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
<f>:
  endbr64
  test   %edi,%edi
  jle    L11
  mov    %edi,%eax
  push   %r13
  mov    %edi,%r9d
  movaps D17(%rip),%xmm1
  shr    $0x2,%eax
  push   %r12
  movss  D17(%rip),%xmm2
  mov    %rsi,%r10
  lea    -0x1(%rax),%r11d
  push   %rbp
  mov    %rsi,%rdx
  lea    -0x1(%rdi),%ebp
  push   %rbx
  add    $0x1,%r11
  movslq %edi,%rbx
  and    $0xfffffffc,%r9d
  movq   D17(%rip),%xmm3
  lea    0x0(,%rbx,4),%r12
  xor    %esi,%esi
  xor    %r8d,%r8d
  shl    $0x4,%r11
  nopw   0x0(%rax,%rax,1)
L12:
  cmp    $0x2,%ebp
  jbe    L77
  lea    (%r11,%rdx,1),%rcx
  mov    %rdx,%rax
  nopl   0x0(%rax)
L17:
  movups (%rax),%xmm0
  add    $0x10,%rax
  addps  %xmm1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L17
  cmp    %r9d,%edi
  je     L61
  mov    %r9d,%ecx
  mov    %r9d,%eax
L43:
  mov    %edi,%r13d
  sub    %ecx,%r13d
  cmp    $0x1,%r13d
  je     L66
  add    %rsi,%rcx
  lea    (%r10,%rcx,4),%rcx
  movq   (%rcx),%xmm0
  addps  %xmm3,%xmm0
  movlps %xmm0,(%rcx)
  mov    %r13d,%ecx
  and    $0xfffffffe,%ecx
  add    %ecx,%eax
  cmp    %ecx,%r13d
  je     L61
L66:
  cltq
  movss  (%rdx,%rax,4),%xmm0
  addss  %xmm2,%xmm0
  movss  %xmm0,(%rdx,%rax,4)
L61:
  add    $0x1,%r8d
  add    %r12,%rdx
  add    %rbx,%rsi
  cmp    %r8d,%edi
  jne    L12
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L77:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L43
L11:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D17
datatype: byte[16]

tool: {"D17": [0x00,0x00,0x80,0x3f,0x00,0x00,0x80,0x3f,0x00,0x00,0x80,0x3f,0x00,0x00,0x80,0x3f]}

assistant: ```c
void f(int n, float m[n][n]) {
  int i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      m[i][j]++;
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fn_ch_120>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <next_utf8@plt>
  cmp    $0x78,%eax
  jne    L29
  mov    -0xc(%rbp),%eax
  jmp    L17
L29:
  mov    $0xffffffff,%eax
L17:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int fn_ch_120(const char *text) {
  int pos = 0;
  if (next_utf8(text, &pos) == 120)
    return pos;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fn_ch_120>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  call   <next_utf8@plt>
  cmp    $0x78,%eax
  mov    $0xffffffff,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn_ch_120(const char *text) {
  int pos = 0;
  if (next_utf8(text, &pos) == 120)
    return pos;
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fn_ch_120>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <next_utf8@plt>
  cmp    $0x78,%eax
  mov    $0xffffffff,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn_ch_120(const char *text) {
  int pos = 0;
  if (next_utf8(text, &pos) == 120)
    return pos;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn_ch_120>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  movl   $0x0,0x4(%rsp)
  call   <next_utf8@plt>
  cmp    $0x78,%eax
  mov    $0xffffffff,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int fn_ch_120(const char *text) {
  int pos = 0;
  if (next_utf8(text, &pos) == 120)
    return pos;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mpow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L10:
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L10
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int mpow(int base, int exponent) {
  for (int i = 0; i <= exponent; i++) {
    base = base * exponent;
  }
  return base;
}
```
user: What is the c source code of the assembly code below:

```asm
<mpow>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  js     L7
  lea    0x1(%rsi),%ecx
  mov    $0x0,%edx
L18:
  imul   %esi,%eax
  add    $0x1,%edx
  cmp    %ecx,%edx
  jne    L18
L7:
  ret
```

assistant: ```c
int mpow(int base, int exponent) {
  for (int i = 0; i <= exponent; i++) {
    base = base * exponent;
  }
  return base;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mpow>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  js     L5
  lea    0x1(%rsi),%ecx
  xor    %edx,%edx
  nop
L12:
  add    $0x1,%edx
  imul   %esi,%eax
  cmp    %ecx,%edx
  jne    L12
L5:
  ret
```

assistant: ```c
int mpow(int base, int exponent) {
  for (int i = 0; i <= exponent; i++) {
    base = base * exponent;
  }
  return base;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mpow>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  js     L9
  lea    0x1(%rsi),%ecx
  xor    %edx,%edx
  nop
L13:
  add    $0x1,%edx
  imul   %esi,%eax
  cmp    %ecx,%edx
  jne    L13
L9:
  ret
```

assistant: ```c
int mpow(int base, int exponent) {
  for (int i = 0; i <= exponent; i++) {
    base = base * exponent;
  }
  return base;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_near_round_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movq   $0x1,-0x10(%rbp)
  movq   $0x1,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L29
L32:
  sarq   -0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L29:
  cmpq   $0x0,-0x18(%rbp)
  jne    L32
  mov    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  shlq   %cl,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%ecx
  shlq   %cl,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jle    L27
  mov    -0x10(%rbp),%rax
  jmp    L14
L27:
  mov    -0x8(%rbp),%rax
L14:
  pop    %rbp
  ret
```

assistant: ```c
long long get_near_round_value(long long size) {
  int c = 0;
  long long up = 1;
  long long down = 1;
  long long temp = size;
  while (temp != 0) {
    temp = temp >> 1;
    c++;
  }
  up = up << c;
  down = down << (c - 1);
  if (size - down > size - up)
    return up;
  else
    return down;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_near_round_value>:
  endbr64
  test   %rdi,%rdi
  je     L4
  mov    $0x0,%edx
L8:
  add    $0x1,%edx
  sar    %rdi
  jne    L8
L10:
  lea    -0x1(%rdx),%ecx
  mov    $0x1,%esi
  mov    %rsi,%rax
  shl    %cl,%rax
  mov    %edx,%ecx
  shl    %cl,%rsi
  cmp    %rsi,%rax
  cmovl  %rsi,%rax
  ret
L4:
  mov    $0x0,%edx
  jmp    L10
```

assistant: ```c
long long get_near_round_value(long long size) {
  int c = 0;
  long long up = 1;
  long long down = 1;
  long long temp = size;
  while (temp != 0) {
    temp = temp >> 1;
    c++;
  }
  up = up << c;
  down = down << (c - 1);
  if (size - down > size - up)
    return up;
  else
    return down;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_near_round_value>:
  endbr64
  test   %rdi,%rdi
  je     L15
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L28:
  sar    %rdi
  mov    %ecx,%edx
  lea    0x1(%rcx),%ecx
  jne    L28
  mov    $0x1,%esi
  shl    %cl,%rsi
L16:
  mov    %edx,%ecx
  mov    $0x1,%eax
  shl    %cl,%rax
  cmp    %rsi,%rax
  cmovl  %rsi,%rax
  ret
  nopl   0x0(%rax)
L15:
  mov    $0x1,%esi
  mov    $0xffffffff,%edx
  jmp    L16
```

assistant: ```c
long long get_near_round_value(long long size) {
  int c = 0;
  long long up = 1;
  long long down = 1;
  long long temp = size;
  while (temp != 0) {
    temp = temp >> 1;
    c++;
  }
  up = up << c;
  down = down << (c - 1);
  if (size - down > size - up)
    return up;
  else
    return down;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_near_round_value>:
  endbr64
  test   %rdi,%rdi
  je     L22
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L5:
  sar    %rdi
  mov    %ecx,%edx
  lea    0x1(%rcx),%ecx
  jne    L5
  mov    $0x1,%esi
  shl    %cl,%rsi
L2:
  mov    %edx,%ecx
  mov    $0x1,%eax
  shl    %cl,%rax
  cmp    %rsi,%rax
  cmovl  %rsi,%rax
  ret
  nopl   0x0(%rax)
L22:
  mov    $0x1,%esi
  mov    $0xffffffff,%edx
  jmp    L2
```

assistant: ```c
long long get_near_round_value(long long size) {
  int c = 0;
  long long up = 1;
  long long down = 1;
  long long temp = size;
  while (temp != 0) {
    temp = temp >> 1;
    c++;
  }
  up = up << c;
  down = down << (c - 1);
  if (size - down > size - up)
    return up;
  else
    return down;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    D74(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D76(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  comiss 0x4(%rsp),%xmm0
  jae    L54
  comiss 0x8(%rsp),%xmm0
  jae    L73
  comiss 0xc(%rsp),%xmm0
  jae    L89
  movss  0x10(%rsp),%xmm1
  comiss %xmm1,%xmm0
  jae    L68
  comiss %xmm0,%xmm1
  jbe    L82
  lea    D49(%rip),%rdi
  call   <printf@plt>
  jmp    L82
L54:
  lea    D77(%rip),%rdi
  call   <printf@plt>
L82:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L73:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L82
L89:
  lea    D62(%rip),%rdi
  call   <printf@plt>
  jmp    L82
L68:
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L82
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D17
type_infered: string
name: retrieve_data
arguments: tag: D74
type_infered: string
name: retrieve_data
arguments: tag: D45
type_infered: string
name: retrieve_data
arguments: tag: D76
type_infered: string
name: retrieve_data
arguments: tag: D54
type_infered: string
name: retrieve_data
arguments: tag: D49
type_infered: string
name: retrieve_data
arguments: tag: D77
type_infered: string
name: retrieve_data
arguments: tag: D10
type_infered: string
name: retrieve_data
arguments: tag: D62
type_infered: string
name: retrieve_data
arguments: tag: D35
type_infered: string

tool: D17: Enter thresholds for A, B, C, D


tool: D74: in that order, decreasing percentages > 

tool: D45: %f%f%f%f

tool: D76: Thank you. Now enter student score (percent) >

tool: D54: %f

tool: D49: Student has failed the course


tool: D77: Student has an A grade


tool: D10: Student has an B grade


tool: D62: Student has an C grade


tool: D35: Student has an D grade


assistant: ```c
int main() {
  float a, b, c, d, score;
  printf("Enter thresholds for A, B, C, D\n");
  printf("in that order, decreasing percentages > ");
  scanf("%f"
        "%f"
        "%f"
        "%f",
        &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%f", &score);
  if (score >= a)
    printf("Student has an A grade\n");
  else if (score >= b)
    printf("Student has an B grade\n");
  else if (score >= c)
    printf("Student has an C grade\n");
  else if (score >= d)
    printf("Student has an D grade\n");
  else if (score < d)
    printf("Student has failed the course\n");
  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
  sub    $0x28,%rsp
  lea    D57(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D64(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D99(%rip),%rdi
  call   <scanf@plt>
  lea    D69(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D88(%rip),%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  comiss 0x4(%rsp),%xmm0
  jae    L4
  comiss 0x8(%rsp),%xmm0
  jae    L9
  comiss 0xc(%rsp),%xmm0
  jae    L1
  movss  0x10(%rsp),%xmm1
  comiss %xmm1,%xmm0
  jae    L49
  comiss %xmm0,%xmm1
  jbe    L10
  lea    D87(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L9:
  lea    D22(%rip),%rdi
  call   <printf@plt>
L10:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L71
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L4:
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L1:
  lea    D44(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L49:
  lea    D60(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L71:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D57
type_guessed: string
name: parse_data
arguments: data_tag: D64
type_guessed: string
name: parse_data
arguments: data_tag: D99
type_guessed: string
name: parse_data
arguments: data_tag: D69
type_guessed: string
name: parse_data
arguments: data_tag: D88
type_guessed: string
name: parse_data
arguments: data_tag: D87
type_guessed: string
name: parse_data
arguments: data_tag: D22
type_guessed: string
name: parse_data
arguments: data_tag: D23
type_guessed: string
name: parse_data
arguments: data_tag: D44
type_guessed: string
name: parse_data
arguments: data_tag: D60
type_guessed: string

tool: D57: Enter thresholds for A, B, C, D


tool: D64: in that order, decreasing percentages > 

tool: D99: %f%f%f%f

tool: D69: Thank you. Now enter student score (percent) >

tool: D88: %f

tool: D87: Student has failed the course


tool: D22: Student has an B grade


tool: D23: Student has an A grade


tool: D44: Student has an C grade


tool: D60: Student has an D grade


assistant: ```c
int main() {
  float a, b, c, d, score;
  printf("Enter thresholds for A, B, C, D\n");
  printf("in that order, decreasing percentages > ");
  scanf("%f"
        "%f"
        "%f"
        "%f",
        &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%f", &score);
  if (score >= a)
    printf("Student has an A grade\n");
  else if (score >= b)
    printf("Student has an B grade\n");
  else if (score >= c)
    printf("Student has an C grade\n");
  else if (score >= d)
    printf("Student has an D grade\n");
  else if (score < d)
    printf("Student has failed the course\n");
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D80(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D44(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    0x10(%rsp),%r8
  lea    D74(%rip),%rdi
  call   <scanf@plt>
  lea    D22(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  movss  0x14(%rsp),%xmm0
  comiss 0x4(%rsp),%xmm0
  jae    L6
  comiss 0x8(%rsp),%xmm0
  jae    L34
  comiss 0xc(%rsp),%xmm0
  jae    L19
  movss  0x10(%rsp),%xmm1
  comiss %xmm1,%xmm0
  jae    L47
  comiss %xmm0,%xmm1
  jbe    L56
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L56
L34:
  lea    D47(%rip),%rdi
  call   <printf@plt>
L56:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L6:
  lea    D72(%rip),%rdi
  call   <printf@plt>
  jmp    L56
L19:
  lea    D41(%rip),%rdi
  call   <printf@plt>
  jmp    L56
L47:
  lea    D40(%rip),%rdi
  call   <printf@plt>
  jmp    L56
L67:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D80
type_guessed: string
name: retrieve_data
arguments: data_label: D44
type_guessed: string
name: retrieve_data
arguments: data_label: D74
type_guessed: string
name: retrieve_data
arguments: data_label: D22
type_guessed: string
name: retrieve_data
arguments: data_label: D34
type_guessed: string
name: retrieve_data
arguments: data_label: D93
type_guessed: string
name: retrieve_data
arguments: data_label: D47
type_guessed: string
name: retrieve_data
arguments: data_label: D72
type_guessed: string
name: retrieve_data
arguments: data_label: D41
type_guessed: string
name: retrieve_data
arguments: data_label: D40
type_guessed: string

tool: D80: Enter thresholds for A, B, C, D


tool: D44: in that order, decreasing percentages > 

tool: D74: %f%f%f%f

tool: D22: Thank you. Now enter student score (percent) >

tool: D34: %f

tool: D93: Student has failed the course


tool: D47: Student has an B grade


tool: D72: Student has an A grade


tool: D41: Student has an C grade


tool: D40: Student has an D grade


assistant: ```c
int main() {
  float a, b, c, d, score;
  printf("Enter thresholds for A, B, C, D\n");
  printf("in that order, decreasing percentages > ");
  scanf("%f"
        "%f"
        "%f"
        "%f",
        &a, &b, &c, &d);
  printf("Thank you. Now enter student score (percent) >");
  scanf("%f", &score);
  if (score >= a)
    printf("Student has an A grade\n");
  else if (score >= b)
    printf("Student has an B grade\n");
  else if (score >= c)
    printf("Student has an C grade\n");
  else if (score >= d)
    printf("Student has an D grade\n");
  else if (score < d)
    printf("Student has failed the course\n");
  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
<lrotate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x3,%edx
  sub    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L28
L39:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L46
L52:
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L46:
  cmpl   $0x5,-0xc(%rbp)
  jle    L52
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L55
L56:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L55:
  cmpl   $0x6,-0xc(%rbp)
  jle    L56
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x10(%rbp)
L28:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L39
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type: string

tool: D2: %d 

assistant: ```c
void lrotate(int seq[], int nr) {
  int try = 0;
  int i = 0;
  int num_rotations = nr % 7;
  for (try = 0; try < num_rotations; try++) {
    int tmp = seq[0];
    for (i = 0; i < 7 - 1; i++) {
      seq[i] = seq[i + 1];
    }
    seq[i] = tmp;
    for (i = 0; i < 7; i++)
      printf("%d ", seq[i]);
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<lrotate>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movslq %esi,%r14
  imul   $0xffffffff92492493,%r14,%r14
  shr    $0x20,%r14
  add    %esi,%r14d
  sar    $0x2,%r14d
  mov    %esi,%eax
  sar    $0x1f,%eax
  sub    %eax,%r14d
  lea    0x0(,%r14,8),%eax
  sub    %r14d,%eax
  sub    %eax,%esi
  test   %esi,%esi
  jle    L54
  mov    %rdi,%r12
  mov    %esi,%r14d
  mov    $0x0,%r13d
  lea    D1(%rip),%rbp
  jmp    L30
L12:
  mov    %esi,0x18(%r12)
  lea    0x1c(%r12),%r15
L19:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%r15
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r13d
  cmp    %r13d,%r14d
  je     L54
L30:
  mov    (%r12),%esi
  mov    %r12,%rbx
  lea    0x18(%r12),%rcx
  mov    %r12,%rax
L36:
  mov    0x4(%rax),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L36
  jmp    L12
L54:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

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

tool: D1: %d 

assistant: ```c
void lrotate(int seq[], int nr) {
  int try = 0;
  int i = 0;
  int num_rotations = nr % 7;
  for (try = 0; try < num_rotations; try++) {
    int tmp = seq[0];
    for (i = 0; i < 7 - 1; i++) {
      seq[i] = seq[i + 1];
    }
    seq[i] = tmp;
    for (i = 0; i < 7; i++)
      printf("%d ", seq[i]);
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lrotate>:
  endbr64
  push   %r15
  mov    %esi,%eax
  push   %r14
  sar    $0x1f,%eax
  push   %r13
  movslq %esi,%r13
  imul   $0xffffffff92492493,%r13,%r13
  push   %r12
  push   %rbp
  push   %rbx
  shr    $0x20,%r13
  add    %esi,%r13d
  sar    $0x2,%r13d
  sub    $0x18,%rsp
  sub    %eax,%r13d
  lea    0x0(,%r13,8),%eax
  sub    %r13d,%eax
  sub    %eax,%esi
  mov    %esi,0xc(%rsp)
  test   %esi,%esi
  jle    L3
  mov    %rdi,%r12
  lea    0x4(%rdi),%r15
  lea    0x1c(%rdi),%rbp
  xor    %r14d,%r14d
  lea    D8(%rip),%rbx
  nopl   0x0(%rax)
L20:
  mov    (%r12),%r13d
  mov    $0x18,%edx
  mov    %r15,%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
  mov    %r13d,0x18(%r12)
  mov    %r12,%r13
  nopl   0x0(%rax)
L22:
  mov    0x0(%r13),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x4,%r13
  call   <printf@plt>
  cmp    %r13,%rbp
  jne    L22
  mov    $0xa,%edi
  add    $0x1,%r14d
  call   <putchar@plt>
  cmp    %r14d,0xc(%rsp)
  jne    L20
L3:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D8: %d 

assistant: ```c
void lrotate(int seq[], int nr) {
  int try = 0;
  int i = 0;
  int num_rotations = nr % 7;
  for (try = 0; try < num_rotations; try++) {
    int tmp = seq[0];
    for (i = 0; i < 7 - 1; i++) {
      seq[i] = seq[i + 1];
    }
    seq[i] = tmp;
    for (i = 0; i < 7; i++)
      printf("%d ", seq[i]);
    printf("\n");
  }
}
```