ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<add_nl>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x2(%rax),%r14d
  mov    %eax,%r13d
  mov    %r14d,%edi
  call   <malloc@plt>
  mov    %r14d,%edx
  xor    %esi,%esi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdi
  lea    D5(%rip),%rsi
  call   <strcat@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: label: D5
type_guessed: string

tool: D5: 


assistant: ```c
char *add_nl(const char *orig) {
  int len = strlen(orig);
  char *new = malloc(sizeof(char) * (len + 2));
  memset(new, '\0', len + 2);
  strncpy(new, orig, len);
  strcat(new, "\n");
  return new;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_isspace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x20,-0x4(%rbp)
  jne    L62
  mov    $0x1,%eax
  jmp    L4
L62:
  cmpb   $0x9,-0x4(%rbp)
  jne    L38
  mov    $0x1,%eax
  jmp    L4
L38:
  cmpb   $0xa,-0x4(%rbp)
  jne    L29
  mov    $0x1,%eax
  jmp    L4
L29:
  cmpb   $0xb,-0x4(%rbp)
  jne    L18
  mov    $0x1,%eax
  jmp    L4
L18:
  cmpb   $0xc,-0x4(%rbp)
  jne    L33
  mov    $0x1,%eax
  jmp    L4
L33:
  cmpb   $0xd,-0x4(%rbp)
  jne    L83
  mov    $0x1,%eax
  jmp    L4
L83:
  mov    $0x0,%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int ft_isspace(char c) {
  if (c == ' ')
    return (1);
  else if (c == '\t')
    return (1);
  else if (c == '\n')
    return (1);
  else if (c == '\v')
    return (1);
  else if (c == '\f')
    return (1);
  else if (c == '\r')
    return (1);
  else
    return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_isspace>:
  endbr64
  cmp    $0xd,%dil
  jle    L1
  cmp    $0x20,%dil
  sete   %al
  movzbl %al,%eax
  ret
L1:
  cmp    $0x8,%dil
  setg   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int ft_isspace(char c) {
  if (c == ' ')
    return (1);
  else if (c == '\t')
    return (1);
  else if (c == '\n')
    return (1);
  else if (c == '\v')
    return (1);
  else if (c == '\f')
    return (1);
  else if (c == '\r')
    return (1);
  else
    return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_isspace>:
  endbr64
  sub    $0x9,%edi
  xor    %eax,%eax
  cmp    $0x17,%dil
  ja     L7
  movzbl %dil,%edi
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L7:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string

tool: D9: 

assistant: ```c
int ft_isspace(char c) {
  if (c == ' ')
    return (1);
  else if (c == '\t')
    return (1);
  else if (c == '\n')
    return (1);
  else if (c == '\v')
    return (1);
  else if (c == '\f')
    return (1);
  else if (c == '\r')
    return (1);
  else
    return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_isspace>:
  endbr64
  sub    $0x9,%edi
  xor    %eax,%eax
  cmp    $0x17,%dil
  ja     L3
  movzbl %dil,%edi
  lea    D1(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L3:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type: string

tool: D1: 

assistant: ```c
int ft_isspace(char c) {
  if (c == ' ')
    return (1);
  else if (c == '\t')
    return (1);
  else if (c == '\n')
    return (1);
  else if (c == '\v')
    return (1);
  else if (c == '\f')
    return (1);
  else if (c == '\r')
    return (1);
  else
    return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0xc(%rbp),%xmm1
  movss  D42(%rip),%xmm0
  mulss  %xmm0,%xmm1
  movss  D21(%rip),%xmm0
  addss  %xmm1,%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm0,%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  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: parse_data
arguments: data_tag: D47
type: string
name: parse_data
arguments: data_tag: D4
type: string
name: parse_data
arguments: data_tag: D42
type: byte[4]
name: parse_data
arguments: data_tag: D21
type: byte[4]
name: parse_data
arguments: data_tag: D1
type: string

tool: D47: Enter x: 

tool: D4: %f

tool: {"D42": [0x00,0x00,0x80,0x40]}

tool: {"D21": [0x00,0x00,0x40,0x40]}

tool: D1: The value of y = 4x + 3 is %.2f

assistant: ```c
int main() {
  float x;
  printf("Enter x: ");
  scanf("%f", &x);
  printf("The value of y = 4x + 3 is %.2f", 4 * x + 3);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  movss  D47(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  addss  D34(%rip),%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D48(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D35
type_infered: string
name: unpack_data
arguments: data_label: D47
type_infered: byte[4]
name: unpack_data
arguments: data_label: D34
type_infered: byte[4]
name: unpack_data
arguments: data_label: D48
type_infered: string

tool: D3: Enter x: 

tool: D35: %f

tool: {"D47": [0x00,0x00,0x80,0x40]}

tool: {"D34": [0x00,0x00,0x40,0x40]}

tool: D48: The value of y = 4x + 3 is %.2f

assistant: ```c
int main() {
  float x;
  printf("Enter x: ");
  scanf("%f", &x);
  printf("The value of y = 4x + 3 is %.2f", 4 * x + 3);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D22(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D49(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movss  D32(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D23(%rip),%rdi
  addss  D41(%rip),%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D22
data_type: string
name: read_data
arguments: tag: D49
data_type: string
name: read_data
arguments: tag: D32
data_type: byte[4]
name: read_data
arguments: tag: D23
data_type: string
name: read_data
arguments: tag: D41
data_type: byte[4]

tool: D22: Enter x: 

tool: D49: %f

tool: {"D32": [0x00,0x00,0x80,0x40]}

tool: D23: The value of y = 4x + 3 is %.2f

tool: {"D41": [0x00,0x00,0x40,0x40]}

assistant: ```c
int main() {
  float x;
  printf("Enter x: ");
  scanf("%f", &x);
  printf("The value of y = 4x + 3 is %.2f", 4 * x + 3);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D48(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movss  D33(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D14(%rip),%rdi
  addss  D18(%rip),%xmm0
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D16
type_infered: string
name: retrieve_data
arguments: label: D48
type_infered: string
name: retrieve_data
arguments: label: D33
type_infered: byte[4]
name: retrieve_data
arguments: label: D14
type_infered: string
name: retrieve_data
arguments: label: D18
type_infered: byte[4]

tool: D16: Enter x: 

tool: D48: %f

tool: {"D33": [0x00,0x00,0x80,0x40]}

tool: D14: The value of y = 4x + 3 is %.2f

tool: {"D18": [0x00,0x00,0x40,0x40]}

assistant: ```c
int main() {
  float x;
  printf("Enter x: ");
  scanf("%f", &x);
  printf("The value of y = 4x + 3 is %.2f", 4 * x + 3);
  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
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  je     L74
  mov    $0x1,%eax
  jmp    L14
L74:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L86
  mov    $0x1,%eax
  jmp    L14
L86:
  orl    $0x3,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L10
  mov    $0x1,%eax
  jmp    L14
L10:
  andl   $0x2,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L43
  mov    $0x1,%eax
  jmp    L14
L43:
  shll   -0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  je     L48
  mov    $0x1,%eax
  jmp    L14
L48:
  mov    $0x0,%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x;
  x = 0;
  x += 1;
  if (x != 1)
    return 1;
  x -= 1;
  if (x != 0)
    return 1;
  x |= 3;
  if (x != 3)
    return 1;
  x &= 2;
  if (x != 2)
    return 1;
  x *= 2;
  if (x != 4)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int main() {
  int x;
  x = 0;
  x += 1;
  if (x != 1)
    return 1;
  x -= 1;
  if (x != 0)
    return 1;
  x |= 3;
  if (x != 3)
    return 1;
  x &= 2;
  if (x != 2)
    return 1;
  x *= 2;
  if (x != 4)
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

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

assistant: ```c
int main() {
  int x;
  x = 0;
  x += 1;
  if (x != 1)
    return 1;
  x -= 1;
  if (x != 0)
    return 1;
  x |= 3;
  if (x != 3)
    return 1;
  x &= 2;
  if (x != 2)
    return 1;
  x *= 2;
  if (x != 4)
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

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

assistant: ```c
int main() {
  int x;
  x = 0;
  x += 1;
  if (x != 1)
    return 1;
  x -= 1;
  if (x != 0)
    return 1;
  x |= 3;
  if (x != 3)
    return 1;
  x &= 2;
  if (x != 2)
    return 1;
  x *= 2;
  if (x != 4)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<prepare_if_request>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,0x4(%rax)
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <strncpy@plt>
  nop
  leave
  ret
```

assistant: ```c
void prepare_if_request(struct ifreq *ifr, const char *iface) {
  memset(ifr, 0, sizeof(struct ifreq));
  ifr->ifr_addr.sa_family = AF_INET;
  strncpy(ifr->ifr_name, iface, IFNAMSIZ - 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<prepare_if_request>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x8,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x4(%rbx)
  mov    D15(%rip),%rax
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  mov    %rbp,%rsi
  mov    (%rbx),%edi
  call   <strncpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void prepare_if_request(struct ifreq *ifr, const char *iface) {
  memset(ifr, 0, sizeof(struct ifreq));
  ifr->ifr_addr.sa_family = AF_INET;
  strncpy(ifr->ifr_name, iface, IFNAMSIZ - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<prepare_if_request>:
  endbr64
  push   %rbp
  mov    $0x8,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <memset@plt>
  mov    D15(%rip),%rax
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  mov    (%rax),%eax
  mov    %eax,0x4(%rbx)
  mov    D11(%rip),%rax
  mov    (%rax),%rdx
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  sub    $0x1,%rdx
  jmp    1060 <strncpy@plt>
```

assistant: ```c
void prepare_if_request(struct ifreq *ifr, const char *iface) {
  memset(ifr, 0, sizeof(struct ifreq));
  ifr->ifr_addr.sa_family = AF_INET;
  strncpy(ifr->ifr_name, iface, IFNAMSIZ - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prepare_if_request>:
  endbr64
  push   %rbp
  mov    $0x8,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <memset@plt>
  mov    D3(%rip),%rax
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  mov    (%rax),%eax
  mov    %eax,0x4(%rbx)
  mov    D4(%rip),%rax
  mov    (%rax),%rdx
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  sub    $0x1,%rdx
  jmp    1060 <strncpy@plt>
```

assistant: ```c
void prepare_if_request(struct ifreq *ifr, const char *iface) {
  memset(ifr, 0, sizeof(struct ifreq));
  ifr->ifr_addr.sa_family = AF_INET;
  strncpy(ifr->ifr_name, iface, IFNAMSIZ - 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_performance_lookup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x186a0,-0x34(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L28
L14:
  mov    -0x38(%rbp),%eax
  cltq
  mov    %rax,%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_get_value_by_key@plt>
  addl   $0x1,-0x38(%rbp)
L28:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L14
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movsd  -0x20(%rbp),%xmm0
  movsd  -0x30(%rbp),%xmm2
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  mov    -0x18(%rbp),%eax
  mov    -0x28(%rbp),%edx
  sub    %edx,%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movsd  D11(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rdx
  mov    -0x34(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D11
type_guessed: byte[8]
name: read_data
arguments: data_label: D8
type_guessed: string

tool: {"D11": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D8: Looked up %d keys in %.2g seconds


assistant: ```c
void test_performance_lookup(struct hashmap *hm) {
  struct timeval t0, t1;
  int i;
  int n = 100000;
  gettimeofday(&t0, NULL);
  for (i = 0; i < n; i++) {
    _get_value_by_key(hm, (void *)i);
  }
  gettimeofday(&t1, NULL);
  printf("Looked up %d keys in %.2g seconds\n", n, t1.tv_sec - t0.tv_sec + 1E-6 * (t1.tv_usec - t0.tv_usec));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test_performance_lookup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x0,%ebx
L19:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <_get_value_by_key@plt>
  add    $0x1,%rbx
  cmp    $0x186a0,%rbx
  jne    L19
  lea    0x10(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    0x18(%rsp),%eax
  sub    0x8(%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D15(%rip),%xmm0
  movsd  0x10(%rsp),%xmm1
  subsd  (%rsp),%xmm1
  addsd  %xmm1,%xmm0
  mov    $0x186a0,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
datatype: byte[8]
name: extract_symbolic_data
arguments: data_tag: D18
datatype: string

tool: {"D15": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D18: Looked up %d keys in %.2g seconds


assistant: ```c
void test_performance_lookup(struct hashmap *hm) {
  struct timeval t0, t1;
  int i;
  int n = 100000;
  gettimeofday(&t0, NULL);
  for (i = 0; i < n; i++) {
    _get_value_by_key(hm, (void *)i);
  }
  gettimeofday(&t1, NULL);
  printf("Looked up %d keys in %.2g seconds\n", n, t1.tv_sec - t0.tv_sec + 1E-6 * (t1.tv_usec - t0.tv_usec));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_performance_lookup>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  nopl   0x0(%rax)
L7:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <_get_value_by_key@plt>
  cmp    $0x186a0,%rbx
  jne    L7
  xor    %esi,%esi
  lea    0x10(%rsp),%rdi
  call   <gettimeofday@plt>
  mov    0x18(%rsp),%eax
  pxor   %xmm0,%xmm0
  sub    0x8(%rsp),%eax
  cvtsi2sd %eax,%xmm0
  mulsd  D5(%rip),%xmm0
  movsd  0x10(%rsp),%xmm1
  mov    $0x186a0,%esi
  subsd  (%rsp),%xmm1
  lea    D19(%rip),%rdi
  addsd  %xmm1,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
data_type: byte[8]
name: unpack_data
arguments: label: D19
data_type: string

tool: {"D5": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D19: Looked up %d keys in %.2g seconds


assistant: ```c
void test_performance_lookup(struct hashmap *hm) {
  struct timeval t0, t1;
  int i;
  int n = 100000;
  gettimeofday(&t0, NULL);
  for (i = 0; i < n; i++) {
    _get_value_by_key(hm, (void *)i);
  }
  gettimeofday(&t1, NULL);
  printf("Looked up %d keys in %.2g seconds\n", n, t1.tv_sec - t0.tv_sec + 1E-6 * (t1.tv_usec - t0.tv_usec));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_performance_lookup>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  nopl   0x0(%rax)
L7:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <_get_value_by_key@plt>
  cmp    $0x186a0,%rbx
  jne    L7
  xor    %esi,%esi
  lea    0x10(%rsp),%rdi
  call   <gettimeofday@plt>
  mov    0x18(%rsp),%eax
  pxor   %xmm0,%xmm0
  sub    0x8(%rsp),%eax
  cvtsi2sd %eax,%xmm0
  mulsd  D14(%rip),%xmm0
  movsd  0x10(%rsp),%xmm1
  mov    $0x186a0,%esi
  subsd  (%rsp),%xmm1
  lea    D1(%rip),%rdi
  addsd  %xmm1,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D14
data_type: byte[8]
name: read_data
arguments: data_tag: D1
data_type: string

tool: {"D14": [0x8d,0xed,0xb5,0xa0,0xf7,0xc6,0xb0,0x3e]}

tool: D1: Looked up %d keys in %.2g seconds


assistant: ```c
void test_performance_lookup(struct hashmap *hm) {
  struct timeval t0, t1;
  int i;
  int n = 100000;
  gettimeofday(&t0, NULL);
  for (i = 0; i < n; i++) {
    _get_value_by_key(hm, (void *)i);
  }
  gettimeofday(&t1, NULL);
  printf("Looked up %d keys in %.2g seconds\n", n, t1.tv_sec - t0.tv_sec + 1E-6 * (t1.tv_usec - t0.tv_usec));
}
```
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
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  movl   $0x5,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  add    $0x14,%rax
  lea    -0x20(%rbp),%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <sum_arr@plt>
  mov    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <sum_arr_with_len@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D17
type: string
name: retrieve_data
arguments: tag: D7
type: string

tool: D17: total = %d 


tool: D7: total_with_len = %d 


assistant: ```c
int main(void) {
  int arr[5] = {1, 2, 3, 4, 5};
  int total = sum_arr(arr, arr + 5);
  printf("total = %d \n", total);
  int total_with_len = sum_arr_with_len(arr, 5);
  printf("total_with_len = %d \n", total_with_len);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  mov    %rsp,%rbx
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <sum_arr@plt>
  mov    %eax,%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %rbx,%rdi
  call   <sum_arr_with_len@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D14
data_type: string
name: read_data
arguments: data_label: D5
data_type: string

tool: D14: total = %d 


tool: D5: total_with_len = %d 


assistant: ```c
int main(void) {
  int arr[5] = {1, 2, 3, 4, 5};
  int total = sum_arr(arr, arr + 5);
  printf("total = %d \n", total);
  int total_with_len = sum_arr_with_len(arr, 5);
  printf("total_with_len = %d \n", total_with_len);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%rbp
  lea    0x14(%rsp),%rsi
  movl   $0x5,0x10(%rsp)
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x8(%rsp)
  call   <sum_arr@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x5,%esi
  mov    %rbp,%rdi
  call   <sum_arr_with_len@plt>
  lea    D17(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

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

tool: D13: total = %d 


tool: D17: total_with_len = %d 


assistant: ```c
int main(void) {
  int arr[5] = {1, 2, 3, 4, 5};
  int total = sum_arr(arr, arr + 5);
  printf("total = %d \n", total);
  int total_with_len = sum_arr_with_len(arr, 5);
  printf("total_with_len = %d \n", total_with_len);
  return 0;
}
```