ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0x44(%rbp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,-0x44(%rbp)
  jle    L126
  mov    $0x0,%r15d
  lea    D31(%rip),%r14
  jmp    L83
L144:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L65
  orq    $0x0,-0x8(%rsp,%rax,1)
L65:
  mov    %rsp,%r13
  mov    %r13,-0x60(%rbp)
  test   %edx,%edx
  jle    L88
  mov    %r13,%rbx
  mov    $0x0,%r12d
L78:
  mov    %r13,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%r12d
  mov    -0x40(%rbp),%eax
  add    $0x4,%r13
  cmp    %r12d,%eax
  jg     L78
  test   %eax,%eax
  jle    L63
  mov    %eax,%eax
  mov    -0x60(%rbp),%rdi
  lea    (%rdi,%rax,4),%rcx
  mov    $0x0,%eax
L5:
  cmpl   $0x0,(%rbx)
  setle  %dl
  movzbl %dl,%edx
  add    %edx,%eax
  add    $0x4,%rbx
  cmp    %rcx,%rbx
  jne    L5
L149:
  cmp    %eax,-0x3c(%rbp)
  jg     L108
  lea    D0(%rip),%rdi
  call   <printf@plt>
L115:
  mov    -0x58(%rbp),%rsp
  add    $0x1,%r15d
  cmp    %r15d,-0x44(%rbp)
  jle    L126
L83:
  mov    %rsp,-0x58(%rbp)
  lea    -0x3c(%rbp),%rdx
  lea    -0x40(%rbp),%rsi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x40(%rbp),%edx
  movslq %edx,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L94:
  cmp    %rcx,%rsp
  je     L144
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L94
L88:
  mov    $0x0,%eax
  jmp    L149
L63:
  mov    $0x0,%eax
  jmp    L149
L108:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L115
L126:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D31
data_type: string
name: parse_data
arguments: tag: D0
data_type: string
name: parse_data
arguments: tag: D9
data_type: string
name: parse_data
arguments: tag: D11
data_type: string

tool: D31: %d

tool: D0: NO


tool: D9: %d %d

tool: D11: YES


assistant: ```c
int main() {
  int t, count = 0;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int k;
    scanf("%d %d", &n, &k);
    int a[n];
    for (int a_i = 0; a_i < n; a_i++) {
      scanf("%d", &a[a_i]);
    }
    count = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] <= 0)
        count++;
    }
    if (count >= k)
      printf("NO\n");
    else
      printf("YES\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x44(%rbp),%rsi
  push   %r13
  push   %r12
  push   %rbx
  lea    D24(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    -0x44(%rbp),%ecx
  test   %ecx,%ecx
  jle    L126
  lea    -0x3c(%rbp),%rax
  movl   $0x0,-0x54(%rbp)
  mov    %rax,-0x68(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,-0x70(%rbp)
  nopl   0x0(%rax)
L71:
  mov    -0x68(%rbp),%rdx
  mov    -0x70(%rbp),%rsi
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  movslq -0x40(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L11
L109:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L109
L11:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L54
  orq    $0x0,-0x8(%rsp,%rdx,1)
L54:
  mov    %rsp,%r13
  mov    %r13,-0x60(%rbp)
  test   %ecx,%ecx
  jle    L96
  mov    %r13,%r15
  xor    %r12d,%r12d
  xchg   %ax,%ax
L92:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r12d
  call   <scanf@plt>
  mov    -0x40(%rbp),%eax
  add    $0x4,%r13
  cmp    %r12d,%eax
  jg     L92
  test   %eax,%eax
  jle    L96
  mov    -0x60(%rbp),%rdx
  lea    (%rdx,%rax,4),%rsi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L10:
  mov    (%r15),%edx
  xor    %ecx,%ecx
  test   %edx,%edx
  setle  %cl
  add    $0x4,%r15
  add    %ecx,%eax
  cmp    %rsi,%r15
  jne    L10
L98:
  cmp    %eax,-0x3c(%rbp)
  jg     L108
  lea    D12(%rip),%rdi
  call   <printf@plt>
L17:
  mov    %r14,%rsp
  addl   $0x1,-0x54(%rbp)
  mov    -0x54(%rbp),%eax
  cmp    %eax,-0x44(%rbp)
  jg     L71
L126:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax)
L108:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L17
L96:
  xor    %eax,%eax
  jmp    L98
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D24
data_type: string
name: disassemble_data
arguments: data_label: D29
data_type: string
name: disassemble_data
arguments: data_label: D12
data_type: string
name: disassemble_data
arguments: data_label: D6
data_type: string

tool: D24: %d

tool: D29: %d %d

tool: D12: NO


tool: D6: YES


assistant: ```c
int main() {
  int t, count = 0;
  scanf("%d", &t);
  for (int a0 = 0; a0 < t; a0++) {
    int n;
    int k;
    scanf("%d %d", &n, &k);
    int a[n];
    for (int a_i = 0; a_i < n; a_i++) {
      scanf("%d", &a[a_i]);
    }
    count = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] <= 0)
        count++;
    }
    if (count >= k)
      printf("NO\n");
    else
      printf("YES\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <first_larger_prime@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type: string
name: extract_symbolic_data
arguments: data_label: D5
type: string

tool: D9: %d

tool: D5: %d - %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int larger_prime = first_larger_prime(n);
  printf("%d - %d = %d\n", larger_prime, n, larger_prime - n);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <first_larger_prime@plt>
  mov    %eax,%esi
  mov    0x4(%rsp),%edx
  mov    %eax,%ecx
  sub    %edx,%ecx
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type: string
name: disassemble_data
arguments: data_tag: D4
type: string

tool: D0: %d

tool: D4: %d - %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int larger_prime = first_larger_prime(n);
  printf("%d - %d = %d\n", larger_prime, n, larger_prime - n);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D17(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <first_larger_prime@plt>
  mov    0x4(%rsp),%edx
  lea    D13(%rip),%rdi
  mov    %eax,%ecx
  mov    %eax,%esi
  sub    %edx,%ecx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D17
type_guessed: string
name: disassemble_data
arguments: label: D13
type_guessed: string

tool: D17: %d

tool: D13: %d - %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int larger_prime = first_larger_prime(n);
  printf("%d - %d = %d\n", larger_prime, n, larger_prime - n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <first_larger_prime@plt>
  mov    0x4(%rsp),%edx
  lea    D8(%rip),%rdi
  mov    %eax,%ecx
  mov    %eax,%esi
  sub    %edx,%ecx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
data_type: string
name: unpack_data
arguments: label: D8
data_type: string

tool: D5: %d

tool: D8: %d - %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int larger_prime = first_larger_prime(n);
  printf("%d - %d = %d\n", larger_prime, n, larger_prime - 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
<forceCpuLoadSlice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getTimeUs@plt>
  mov    -0x34(%rbp),%eax
  imul   -0x38(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
L0:
  movl   $0x0,-0x2c(%rbp)
  jmp    L5
L32:
  addl   $0x1,-0x2c(%rbp)
L5:
  cmpl   $0x3e7,-0x2c(%rbp)
  jle    L32
  mov    -0x38(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <getTimeUs@plt>
  mov    -0x20(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jg     L0
  mov    -0x38(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <usleep@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
int forceCpuLoadSlice(int cpuLoad, int durationUs) {
  uint64_t workDuration;
  uint64_t begin;
  uint64_t end;
  uint64_t current;
  int i;
  getTimeUs(&begin);
  workDuration = (cpuLoad * durationUs) / 100;
  end = begin + workDuration;
  do {
    for (i = 0; i < 1000; i++)
      ;
    sqrt(durationUs);
    getTimeUs(&current);
  } while (current < end);
  usleep(durationUs - workDuration);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<forceCpuLoadSlice>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  call   <getTimeUs@plt>
  mov    %ebp,%eax
  imul   %ebx,%eax
  movslq %eax,%r12
  imul   $0x51eb851f,%r12,%r12
  sar    $0x25,%r12
  sar    $0x1f,%eax
  sub    %eax,%r12d
  movslq %r12d,%rbp
  add    0x8(%rsp),%rbp
  lea    0x10(%rsp),%r13
L28:
  mov    $0x3e8,%eax
L21:
  sub    $0x1,%eax
  jne    L21
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r13,%rdi
  call   <getTimeUs@plt>
  cmp    %rbp,0x10(%rsp)
  jl     L28
  sub    %r12d,%ebx
  mov    %ebx,%edi
  call   <usleep@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int forceCpuLoadSlice(int cpuLoad, int durationUs) {
  uint64_t workDuration;
  uint64_t begin;
  uint64_t end;
  uint64_t current;
  int i;
  getTimeUs(&begin);
  workDuration = (cpuLoad * durationUs) / 100;
  end = begin + workDuration;
  do {
    for (i = 0; i < 1000; i++)
      ;
    sqrt(durationUs);
    getTimeUs(&current);
  } while (current < end);
  usleep(durationUs - workDuration);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<forceCpuLoadSlice>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  lea    0x10(%rsp),%r12
  call   <getTimeUs@plt>
  mov    %ebp,%edi
  imul   %ebx,%edi
  movslq %edi,%r13
  sar    $0x1f,%edi
  imul   $0x51eb851f,%r13,%r13
  sar    $0x25,%r13
  sub    %edi,%r13d
  movslq %r13d,%rbp
  add    0x8(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r12,%rdi
  call   <getTimeUs@plt>
  cmp    %rbp,0x10(%rsp)
  jl     L6
  sub    %r13d,%ebx
  mov    %ebx,%edi
  call   <usleep@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int forceCpuLoadSlice(int cpuLoad, int durationUs) {
  uint64_t workDuration;
  uint64_t begin;
  uint64_t end;
  uint64_t current;
  int i;
  getTimeUs(&begin);
  workDuration = (cpuLoad * durationUs) / 100;
  end = begin + workDuration;
  do {
    for (i = 0; i < 1000; i++)
      ;
    sqrt(durationUs);
    getTimeUs(&current);
  } while (current < end);
  usleep(durationUs - workDuration);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<forceCpuLoadSlice>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdi
  lea    0x10(%rsp),%r12
  call   <getTimeUs@plt>
  mov    %ebp,%edi
  imul   %ebx,%edi
  movslq %edi,%r13
  sar    $0x1f,%edi
  imul   $0x51eb851f,%r13,%r13
  sar    $0x25,%r13
  sub    %edi,%r13d
  movslq %r13d,%rbp
  add    0x8(%rsp),%rbp
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r12,%rdi
  call   <getTimeUs@plt>
  cmp    %rbp,0x10(%rsp)
  jl     L14
  sub    %r13d,%ebx
  mov    %ebx,%edi
  call   <usleep@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int forceCpuLoadSlice(int cpuLoad, int durationUs) {
  uint64_t workDuration;
  uint64_t begin;
  uint64_t end;
  uint64_t current;
  int i;
  getTimeUs(&begin);
  workDuration = (cpuLoad * durationUs) / 100;
  end = begin + workDuration;
  do {
    for (i = 0; i < 1000; i++)
      ;
    sqrt(durationUs);
    getTimeUs(&current);
  } while (current < end);
  usleep(durationUs - workDuration);
  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
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  jle    L7
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <f>
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
  jmp    L10
L7:
  nop
L10:
  leave
  ret
```

assistant: ```c
void f(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<f>:
  endbr64
  cmp    $0x2,%edi
  jg     L6
  ret
L6:
  sub    $0x8,%rsp
  sub    $0x1,%edi
  call   <f>
  mov    $0x0,%eax
  call   <__VERIFIER_error@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void f(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<f>:
  endbr64
  cmp    $0x2,%edi
  jg     L46
  ret
  nopw   0x0(%rax,%rax,1)
L46:
  cmp    $0x3,%edi
  je     L35
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jne    L19
L39:
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <__VERIFIER_error@plt>
L35:
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_error@plt>
  nopw   0x0(%rax,%rax,1)
L19:
  cmp    $0x5,%edi
  je     L40
  sub    $0x4,%edi
  call   <f>
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
L40:
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
  jmp    L39
```

assistant: ```c
void f(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f>:
  endbr64
  cmp    $0x2,%edi
  jg     L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  cmp    $0x3,%edi
  je     L46
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jne    L11
L48:
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1050 <__VERIFIER_error@plt>
L46:
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_error@plt>
  nopw   0x0(%rax,%rax,1)
L11:
  cmp    $0x5,%edi
  je     L1
  sub    $0x4,%edi
  call   <f>
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
L1:
  xor    %eax,%eax
  call   <__VERIFIER_error@plt>
  jmp    L48
```

assistant: ```c
void f(int n) {
  if (n < 3)
    return;
  n--;
  f(n);
  __VERIFIER_error();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aLegendre>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  movsd  %xmm0,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jge    L57
  pxor   %xmm0,%xmm0
  jmp    L34
L57:
  movsd  D3(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  cmpl   $0x0,-0x38(%rbp)
  je     L48
  movsd  -0x40(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm0,%xmm1
  movsd  D3(%rip),%xmm0
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   D15(%rip),%xmm0
  movq   %rax,%xmm2
  xorpd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x1,-0x24(%rbp)
  jmp    L65
L18:
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  movsd  -0x20(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  addl   $0x1,-0x24(%rbp)
L65:
  mov    -0x24(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jle    L18
L48:
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x24(%rbp)
  jmp    L62
L47:
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  -0x40(%rbp),%xmm0
  mulsd  -0x18(%rbp),%xmm0
  mov    -0x24(%rbp),%edx
  mov    -0x38(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x24(%rbp),%eax
  sub    -0x38(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  addl   $0x1,-0x24(%rbp)
L62:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jle    L47
  movsd  -0x20(%rbp),%xmm0
L34:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D3
datatype: f64
name: retrieve_data
arguments: data_label: D15
datatype: f32[2]

tool: D3: 1

tool: {"D15": [0.0,-0.0]}

assistant: ```c
double aLegendre(int l, int m, double x) {
  double p, pm1, pm2, sqx;
  int i;
  if (l < m)
    return 0e0;
  p = 1e0;
  pm1 = 0e0;
  if (m) {
    sqx = -sqrt(1e0 - x * x);
    for (i = 1; i <= m; i++)
      p *= (2 * i - 1) * sqx;
  }
  for (i = m + 1; i <= l; i++) {
    pm2 = pm1;
    pm1 = p;
    p = ((2 * i - 1) * x * pm1 - (i + m - 1) * pm2) / (i - m);
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<aLegendre>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  movsd  %xmm0,0x8(%rsp)
  pxor   %xmm0,%xmm0
  cmp    %esi,%edi
  jl     L95
  mov    %edi,%ebp
  mov    %esi,%ebx
  movsd  D1(%rip),%xmm2
  test   %esi,%esi
  jne    L65
L98:
  lea    0x1(%rbx),%eax
  cmp    %eax,%ebp
  jl     L30
  lea    -0x1(%rax,%rax,1),%edx
  mov    %ebp,%esi
  sub    %ebx,%esi
  mov    $0x1,%eax
  pxor   %xmm3,%xmm3
  lea    -0x1(%rbx,%rbx,1),%edi
  jmp    L46
L65:
  movapd %xmm2,%xmm0
  movsd  0x8(%rsp),%xmm5
  movapd %xmm5,%xmm1
  mulsd  %xmm5,%xmm1
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%edi
  call   <sqrt@plt>
  xorpd  D28(%rip),%xmm0
  test   %ebx,%ebx
  jle    L31
  lea    0x1(%rbx,%rbx,1),%edx
  mov    $0x1,%eax
  movsd  D1(%rip),%xmm2
L4:
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mulsd  %xmm0,%xmm1
  mulsd  %xmm1,%xmm2
  add    $0x2,%eax
  cmp    %edx,%eax
  jne    L4
  jmp    L98
L31:
  movsd  D1(%rip),%xmm2
  jmp    L98
L14:
  movapd %xmm0,%xmm2
L46:
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  mulsd  0x8(%rsp),%xmm0
  mulsd  %xmm2,%xmm0
  lea    (%rdi,%rax,1),%ecx
  pxor   %xmm1,%xmm1
  cvtsi2sd %ecx,%xmm1
  mulsd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  add    $0x2,%edx
  mov    %eax,%ecx
  add    $0x1,%eax
  movapd %xmm2,%xmm3
  cmp    %esi,%ecx
  jne    L14
L95:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  movapd %xmm2,%xmm0
  jmp    L95
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D28
data_type: f32[4]

tool: D1: 1

tool: {"D28": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double aLegendre(int l, int m, double x) {
  double p, pm1, pm2, sqx;
  int i;
  if (l < m)
    return 0e0;
  p = 1e0;
  pm1 = 0e0;
  if (m) {
    sqx = -sqrt(1e0 - x * x);
    for (i = 1; i <= m; i++)
      p *= (2 * i - 1) * sqx;
  }
  for (i = m + 1; i <= l; i++) {
    pm2 = pm1;
    pm1 = p;
    p = ((2 * i - 1) * x * pm1 - (i + m - 1) * pm2) / (i - m);
  }
  return p;
}
```