ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<markVisit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  movslq %ecx,%rcx
  movl   $0x1,(%r8,%rcx,4)
  test   %edx,%edx
  jle    L17
  mov    %rdi,%r15
  mov    %edx,%r12d
  mov    %r8,%rbp
  lea    (%rdi,%rcx,8),%r14
  mov    %edx,%r13d
  mov    $0x0,%ebx
  jmp    L55
L36:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  je     L17
L55:
  mov    %ebx,%ecx
  mov    (%r14),%rax
  cmpl   $0x1,(%rax,%rbx,4)
  jne    L36
  cmpl   $0x0,0x0(%rbp,%rbx,4)
  jne    L36
  mov    %rbp,%r8
  mov    %r12d,%edx
  mov    0xc(%rsp),%esi
  mov    %r15,%rdi
  call   <markVisit>
  jmp    L36
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void markVisit(int **M, int r, int c, int m, int *visit) {
  visit[m] = 1;
  for (int i = 0; i < c; i++) {
    if (1 == M[m][i] && 0 == visit[i]) {
      markVisit(M, r, c, i, visit);
    }
  }
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<markVisit>:
  endbr64
  movslq %ecx,%rcx
  movl   $0x1,(%r8,%rcx,4)
  test   %edx,%edx
  jle    L16
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    (%rdi,%rcx,8),%r14
L2:
  cmpl   $0x1,(%r14,%rbx,4)
  mov    %ebx,%ecx
  je     L19
L23:
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L2
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L19:
  mov    0x0(%rbp,%rbx,4),%eax
  test   %eax,%eax
  jne    L23
  mov    %rbp,%r8
  mov    %r12d,%edx
  mov    %r15,%rdi
  mov    %esi,0xc(%rsp)
  call   <markVisit>
  mov    0xc(%rsp),%esi
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L16:
  ret
```

assistant: ```c
void markVisit(int **M, int r, int c, int m, int *visit) {
  visit[m] = 1;
  for (int i = 0; i < c; i++) {
    if (1 == M[m][i] && 0 == visit[i]) {
      markVisit(M, r, c, i, visit);
    }
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<markVisit>:
  endbr64
  movslq %ecx,%rcx
  movl   $0x1,(%r8,%rcx,4)
  test   %edx,%edx
  jle    L103
  push   %r15
  push   %r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    (%rdi,%rcx,8),%r14
L99:
  cmpl   $0x1,(%r14,%rbx,4)
  je     L40
L14:
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L99
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L40:
  mov    (%r8,%rbx,4),%r11d
  test   %r11d,%r11d
  jne    L14
  mov    (%rdi,%rbx,8),%r15
  xor    %ebp,%ebp
  movl   $0x1,(%r8,%rbx,4)
  cmpl   $0x1,(%r15,%rbp,4)
  je     L79
  xchg   %ax,%ax
L23:
  add    $0x1,%rbp
  cmp    %rbp,%r13
  je     L14
  cmpl   $0x1,(%r15,%rbp,4)
  jne    L23
L79:
  mov    (%r8,%rbp,4),%r10d
  test   %r10d,%r10d
  jne    L23
  mov    (%rdi,%rbp,8),%rax
  xor    %r12d,%r12d
  movl   $0x1,(%r8,%rbp,4)
  mov    %r12d,%ecx
  cmpl   $0x1,(%rax,%r12,4)
  je     L7
  nopw   0x0(%rax,%rax,1)
L30:
  add    $0x1,%r12
  cmp    %r12,%r13
  je     L23
  cmpl   $0x1,(%rax,%r12,4)
  mov    %r12d,%ecx
  jne    L30
L7:
  mov    (%r8,%r12,4),%r9d
  test   %r9d,%r9d
  jne    L30
  mov    %rax,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %esi,0x8(%rsp)
  mov    %rdi,(%rsp)
  call   <markVisit>
  mov    0x18(%rsp),%rax
  mov    0x10(%rsp),%r8
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    (%rsp),%rdi
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L103:
  ret
```

assistant: ```c
void markVisit(int **M, int r, int c, int m, int *visit) {
  visit[m] = 1;
  for (int i = 0; i < c; i++) {
    if (1 == M[m][i] && 0 == visit[i]) {
      markVisit(M, r, c, i, visit);
    }
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fibnacci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L26
  mov    $0x1,%eax
  jmp    L33
L26:
  cmpl   $0x1,-0x14(%rbp)
  jne    L39
  mov    $0x1,%eax
  jmp    L33
L39:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibnacci>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibnacci>
  add    %ebx,%eax
L33:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fibnacci(int n) {
  if (0 == n)
    return 1;
  if (1 == n)
    return 1;
  return fibnacci(n - 1) + fibnacci(n - 2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fibnacci>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  ja     L5
  ret
L5:
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fibnacci>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fibnacci>
  add    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fibnacci(int n) {
  if (0 == n)
    return 1;
  if (1 == n)
    return 1;
  return fibnacci(n - 1) + fibnacci(n - 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bisquare>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movq   D2(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  D14(%rip),%xmm0
  comisd %xmm1,%xmm0
  jb     L13
  movsd  -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D14(%rip),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  jmp    L12
L13:
  pxor   %xmm0,%xmm0
L12:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
data_type: qword
name: parse_data
arguments: data_label: D14
data_type: f64

tool: D2: 9223372036854775807

tool: D14: 1

assistant: ```c
double bisquare(const double x) {
  if (fabs(x) <= 1.0) {
    double f = 1.0 - x * x;
    return (f * f);
  } else
    return 0.0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bisquare>:
  endbr64
  movapd %xmm0,%xmm1
  andpd  D11(%rip),%xmm1
  movsd  D1(%rip),%xmm2
  comisd %xmm1,%xmm2
  jae    L2
  pxor   %xmm0,%xmm0
  ret
L2:
  mulsd  %xmm0,%xmm0
  movapd %xmm2,%xmm1
  subsd  %xmm0,%xmm1
  mulsd  %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D11
type_infered: qword[2]
name: retrieve_data
arguments: label: D1
type_infered: f64

tool: {"D11": [0x7fffffffffffffff,0x0000000000000000]}

tool: D1: 1

assistant: ```c
double bisquare(const double x) {
  if (fabs(x) <= 1.0) {
    double f = 1.0 - x * x;
    return (f * f);
  } else
    return 0.0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bisquare>:
  endbr64
  movsd  D18(%rip),%xmm1
  movapd %xmm0,%xmm2
  andpd  D12(%rip),%xmm2
  comisd %xmm2,%xmm1
  jae    L1
  pxor   %xmm0,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_label: D18
datatype: f64
name: read_data
arguments: data_label: D12
datatype: qword[2]

tool: D18: 1

tool: {"D12": [0x7fffffffffffffff,0x0000000000000000]}

assistant: ```c
double bisquare(const double x) {
  if (fabs(x) <= 1.0) {
    double f = 1.0 - x * x;
    return (f * f);
  } else
    return 0.0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bisquare>:
  endbr64
  movsd  D17(%rip),%xmm1
  movapd %xmm0,%xmm2
  andpd  D19(%rip),%xmm2
  comisd %xmm2,%xmm1
  jae    L4
  pxor   %xmm0,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
type_infered: f64
name: extract_symbolic_data
arguments: label: D19
type_infered: qword[2]

tool: D17: 1

tool: {"D19": [0x7fffffffffffffff,0x0000000000000000]}

assistant: ```c
double bisquare(const double x) {
  if (fabs(x) <= 1.0) {
    double f = 1.0 - x * x;
    return (f * f);
  } else
    return 0.0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L26
L19:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L38
  mov    $0x0,%eax
  jmp    L16
L38:
  addl   $0x1,-0x4(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L19
  mov    $0x1,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int cmp(unsigned char *str1, unsigned char *str2, int l) {
  int i = 0;
  for (i = 0; i < l; i++) {
    if (str1[i] != str2[i])
      return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cmp>:
  endbr64
  test   %edx,%edx
  jle    L29
  mov    %edx,%edx
  mov    $0x0,%eax
L14:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  jne    L19
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L14
  mov    $0x1,%eax
  ret
L29:
  mov    $0x1,%eax
  ret
L19:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int cmp(unsigned char *str1, unsigned char *str2, int l) {
  int i = 0;
  for (i = 0; i < l; i++) {
    if (str1[i] != str2[i])
      return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmp>:
  endbr64
  test   %edx,%edx
  jle    L39
  movslq %edx,%rdx
  xor    %eax,%eax
  jmp    L10
  nop
L6:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L39
L10:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L6
  xor    %eax,%eax
  ret
  nopl   (%rax)
L39:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int cmp(unsigned char *str1, unsigned char *str2, int l) {
  int i = 0;
  for (i = 0; i < l; i++) {
    if (str1[i] != str2[i])
      return 0;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cmp>:
  endbr64
  test   %edx,%edx
  jle    L30
  movslq %edx,%rdx
  xor    %eax,%eax
  jmp    L20
  nop
L5:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L30
L20:
  movzbl (%rsi,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L5
  xor    %eax,%eax
  ret
  nopl   (%rax)
L30:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int cmp(unsigned char *str1, unsigned char *str2, int l) {
  int i = 0;
  for (i = 0; i < l; i++) {
    if (str1[i] != str2[i])
      return 0;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<print_res>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x4(%rbp)
  jne    L4
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L19
L4:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L19:
  cmpl   $0x0,-0x8(%rbp)
  jne    L10
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L14
L10:
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L14:
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D41
type: string
name: read_data
arguments: data_tag: D26
type: string
name: read_data
arguments: data_tag: D25
type: string
name: read_data
arguments: data_tag: D34
type: string
name: read_data
arguments: data_tag: D44
type: string

tool: D41: 
Is '%s' a Palindrome?


tool: D26: 
Iterativ function says Yes!

tool: D25: 
Iterativ function says No!

tool: D34: Rekursiv function says Yes!


tool: D44: Rekursiv function says No!


assistant: ```c
void print_res(int res1, int res2, char *input) {
  printf("\nIs '%s' a Palindrome?\n", input);
  if (res1 == 0)
    printf("\nIterativ function says Yes!\n");
  else
    printf("\nIterativ function says No!\n");
  if (res2 == 0)
    printf("Rekursiv function says Yes!\n\n");
  else
    printf("Rekursiv function says No!\n\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_res>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %rdx,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jne    L18
  lea    D15(%rip),%rdi
  call   <puts@plt>
L6:
  test   %ebx,%ebx
  jne    L2
  lea    D8(%rip),%rdi
  call   <puts@plt>
L27:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  lea    D27(%rip),%rdi
  call   <puts@plt>
  jmp    L6
L2:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L27
```

tool_calls: name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D15
datatype: string
name: unpack_data
arguments: tag: D8
datatype: string
name: unpack_data
arguments: tag: D27
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string

tool: D16: 
Is '%s' a Palindrome?


tool: D15: 
Iterativ function says Yes!

tool: D8: Rekursiv function says Yes!


tool: D27: 
Iterativ function says No!

tool: D23: Rekursiv function says No!


assistant: ```c
void print_res(int res1, int res2, char *input) {
  printf("\nIs '%s' a Palindrome?\n", input);
  if (res1 == 0)
    printf("\nIterativ function says Yes!\n");
  else
    printf("\nIterativ function says No!\n");
  if (res2 == 0)
    printf("Rekursiv function says Yes!\n\n");
  else
    printf("Rekursiv function says No!\n\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_res>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  lea    D33(%rip),%rdi
  push   %rbx
  mov    %esi,%ebx
  mov    %rdx,%rsi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jne    L2
  lea    D38(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jne    L24
L0:
  add    $0x8,%rsp
  lea    D12(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <puts@plt>
  nopl   (%rax)
L2:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  je     L0
L24:
  add    $0x8,%rsp
  lea    D25(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <puts@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D33
type_infered: string
name: extract_symbolic_data
arguments: label: D38
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string
name: extract_symbolic_data
arguments: label: D15
type_infered: string
name: extract_symbolic_data
arguments: label: D25
type_infered: string

tool: D33: 
Is '%s' a Palindrome?


tool: D38: 
Iterativ function says Yes!

tool: D12: Rekursiv function says Yes!


tool: D15: 
Iterativ function says No!

tool: D25: Rekursiv function says No!


assistant: ```c
void print_res(int res1, int res2, char *input) {
  printf("\nIs '%s' a Palindrome?\n", input);
  if (res1 == 0)
    printf("\nIterativ function says Yes!\n");
  else
    printf("\nIterativ function says No!\n");
  if (res2 == 0)
    printf("Rekursiv function says Yes!\n\n");
  else
    printf("Rekursiv function says No!\n\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<print_res>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  lea    D15(%rip),%rdi
  push   %rbx
  mov    %esi,%ebx
  mov    %rdx,%rsi
  sub    $0x8,%rsp
  call   <printf@plt>
  test   %ebp,%ebp
  jne    L26
  lea    D4(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  jne    L11
L24:
  add    $0x8,%rsp
  lea    D31(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <puts@plt>
  nopl   (%rax)
L26:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  test   %ebx,%ebx
  je     L24
L11:
  add    $0x8,%rsp
  lea    D6(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <puts@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
type: string
name: read_data
arguments: data_label: D4
type: string
name: read_data
arguments: data_label: D31
type: string
name: read_data
arguments: data_label: D35
type: string
name: read_data
arguments: data_label: D6
type: string

tool: D15: 
Is '%s' a Palindrome?


tool: D4: 
Iterativ function says Yes!

tool: D31: Rekursiv function says Yes!


tool: D35: 
Iterativ function says No!

tool: D6: Rekursiv function says No!


assistant: ```c
void print_res(int res1, int res2, char *input) {
  printf("\nIs '%s' a Palindrome?\n", input);
  if (res1 == 0)
    printf("\nIterativ function says Yes!\n");
  else
    printf("\nIterativ function says No!\n");
  if (res2 == 0)
    printf("Rekursiv function says Yes!\n\n");
  else
    printf("Rekursiv function says No!\n\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<print_raw_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L92:
  movl   $0x0,-0x4(%rbp)
  jmp    L95
L0:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L24
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L84
L24:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L84:
  addl   $0x1,-0x4(%rbp)
L95:
  cmpl   $0xf,-0x4(%rbp)
  jle    L0
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L82:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x1c(%rbp)
  jle    L56
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L37
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L56
L37:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L56:
  addl   $0x1,-0x4(%rbp)
L9:
  cmpl   $0xf,-0x4(%rbp)
  jle    L82
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x10,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L92
  nop
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D16
data_type: string
name: retrieve_data
arguments: data_label: D11
data_type: string

tool: D16: %02X 

tool: D11:    

assistant: ```c
void print_raw_data(char *data, int len) {
  int i;
  int j;
  for (i = 0; i < len; i += 16) {
    for (j = 0; j < 16; j++) {
      if (i + j < len)
        printf("%02X ", data[i + j]);
      else
        printf("   ");
    }
    for (j = 0; j < 16; j++) {
      if (i + j < len) {
        if (data[i + j] == '\0')
          printf(".");
        else
          printf("%c", data[i + j]);
      }
    }
    printf("\n");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_raw_data>:
  endbr64
  test   %esi,%esi
  jle    L72
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,%r13d
  mov    %rdi,(%rsp)
  lea    0x10(%rdi),%rbp
  lea    -0x1(%rsi),%eax
  and    $0xfffffff0,%eax
  add    $0x10,%eax
  mov    %eax,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  lea    D19(%rip),%r15
  lea    D3(%rip),%r14
  jmp    L15
L35:
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L7:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L99
L86:
  lea    (%r12,%rbx,1),%eax
  cmp    %eax,%r13d
  jle    L35
  movsbl (%rbx),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
L99:
  mov    (%rsp),%rbx
  jmp    L113
L75:
  movsbl %dil,%edi
  call   <putchar@plt>
L32:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L106
L113:
  lea    (%r12,%rbx,1),%eax
  cmp    %eax,%r13d
  jle    L32
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L75
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L32
L106:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x10,0x8(%rsp)
  mov    0x8(%rsp),%eax
  add    $0x10,%rbp
  addq   $0x10,(%rsp)
  cmp    0xc(%rsp),%eax
  je     L41
L15:
  mov    (%rsp),%rax
  mov    %rax,%rbx
  mov    0x8(%rsp),%r12d
  sub    %eax,%r12d
  jmp    L86
L41:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L72:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D19
type_guessed: string
name: retrieve_data
arguments: data_tag: D3
type_guessed: string

tool: D19:    

tool: D3: %02X 

assistant: ```c
void print_raw_data(char *data, int len) {
  int i;
  int j;
  for (i = 0; i < len; i += 16) {
    for (j = 0; j < 16; j++) {
      if (i + j < len)
        printf("%02X ", data[i + j]);
      else
        printf("   ");
    }
    for (j = 0; j < 16; j++) {
      if (i + j < len) {
        if (data[i + j] == '\0')
          printf(".");
        else
          printf("%c", data[i + j]);
      }
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_raw_data>:
  endbr64
  test   %esi,%esi
  jle    L37
  push   %r15
  lea    -0x1(%rsi),%eax
  push   %r14
  and    $0xfffffff0,%eax
  lea    D11(%rip),%r14
  push   %r13
  add    $0x10,%eax
  lea    D17(%rip),%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  lea    0x10(%rdi),%rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %eax,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L6:
  mov    (%rsp),%rax
  mov    0x8(%rsp),%ebp
  mov    %rax,%r15
  sub    %eax,%ebp
  jmp    L77
  nop
L12:
  movsbl (%r15),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  je     L4
L77:
  lea    0x0(%rbp,%r15,1),%eax
  cmp    %eax,%r12d
  jg     L12
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L77
L4:
  mov    (%rsp),%r15
  jmp    L29
  nopw   0x0(%rax,%rax,1)
L7:
  mov    $0x2e,%edi
  call   <putchar@plt>
L31:
  add    $0x1,%r15
  cmp    %rbx,%r15
  je     L49
L29:
  lea    0x0(%rbp,%r15,1),%eax
  cmp    %eax,%r12d
  jle    L31
  movsbl (%r15),%edi
  test   %dil,%dil
  je     L7
  call   <putchar@plt>
  add    $0x1,%r15
  cmp    %rbx,%r15
  jne    L29
L49:
  mov    $0xa,%edi
  add    $0x10,%rbx
  call   <putchar@plt>
  addl   $0x10,0x8(%rsp)
  mov    0x8(%rsp),%eax
  addq   $0x10,(%rsp)
  cmp    0xc(%rsp),%eax
  jne    L6
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  ret
```

tool_calls: name: disassemble_data
arguments: label: D11
type_guessed: string
name: disassemble_data
arguments: label: D17
type_guessed: string

tool: D11:    

tool: D17: %02X 

assistant: ```c
void print_raw_data(char *data, int len) {
  int i;
  int j;
  for (i = 0; i < len; i += 16) {
    for (j = 0; j < 16; j++) {
      if (i + j < len)
        printf("%02X ", data[i + j]);
      else
        printf("   ");
    }
    for (j = 0; j < 16; j++) {
      if (i + j < len) {
        if (data[i + j] == '\0')
          printf(".");
        else
          printf("%c", data[i + j]);
      }
    }
    printf("\n");
  }
}
```