ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<parse_channels>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L26
  lea    0x1(%rdi),%rbx
  mov    $0x0,%esi
  mov    $0x0,%ebp
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rsp,%r12
  jmp    L157
L106:
  mov    %rbx,%rax
  mov    $0x1,%esi
  mov    $0x1,%ebp
  jmp    L132
L20:
  movslq %ecx,%rax
  movb   $0x0,0x70(%rsp,%rax,1)
  mov    %rbx,%rax
  mov    $0x0,%ebp
  jmp    L132
L155:
  movslq %ecx,%rdi
  mov    %al,0x70(%rsp,%rdi,1)
  mov    %rbx,%rax
  lea    0x1(%rcx),%ecx
  jmp    L132
L0:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L4
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <insert_channels@plt>
L64:
  movb   $0x0,0x70(%rsp)
  mov    %ebp,%esi
  mov    %ebp,%ecx
  mov    %ebp,%edx
L72:
  mov    %rbx,%rax
L132:
  movzbl (%rax),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L15
L157:
  cmp    $0x28,%al
  je     L106
  cmp    $0x29,%al
  je     L20
  test   %ebp,%ebp
  jne    L155
  cmp    $0x2c,%al
  je     L0
  movslq %edx,%rdi
  mov    %al,(%rsp,%rdi,1)
  lea    0x1(%rdx),%edx
  jmp    L72
L4:
  lea    0x70(%rsp),%rsi
  mov    %r12,%rdi
  call   <insert_channels@plt>
  jmp    L64
L15:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L75
L169:
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <insert_channels@plt>
L153:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  add    $0xe0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L75:
  lea    0x70(%rsp),%rsi
  mov    %rsp,%rdi
  call   <insert_channels@plt>
  jmp    L153
L26:
  movb   $0x0,(%rsp)
  jmp    L169
L92:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_channels(char *line) {
  char temp[100];
  char key[100];
  int x = 0;
  int y = 0;
  int keyflag = 0;
  int haskey = 0;
  while (*line != '\0') {
    if (*line == '(') {
      keyflag = 1;
      haskey = 1;
      line++;
      continue;
    }
    if (*line == ')') {
      keyflag = 0;
      key[y] = '\0';
      line++;
      continue;
    }
    if (keyflag) {
      key[y++] = *line;
      line++;
      continue;
    }
    if (*line == ',') {
      temp[x] = '\0';
      if (haskey == 1) {
        insert_channels(temp, key);
      } else {
        insert_channels(temp, NULL);
      }
      y = 0;
      key[y] = '\0';
      x = 0;
      haskey = 0;
    } else {
      temp[x++] = *line;
    }
    line++;
  }
  temp[x] = '\0';
  if (haskey == 1) {
    insert_channels(temp, key);
  } else {
    insert_channels(temp, NULL);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<parse_channels>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L109
  mov    %rdi,%rbx
  xor    %esi,%esi
  xor    %ebp,%ebp
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rsp,%r12
  jmp    L137
  nopl   (%rax)
L68:
  cmp    $0x2c,%al
  je     L141
  movslq %edx,%rdi
  add    $0x1,%edx
  mov    %al,(%rsp,%rdi,1)
L20:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L37
L137:
  add    $0x1,%rbx
  cmp    $0x28,%al
  je     L94
  cmp    $0x29,%al
  je     L150
  test   %ebp,%ebp
  je     L68
  movslq %ecx,%rdi
  add    $0x1,%ecx
  mov    %al,0x70(%rsp,%rdi,1)
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L137
L37:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L1
L101:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <insert_channels@plt>
L111:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0xe0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L94:
  mov    $0x1,%esi
  mov    $0x1,%ebp
  jmp    L20
  nopl   0x0(%rax)
L150:
  movslq %ecx,%rax
  xor    %ebp,%ebp
  movb   $0x0,0x70(%rsp,%rax,1)
  jmp    L20
  nop
L141:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L118
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <insert_channels@plt>
L126:
  movb   $0x0,0x70(%rsp)
  xor    %esi,%esi
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L118:
  lea    0x70(%rsp),%rsi
  mov    %r12,%rdi
  call   <insert_channels@plt>
  jmp    L126
  nop
L1:
  lea    0x70(%rsp),%rsi
  mov    %r12,%rdi
  call   <insert_channels@plt>
  jmp    L111
  nopw   0x0(%rax,%rax,1)
L109:
  movb   $0x0,(%rsp)
  mov    %rsp,%r12
  jmp    L101
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_channels(char *line) {
  char temp[100];
  char key[100];
  int x = 0;
  int y = 0;
  int keyflag = 0;
  int haskey = 0;
  while (*line != '\0') {
    if (*line == '(') {
      keyflag = 1;
      haskey = 1;
      line++;
      continue;
    }
    if (*line == ')') {
      keyflag = 0;
      key[y] = '\0';
      line++;
      continue;
    }
    if (keyflag) {
      key[y++] = *line;
      line++;
      continue;
    }
    if (*line == ',') {
      temp[x] = '\0';
      if (haskey == 1) {
        insert_channels(temp, key);
      } else {
        insert_channels(temp, NULL);
      }
      y = 0;
      key[y] = '\0';
      x = 0;
      haskey = 0;
    } else {
      temp[x++] = *line;
    }
    line++;
  }
  temp[x] = '\0';
  if (haskey == 1) {
    insert_channels(temp, key);
  } else {
    insert_channels(temp, NULL);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parse_channels>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L16
  mov    %rdi,%rbx
  xor    %esi,%esi
  xor    %ebp,%ebp
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rsp,%r12
  jmp    L144
  nopl   (%rax)
L25:
  cmp    $0x2c,%al
  je     L24
  movslq %edx,%rdi
  add    $0x1,%edx
  mov    %al,(%rsp,%rdi,1)
L130:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L166
L144:
  add    $0x1,%rbx
  cmp    $0x28,%al
  je     L167
  cmp    $0x29,%al
  je     L111
  test   %ebp,%ebp
  je     L25
  movslq %ecx,%rdi
  add    $0x1,%ecx
  mov    %al,0x70(%rsp,%rdi,1)
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L144
L166:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L36
L32:
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <insert_channels@plt>
L63:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L159
  add    $0xe0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L167:
  mov    $0x1,%esi
  mov    $0x1,%ebp
  jmp    L130
  nopl   0x0(%rax)
L111:
  movslq %ecx,%rax
  xor    %ebp,%ebp
  movb   $0x0,0x70(%rsp,%rax,1)
  jmp    L130
  nop
L24:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  cmp    $0x1,%esi
  je     L59
  xor    %esi,%esi
  mov    %r12,%rdi
  call   <insert_channels@plt>
L133:
  movb   $0x0,0x70(%rsp)
  xor    %esi,%esi
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L130
  cs nopw 0x0(%rax,%rax,1)
L59:
  lea    0x70(%rsp),%rsi
  mov    %r12,%rdi
  call   <insert_channels@plt>
  jmp    L133
  nop
L36:
  lea    0x70(%rsp),%rsi
  mov    %r12,%rdi
  call   <insert_channels@plt>
  jmp    L63
  nopw   0x0(%rax,%rax,1)
L16:
  movb   $0x0,(%rsp)
  mov    %rsp,%r12
  jmp    L32
L159:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_channels(char *line) {
  char temp[100];
  char key[100];
  int x = 0;
  int y = 0;
  int keyflag = 0;
  int haskey = 0;
  while (*line != '\0') {
    if (*line == '(') {
      keyflag = 1;
      haskey = 1;
      line++;
      continue;
    }
    if (*line == ')') {
      keyflag = 0;
      key[y] = '\0';
      line++;
      continue;
    }
    if (keyflag) {
      key[y++] = *line;
      line++;
      continue;
    }
    if (*line == ',') {
      temp[x] = '\0';
      if (haskey == 1) {
        insert_channels(temp, key);
      } else {
        insert_channels(temp, NULL);
      }
      y = 0;
      key[y] = '\0';
      x = 0;
      haskey = 0;
    } else {
      temp[x++] = *line;
    }
    line++;
  }
  temp[x] = '\0';
  if (haskey == 1) {
    insert_channels(temp, key);
  } else {
    insert_channels(temp, NULL);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<record_execution>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L12
L6:
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L6
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void record_execution(int start, int end, int process_number, int execution[]) {
  int i = start, j = start + end;
  while (i < j) {
    execution[i++] = process_number;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<record_execution>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  cmp    %eax,%edi
  jge    L18
  movslq %edi,%rdi
  lea    (%rcx,%rdi,4),%rax
  lea    -0x1(%rsi),%esi
  add    %rdi,%rsi
  lea    0x4(%rcx,%rsi,4),%rcx
L12:
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L12
L18:
  ret
```

assistant: ```c
void record_execution(int start, int end, int process_number, int execution[]) {
  int i = start, j = start + end;
  while (i < j) {
    execution[i++] = process_number;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<record_execution>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  cmp    %eax,%edi
  jge    L10
  movslq %edi,%rdi
  sub    $0x1,%esi
  add    %rdi,%rsi
  lea    (%rcx,%rdi,4),%rax
  lea    0x4(%rcx,%rsi,4),%rcx
  nopl   (%rax)
L5:
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L5
L10:
  ret
```

assistant: ```c
void record_execution(int start, int end, int process_number, int execution[]) {
  int i = start, j = start + end;
  while (i < j) {
    execution[i++] = process_number;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<record_execution>:
  endbr64
  lea    (%rdi,%rsi,1),%r9d
  mov    %edx,%r10d
  mov    %rcx,%r8
  cmp    %r9d,%edi
  jge    L53
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L49
  movslq %edi,%rax
  movd   %edx,%xmm1
  lea    (%rcx,%rax,4),%rdx
  mov    %esi,%eax
  pshufd $0x0,%xmm1,%xmm0
  shr    $0x2,%eax
  lea    -0x1(%rax),%ecx
  lea    0x10(%rdx),%rax
  shl    $0x4,%rcx
  add    %rax,%rcx
  jmp    L69
L55:
  add    $0x10,%rax
L69:
  movups %xmm0,(%rdx)
  mov    %rax,%rdx
  cmp    %rcx,%rax
  jne    L55
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  add    %eax,%edi
  cmp    %eax,%esi
  je     L53
L49:
  movslq %edi,%rax
  mov    %r10d,(%r8,%rax,4)
  lea    0x0(,%rax,4),%rdx
  lea    0x1(%rdi),%eax
  cmp    %r9d,%eax
  jge    L53
  add    $0x2,%edi
  mov    %r10d,0x4(%r8,%rdx,1)
  cmp    %edi,%r9d
  jle    L53
  mov    %r10d,0x8(%r8,%rdx,1)
L53:
  ret
```

assistant: ```c
void record_execution(int start, int end, int process_number, int execution[]) {
  int i = start, j = start + end;
  while (i < j) {
    execution[i++] = process_number;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ln_comb>:
  endbr64
  cmp    %edi,%esi
  je     L1
  push   %rbx
  sub    $0x20,%rsp
  mov    %esi,%ebx
  pxor   %xmm0,%xmm0
  test   %esi,%esi
  je     L13
  mov    %edi,%eax
  sub    %esi,%eax
  pxor   %xmm7,%xmm7
  cvtsi2sd %rax,%xmm7
  mov    %edi,%edi
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdi,%xmm0
  movsd  %xmm0,0x8(%rsp)
  movsd  %xmm7,(%rsp)
  divsd  %xmm7,%xmm0
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movapd %xmm0,%xmm3
  mulsd  (%rsp),%xmm3
  movsd  %xmm3,0x18(%rsp)
  mov    %ebx,%ebx
  pxor   %xmm5,%xmm5
  cvtsi2sd %rbx,%xmm5
  movsd  0x8(%rsp),%xmm0
  movsd  %xmm5,0x10(%rsp)
  divsd  %xmm5,%xmm0
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movsd  0x10(%rsp),%xmm5
  mulsd  %xmm5,%xmm0
  movapd %xmm0,%xmm2
  addsd  0x18(%rsp),%xmm2
  movq   %xmm2,%rbx
  movapd %xmm5,%xmm6
  mulsd  D23(%rip),%xmm6
  movapd %xmm6,%xmm0
  mulsd  (%rsp),%xmm0
  divsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movapd %xmm0,%xmm2
  movsd  D25(%rip),%xmm1
  movapd %xmm1,%xmm3
  divsd  0x8(%rsp),%xmm3
  movapd %xmm3,%xmm0
  movapd %xmm1,%xmm3
  divsd  (%rsp),%xmm3
  subsd  %xmm3,%xmm0
  divsd  0x10(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  mulsd  D39(%rip),%xmm0
  mulsd  D34(%rip),%xmm2
  movq   %rbx,%xmm1
  subsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
L13:
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D23
type_infered: byte[8]
name: parse_data
arguments: data_label: D25
type_infered: f64
name: parse_data
arguments: data_label: D39
type_infered: byte[8]
name: parse_data
arguments: data_label: D34
type_infered: f64

tool: {"D23": [0x4a,0xd8,0x12,0x4d,0xfb,0x21,0x19,0x40]}

tool: D25: 1

tool: {"D39": [0x55,0x55,0x55,0x55,0x55,0x55,0xb5,0x3f]}

tool: D34: 0.5

assistant: ```c
double ln_comb(unsigned int n, unsigned int r) {
  if (r == n || r == 0)
    return 0;
  return (n - r) * log((double)n / (n - r)) + r * log((double)n / r) - 0.5 * log(2 * 3.1415926 * r * (n - r) / n) + 1.0 / 12 * (1.0 / n - 1.0 / (n - r) - 1.0 / r);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ln_comb>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x20,%rsp
  cmp    %edi,%ebx
  je     L0
  test   %ebx,%ebx
  je     L0
  mov    %edi,%eax
  pxor   %xmm7,%xmm7
  pxor   %xmm0,%xmm0
  mov    %edi,%edi
  sub    %ebx,%eax
  cvtsi2sd %rdi,%xmm0
  cvtsi2sd %rax,%xmm7
  movsd  %xmm0,0x8(%rsp)
  divsd  %xmm7,%xmm0
  movsd  %xmm7,(%rsp)
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  pxor   %xmm3,%xmm3
  movsd  (%rsp),%xmm4
  cvtsi2sd %rbx,%xmm3
  mulsd  %xmm0,%xmm4
  movsd  0x8(%rsp),%xmm0
  divsd  %xmm3,%xmm0
  movsd  %xmm3,0x18(%rsp)
  movsd  %xmm4,0x10(%rsp)
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movsd  0x18(%rsp),%xmm3
  movapd %xmm0,%xmm1
  movsd  (%rsp),%xmm0
  mulsd  %xmm3,%xmm1
  movapd %xmm3,%xmm5
  mulsd  D17(%rip),%xmm5
  addsd  0x10(%rsp),%xmm1
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm1,0x18(%rsp)
  mulsd  %xmm5,%xmm0
  divsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movsd  D10(%rip),%xmm2
  movsd  0x10(%rsp),%xmm3
  movapd %xmm0,%xmm4
  movsd  0x18(%rsp),%xmm1
  mulsd  D31(%rip),%xmm4
  movapd %xmm2,%xmm5
  divsd  0x8(%rsp),%xmm5
  movapd %xmm5,%xmm0
  movapd %xmm2,%xmm5
  divsd  (%rsp),%xmm5
  subsd  %xmm5,%xmm0
  divsd  %xmm3,%xmm2
  subsd  %xmm4,%xmm1
  subsd  %xmm2,%xmm0
  mulsd  D19(%rip),%xmm0
  add    $0x20,%rsp
  pop    %rbx
  addsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L0:
  add    $0x20,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D17
data_type: byte[8]
name: retrieve_data
arguments: tag: D10
data_type: f64
name: retrieve_data
arguments: tag: D31
data_type: f64
name: retrieve_data
arguments: tag: D19
data_type: byte[8]

tool: {"D17": [0x4a,0xd8,0x12,0x4d,0xfb,0x21,0x19,0x40]}

tool: D10: 1

tool: D31: 0.5

tool: {"D19": [0x55,0x55,0x55,0x55,0x55,0x55,0xb5,0x3f]}

assistant: ```c
double ln_comb(unsigned int n, unsigned int r) {
  if (r == n || r == 0)
    return 0;
  return (n - r) * log((double)n / (n - r)) + r * log((double)n / r) - 0.5 * log(2 * 3.1415926 * r * (n - r) / n) + 1.0 / 12 * (1.0 / n - 1.0 / (n - r) - 1.0 / r);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ln_comb>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  sub    $0x20,%rsp
  cmp    %edi,%ebx
  je     L15
  test   %ebx,%ebx
  je     L15
  mov    %edi,%eax
  pxor   %xmm7,%xmm7
  pxor   %xmm0,%xmm0
  mov    %edi,%edi
  sub    %ebx,%eax
  cvtsi2sd %rdi,%xmm0
  cvtsi2sd %rax,%xmm7
  movsd  %xmm0,0x8(%rsp)
  divsd  %xmm7,%xmm0
  movsd  %xmm7,(%rsp)
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  pxor   %xmm3,%xmm3
  movsd  (%rsp),%xmm4
  cvtsi2sd %rbx,%xmm3
  mulsd  %xmm0,%xmm4
  movsd  0x8(%rsp),%xmm0
  divsd  %xmm3,%xmm0
  movsd  %xmm3,0x18(%rsp)
  movsd  %xmm4,0x10(%rsp)
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movsd  0x18(%rsp),%xmm3
  movapd %xmm0,%xmm1
  movsd  (%rsp),%xmm0
  mulsd  %xmm3,%xmm1
  movapd %xmm3,%xmm5
  mulsd  D4(%rip),%xmm5
  addsd  0x10(%rsp),%xmm1
  movsd  %xmm3,0x10(%rsp)
  movsd  %xmm1,0x18(%rsp)
  mulsd  %xmm5,%xmm0
  divsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%edi
  call   <log@plt>
  movsd  D26(%rip),%xmm2
  movsd  0x10(%rsp),%xmm3
  movapd %xmm0,%xmm4
  movsd  0x18(%rsp),%xmm1
  mulsd  D19(%rip),%xmm4
  movapd %xmm2,%xmm5
  divsd  0x8(%rsp),%xmm5
  movapd %xmm5,%xmm0
  movapd %xmm2,%xmm5
  divsd  (%rsp),%xmm5
  subsd  %xmm5,%xmm0
  divsd  %xmm3,%xmm2
  subsd  %xmm4,%xmm1
  subsd  %xmm2,%xmm0
  mulsd  D21(%rip),%xmm0
  add    $0x20,%rsp
  pop    %rbx
  addsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax)
L15:
  add    $0x20,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
data_type: byte[8]
name: unpack_data
arguments: label: D26
data_type: f64
name: unpack_data
arguments: label: D19
data_type: f64
name: unpack_data
arguments: label: D21
data_type: byte[8]

tool: {"D4": [0x4a,0xd8,0x12,0x4d,0xfb,0x21,0x19,0x40]}

tool: D26: 1

tool: D19: 0.5

tool: {"D21": [0x55,0x55,0x55,0x55,0x55,0x55,0xb5,0x3f]}

assistant: ```c
double ln_comb(unsigned int n, unsigned int r) {
  if (r == n || r == 0)
    return 0;
  return (n - r) * log((double)n / (n - r)) + r * log((double)n / r) - 0.5 * log(2 * 3.1415926 * r * (n - r) / n) + 1.0 / 12 * (1.0 / n - 1.0 / (n - r) - 1.0 / r);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<myexp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x38(%rbp)
  movsd  D40(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D11(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D11(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x1,-0x24(%rbp)
  jmp    L17
L11:
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x38(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x24(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x20(%rbp),%xmm0
  addsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  subsd  -0x8(%rbp),%xmm0
  movq   D0(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  -0x8(%rbp),%xmm0
  movq   D0(%rip),%xmm2
  andpd  %xmm2,%xmm0
  mulsd  -0x10(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jbe    L16
  movsd  -0x20(%rbp),%xmm0
  jmp    L24
L16:
  addl   $0x1,-0x24(%rbp)
L17:
  cmpl   $0xc8,-0x24(%rbp)
  jle    L11
  pxor   %xmm0,%xmm0
L24:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D40
data_type: byte[8]
name: unpack_data
arguments: tag: D11
data_type: f64
name: unpack_data
arguments: tag: D0
data_type: qword

tool: {"D40": [0x3a,0x8c,0x30,0xe2,0x8e,0x79,0x45,0x3e]}

tool: D11: 1

tool: D0: 9223372036854775807

assistant: ```c
double myexp(double x) {
  double EPS = 1e-08;
  double s = 1.0, e = 1.0, d;
  int k;
  for (k = 1; k <= 200; k++) {
    d = s;
    e = e * x / k;
    s = s + e;
    if (fabs(s - d) < EPS * fabs(d)) {
      return s;
    }
  }
  return 0.0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<myexp>:
  endbr64
  movapd %xmm0,%xmm5
  mov    $0x1,%eax
  movsd  D16(%rip),%xmm2
  movapd %xmm2,%xmm0
  movq   D0(%rip),%xmm4
  movsd  D13(%rip),%xmm6
L14:
  mulsd  %xmm5,%xmm2
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm2
  movapd %xmm0,%xmm1
  addsd  %xmm2,%xmm0
  movapd %xmm0,%xmm3
  subsd  %xmm1,%xmm3
  andpd  %xmm4,%xmm3
  andpd  %xmm4,%xmm1
  mulsd  %xmm6,%xmm1
  comisd %xmm3,%xmm1
  ja     L3
  add    $0x1,%eax
  cmp    $0xc9,%eax
  jne    L14
  pxor   %xmm0,%xmm0
L3:
  ret
```

tool_calls: name: read_data
arguments: label: D16
type_guessed: f64
name: read_data
arguments: label: D0
type_guessed: qword
name: read_data
arguments: label: D13
type_guessed: byte[8]

tool: D16: 1

tool: D0: 9223372036854775807

tool: {"D13": [0x3a,0x8c,0x30,0xe2,0x8e,0x79,0x45,0x3e]}

assistant: ```c
double myexp(double x) {
  double EPS = 1e-08;
  double s = 1.0, e = 1.0, d;
  int k;
  for (k = 1; k <= 200; k++) {
    d = s;
    e = e * x / k;
    s = s + e;
    if (fabs(s - d) < EPS * fabs(d)) {
      return s;
    }
  }
  return 0.0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<myexp>:
  endbr64
  movsd  D1(%rip),%xmm2
  movapd %xmm0,%xmm5
  movq   D10(%rip),%xmm4
  mov    $0x1,%eax
  movsd  D14(%rip),%xmm6
  movapd %xmm2,%xmm0
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L6:
  add    $0x1,%eax
  cmp    $0xc9,%eax
  je     L24
L27:
  mulsd  %xmm5,%xmm2
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm2
  movapd %xmm0,%xmm1
  addsd  %xmm2,%xmm0
  movapd %xmm0,%xmm3
  subsd  %xmm1,%xmm3
  andpd  %xmm4,%xmm1
  mulsd  %xmm6,%xmm1
  andpd  %xmm4,%xmm3
  comisd %xmm3,%xmm1
  jbe    L6
  ret
  nopl   (%rax)
L24:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
data_type: f64
name: disassemble_data
arguments: data_label: D10
data_type: qword
name: disassemble_data
arguments: data_label: D14
data_type: byte[8]

tool: D1: 1

tool: D10: 9223372036854775807

tool: {"D14": [0x3a,0x8c,0x30,0xe2,0x8e,0x79,0x45,0x3e]}

assistant: ```c
double myexp(double x) {
  double EPS = 1e-08;
  double s = 1.0, e = 1.0, d;
  int k;
  for (k = 1; k <= 200; k++) {
    d = s;
    e = e * x / k;
    s = s + e;
    if (fabs(s - d) < EPS * fabs(d)) {
      return s;
    }
  }
  return 0.0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<myexp>:
  endbr64
  movsd  D2(%rip),%xmm2
  movapd %xmm0,%xmm5
  movq   D20(%rip),%xmm4
  mov    $0x1,%eax
  movsd  D8(%rip),%xmm6
  movapd %xmm2,%xmm0
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L25:
  add    $0x1,%eax
  cmp    $0xc9,%eax
  je     L14
L1:
  mulsd  %xmm5,%xmm2
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm2
  movapd %xmm0,%xmm1
  addsd  %xmm2,%xmm0
  movapd %xmm0,%xmm3
  subsd  %xmm1,%xmm3
  andpd  %xmm4,%xmm1
  mulsd  %xmm6,%xmm1
  andpd  %xmm4,%xmm3
  comisd %xmm3,%xmm1
  jbe    L25
  ret
  nopl   (%rax)
L14:
  pxor   %xmm0,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_guessed: f64
name: parse_data
arguments: data_tag: D20
type_guessed: qword
name: parse_data
arguments: data_tag: D8
type_guessed: byte[8]

tool: D2: 1

tool: D20: 9223372036854775807

tool: {"D8": [0x3a,0x8c,0x30,0xe2,0x8e,0x79,0x45,0x3e]}

assistant: ```c
double myexp(double x) {
  double EPS = 1e-08;
  double s = 1.0, e = 1.0, d;
  int k;
  for (k = 1; k <= 200; k++) {
    d = s;
    e = e * x / k;
    s = s + e;
    if (fabs(s - d) < EPS * fabs(d)) {
      return s;
    }
  }
  return 0.0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <gen_tmp_file@plt>
  mov    $0x0,%eax
  call   <test@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
data_type: string

tool: D9: lock.dat

assistant: ```c
int main() {
  gen_tmp_file("lock.dat");
  test();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <gen_tmp_file@plt>
  mov    $0x0,%eax
  call   <test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D3
type: string

tool: D3: lock.dat

assistant: ```c
int main() {
  gen_tmp_file("lock.dat");
  test();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <gen_tmp_file@plt>
  xor    %eax,%eax
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
type: string

tool: D8: lock.dat

assistant: ```c
int main() {
  gen_tmp_file("lock.dat");
  test();
}
```