %xmm0,%xmm1
  cmp    %rbp,%rax
  jne    L7
  movsd  D6(%rip),%xmm0
  movslq %r13d,%rsi
  shl    $0x3,%rsi
  addsd  %xmm0,%xmm1
  lea    (%r15,%rsi,1),%rdx
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx)
  movsd  (%r15),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%r15)
  cmp    $0x1,%r13d
  je     L6
  lea    0x8(%r15,%rsi,1),%rdi
  mov    0x18(%rsp),%rcx
  lea    0x8(%r15),%rax
  cmp    %rdi,%rax
  setae  %dil
  shl    $0x3,%rcx
  cmp    %rcx,%rsi
  setge  %sil
  or     %sil,%dil
  je     L16
  cmp    $0x3,%r12d
  je     L16
  mov    0x14(%rsp),%esi
  movsd  (%rdx),%xmm1
  shr    %esi
  unpcklpd %xmm1,%xmm1
  sub    $0x1,%esi
  shl    $0x4,%rsi
  lea    0x18(%r15,%rsi,1),%rsi
  nopl   (%rax)
L54:
  movupd (%rax),%xmm0
  add    $0x10,%rax
  divpd  %xmm1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rsi,%rax
  jne    L54
  mov    0x14(%rsp),%ecx
  mov    %ecx,%eax
  or     $0x1,%eax
  and    $0x1,%ecx
  je     L6
  cltq
  lea    (%r15,%rax,8),%rax
  movsd  (%rax),%xmm0
  divsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
L6:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L16:
  lea    -0x3(%r12),%esi
  lea    0x10(%r15,%rsi,8),%rcx
  nopw   0x0(%rax,%rax,1)
L10:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  divsd  (%rdx),%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %rax,%rcx
  jne    L10
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L61:
  mov    D6(%rip),%rax
  mov    %rax,(%rdi)
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
datatype: f64

tool: D6: 1

assistant: ```c
void inverse_transform_msimplex(double *x, const double *y, int K) {
  double sum = 0;
  int k = 0;
  for (; k < K - 1; k++) {
    x[k] = exp(y[k]);
    sum += x[k];
  }
  x[k] = 1.0 / (1.0 + sum);
  for (int i = 0; i < K - 1; i++) {
    x[i] /= x[k];
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<msx_writeproc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x40(%rbp),%rcx
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  lea    D8(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  mov    -0x38(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <msx_write@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: string

tool: D8: /proc/%d/%s

assistant: ```c
int msx_writeproc(int pid, char *item, int val) {
  char fn[30];
  sprintf(fn, "/proc/%d/%s", pid, item);
  return (msx_write(fn, val));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<msx_writeproc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rsi,%rcx
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %edi,%edx
  lea    D2(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <msx_write@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string

tool: D2: /proc/%d/%s

assistant: ```c
int msx_writeproc(int pid, char *item, int val) {
  char fn[30];
  sprintf(fn, "/proc/%d/%s", pid, item);
  return (msx_write(fn, val));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<msx_writeproc>:
  endbr64
  push   %r12
  mov    %rsi,%rcx
  lea    D1(%rip),%rsi
  push   %rbp
  mov    %edx,%ebp
  mov    %edi,%edx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <msx_write@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
datatype: string

tool: D1: /proc/%d/%s

assistant: ```c
int msx_writeproc(int pid, char *item, int val) {
  char fn[30];
  sprintf(fn, "/proc/%d/%s", pid, item);
  return (msx_write(fn, val));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<msx_writeproc>:
  endbr64
  push   %r12
  mov    %rsi,%rcx
  lea    D2(%rip),%rsi
  push   %rbp
  mov    %edx,%ebp
  mov    %edi,%edx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <msx_write@plt>
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_guessed: string

tool: D2: /proc/%d/%s

assistant: ```c
int msx_writeproc(int pid, char *item, int val) {
  char fn[30];
  sprintf(fn, "/proc/%d/%s", pid, item);
  return (msx_write(fn, val));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ins_strstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L18
L3:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <ins_memcmp@plt>
  test   %rax,%rax
  jne    L6
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  jmp    L29
L6:
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L3
  mov    $0x0,%eax
L29:
  leave
  ret
```

assistant: ```c
char *ins_strstr(char *buf, char *str) {
  int lenb;
  int lens;
  int x;
  lenb = strlen(buf);
  lens = strlen(str);
  for (x = 0; x <= (lenb - lens); x++) {
    if (ins_memcmp(buf + x, str, lens) == 0) {
      return (buf + x);
    }
  }
  return (NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ins_strstr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r15
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %r15,%rdi
  call   <strlen@plt>
  sub    %eax,%r12d
  js     L35
  mov    %eax,%r13d
  mov    $0x0,%ebx
L21:
  lea    (%r14,%rbx,1),%rbp
  mov    %r13d,%edx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <ins_memcmp@plt>
  test   %rax,%rax
  je     L33
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jge    L21
  mov    $0x0,%ebp
L33:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L35:
  mov    $0x0,%ebp
  jmp    L33
```

assistant: ```c
char *ins_strstr(char *buf, char *str) {
  int lenb;
  int lens;
  int x;
  lenb = strlen(buf);
  lens = strlen(str);
  for (x = 0; x <= (lenb - lens); x++) {
    if (ins_memcmp(buf + x, str, lens) == 0) {
      return (buf + x);
    }
  }
  return (NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ins_strstr>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  sub    %eax,%ebp
  js     L29
  mov    %eax,%r13d
  xor    %ebx,%ebx
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L22:
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jl     L29
L14:
  lea    (%r14,%rbx,1),%r12
  mov    %r13d,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <ins_memcmp@plt>
  test   %rax,%rax
  jne    L22
L25:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L29:
  xor    %r12d,%r12d
  jmp    L25
```

assistant: ```c
char *ins_strstr(char *buf, char *str) {
  int lenb;
  int lens;
  int x;
  lenb = strlen(buf);
  lens = strlen(str);
  for (x = 0; x <= (lenb - lens); x++) {
    if (ins_memcmp(buf + x, str, lens) == 0) {
      return (buf + x);
    }
  }
  return (NULL);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ins_strstr>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  sub    %eax,%ebp
  js     L36
  mov    %eax,%r13d
  xor    %ebx,%ebx
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jl     L36
L27:
  lea    (%r14,%rbx,1),%r12
  mov    %r13d,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <ins_memcmp@plt>
  test   %rax,%rax
  jne    L21
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L36:
  xor    %r12d,%r12d
  jmp    L7
```

assistant: ```c
char *ins_strstr(char *buf, char *str) {
  int lenb;
  int lens;
  int x;
  lenb = strlen(buf);
  lens = strlen(str);
  for (x = 0; x <= (lenb - lens); x++) {
    if (ins_memcmp(buf + x, str, lens) == 0) {
      return (buf + x);
    }
  }
  return (NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<in_hex_to_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
L22:
  mov    -0x18(%rbp),%rax
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strspn@plt>
  cltq
  add    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L37
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x2f,-0xc(%rbp)
  jle    L46
  cmpl   $0x39,-0xc(%rbp)
  jg     L46
  subl   $0x30,-0xc(%rbp)
  jmp    L40
L46:
  cmpl   $0x40,-0xc(%rbp)
  jle    L71
  cmpl   $0x46,-0xc(%rbp)
  jg     L71
  subl   $0x37,-0xc(%rbp)
L40:
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  or     -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x100,%eax
  test   %eax,%eax
  je     L22
  cmpq   $0x0,-0x20(%rbp)
  je     L50
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L50:
  addl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L22
L37:
  nop
L71:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D1:  	


assistant: ```c
int in_hex_to_data(const char *p, uint8_t *data) {
  int c, len, v;
  len = 0;
  v = 1;
  for (;;) {
    p += strspn(p, " \t\r\n");
    if (*p == '\0')
      break;
    c = toupper((unsigned char)*p++);
    if (c >= '0' && c <= '9')
      c = c - '0';
    else if (c >= 'A' && c <= 'F')
      c = c - 'A' + 10;
    else
      break;
    v = (v << 4) | c;
    if (v & 0x100) {
      if (data)
        data[len] = v;
      len++;
      v = 1;
    }
  }
  return len;
}
```