ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<bgpdump_test_utils>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <ti2s@plt>
  mov    $0x1869f,%edi
  call   <ti2s@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1050 <ti2s@plt>
```

assistant: ```c
void bgpdump_test_utils() {
  ti2s(0);
  ti2s(99999);
  ti2s(4294967295L);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_count_digits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  lea    -0x14(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ft_pop_sign@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L38
  mov    $0x1,%eax
  jmp    L32
L3:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L38:
  cmpl   $0x0,-0xc(%rbp)
  jne    L3
  mov    -0x10(%rbp),%eax
L32:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
int ft_count_digits(int n) {
  int i;
  unsigned int cur;
  int sign;
  i = 0;
  cur = ft_pop_sign(n, &sign);
  if (n == 0)
    return (1);
  while (cur >= 1) {
    cur /= 10;
    i++;
  }
  return (i);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ft_count_digits>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <ft_pop_sign@plt>
  mov    $0x1,%edx
  test   %ebx,%ebx
  je     L22
  test   %eax,%eax
  je     L12
  mov    $0x0,%edx
  mov    $0xcccccccd,%esi
L5:
  mov    %eax,%ecx
  mov    %eax,%eax
  imul   %rsi,%rax
  shr    $0x23,%rax
  add    $0x1,%edx
  cmp    $0x9,%ecx
  ja     L5
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    %edx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L12:
  mov    $0x0,%edx
  jmp    L22
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_count_digits(int n) {
  int i;
  unsigned int cur;
  int sign;
  i = 0;
  cur = ft_pop_sign(n, &sign);
  if (n == 0)
    return (1);
  while (cur >= 1) {
    cur /= 10;
    i++;
  }
  return (i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_count_digits>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <ft_pop_sign@plt>
  mov    $0x1,%r8d
  test   %ebx,%ebx
  je     L32
  xor    %r8d,%r8d
  test   %eax,%eax
  je     L32
  mov    $0xcccccccd,%ecx
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %eax,%eax
  add    $0x1,%r8d
  mov    %rax,%rdx
  imul   %rcx,%rax
  shr    $0x23,%rax
  cmp    $0x9,%edx
  ja     L27
L32:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x10,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_count_digits(int n) {
  int i;
  unsigned int cur;
  int sign;
  i = 0;
  cur = ft_pop_sign(n, &sign);
  if (n == 0)
    return (1);
  while (cur >= 1) {
    cur /= 10;
    i++;
  }
  return (i);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ft_count_digits>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <ft_pop_sign@plt>
  mov    $0x1,%r8d
  test   %ebx,%ebx
  je     L34
  xor    %r8d,%r8d
  test   %eax,%eax
  je     L34
  mov    $0xcccccccd,%ecx
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %eax,%eax
  add    $0x1,%r8d
  mov    %rax,%rdx
  imul   %rcx,%rax
  shr    $0x23,%rax
  cmp    $0x9,%edx
  ja     L2
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_count_digits(int n) {
  int i;
  unsigned int cur;
  int sign;
  i = 0;
  cur = ft_pop_sign(n, &sign);
  if (n == 0)
    return (1);
  while (cur >= 1) {
    cur /= 10;
    i++;
  }
  return (i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uart_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L11
L16:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <uart_putchar@plt>
L11:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L16
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int uart_puts(const char *str) {
  while (*str) {
    uart_putchar(*str++);
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L5
L11:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <uart_putchar@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L11
L5:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int uart_puts(const char *str) {
  while (*str) {
    uart_putchar(*str++);
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L19
L5:
  call   <uart_putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L5
L19:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int uart_puts(const char *str) {
  while (*str) {
    uart_putchar(*str++);
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L5
L14:
  call   <uart_putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L14
L5:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int uart_puts(const char *str) {
  while (*str) {
    uart_putchar(*str++);
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  mov    -0x8(%rbp),%rax
  jmp    L27
L20:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L10
  mov    -0x8(%rbp),%rax
  jmp    L27
L10:
  addq   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L20
  mov    $0x0,%eax
L27:
  pop    %rbp
  ret
```

assistant: ```c
char *strchr(const char *string, int character) {
  if (!string)
    return ((char *)string);
  while (string[0] != '\0') {
    if (string[0] == (char)character)
      return ((char *)string);
    string += 1;
  }
  return (NULL);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L15
L22:
  cmp    %sil,%al
  je     L11
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L22
  mov    $0x0,%eax
  ret
L15:
  mov    $0x0,%eax
  ret
L11:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *string, int character) {
  if (!string)
    return ((char *)string);
  while (string[0] != '\0') {
    if (string[0] == (char)character)
      return ((char *)string);
    string += 1;
  }
  return (NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  jne    L22
  jmp    L33
L5:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L33
L22:
  cmp    %dl,%sil
  jne    L5
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *strchr(const char *string, int character) {
  if (!string)
    return ((char *)string);
  while (string[0] != '\0') {
    if (string[0] == (char)character)
      return ((char *)string);
    string += 1;
  }
  return (NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  jne    L23
  jmp    L21
L22:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  test   %dl,%dl
  je     L21
L23:
  cmp    %dl,%sil
  jne    L22
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *strchr(const char *string, int character) {
  if (!string)
    return ((char *)string);
  while (string[0] != '\0') {
    if (string[0] == (char)character)
      return ((char *)string);
    string += 1;
  }
  return (NULL);
}
```
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
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L29
L11:
  mov    -0x20(%rbp),%ecx
  mov    -0x1c(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  imul   -0x10(%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
  mov    %eax,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  addss  -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0xc(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L29:
  mov    -0x28(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L11
  mov    -0x18(%rbp),%eax
  movd   %eax,%xmm0
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D19
type_guessed: string
name: unpack_data
arguments: data_tag: D22
type_guessed: string

tool: D19: %d

tool: D22: %.f


assistant: ```c
int main() {
  int n, m, k, i = 0, a, b, c;
  float sum = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  while (i < n) {
    a = i / k;
    b = a * 2;
    c = m * b / 100;
    sum = sum + m - c;
    i++;
  }
  printf("%.f\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D8(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  test   %edi,%edi
  jle    L25
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%r8d
  pxor   %xmm2,%xmm2
  cvtsi2ss %r8d,%xmm2
  pxor   %xmm0,%xmm0
  mov    $0x0,%ecx
L28:
  addss  %xmm2,%xmm0
  mov    %ecx,%eax
  cltd
  idiv   %esi
  imul   %r8d,%eax
  add    %eax,%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sar    $0x1f,%eax
  sub    %eax,%edx
  pxor   %xmm1,%xmm1
  cvtsi2ss %edx,%xmm1
  subss  %xmm1,%xmm0
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jne    L28
L30:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L25:
  pxor   %xmm0,%xmm0
  jmp    L30
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D11
type_guessed: string

tool: D8: %d

tool: D11: %.f


assistant: ```c
int main() {
  int n, m, k, i = 0, a, b, c;
  float sum = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  while (i < n) {
    a = i / k;
    b = a * 2;
    c = m * b / 100;
    sum = sum + m - c;
    i++;
  }
  printf("%.f\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  test   %edi,%edi
  jle    L23
  mov    0x10(%rsp),%r8d
  pxor   %xmm2,%xmm2
  mov    0x14(%rsp),%esi
  xor    %ecx,%ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %r8d,%xmm2
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ecx,%eax
  pxor   %xmm1,%xmm1
  addss  %xmm2,%xmm0
  add    $0x1,%ecx
  cltd
  idiv   %esi
  imul   %r8d,%eax
  add    %eax,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sub    %eax,%edx
  cvtsi2ss %edx,%xmm1
  subss  %xmm1,%xmm0
  cmp    %edi,%ecx
  jne    L6
L7:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L23:
  pxor   %xmm0,%xmm0
  jmp    L7
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D17
type: string
name: retrieve_data
arguments: label: D13
type: string

tool: D17: %d

tool: D13: %.f


assistant: ```c
int main() {
  int n, m, k, i = 0, a, b, c;
  float sum = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  while (i < n) {
    a = i / k;
    b = a * 2;
    c = m * b / 100;
    sum = sum + m - c;
    i++;
  }
  printf("%.f\n", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  test   %edi,%edi
  jle    L18
  mov    0x10(%rsp),%r8d
  pxor   %xmm2,%xmm2
  mov    0x14(%rsp),%esi
  xor    %ecx,%ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %r8d,%xmm2
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %ecx,%eax
  pxor   %xmm1,%xmm1
  addss  %xmm2,%xmm0
  add    $0x1,%ecx
  cltd
  idiv   %esi
  imul   %r8d,%eax
  add    %eax,%eax
  movslq %eax,%rdx
  sar    $0x1f,%eax
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  sub    %eax,%edx
  cvtsi2ss %edx,%xmm1
  subss  %xmm1,%xmm0
  cmp    %edi,%ecx
  jne    L26
L29:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L18:
  pxor   %xmm0,%xmm0
  jmp    L29
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type: string
name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D17: %d

tool: D1: %.f


assistant: ```c
int main() {
  int n, m, k, i = 0, a, b, c;
  float sum = 0;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  while (i < n) {
    a = i / k;
    b = a * 2;
    c = m * b / 100;
    sum = sum + m - c;
    i++;
  }
  printf("%.f\n", sum);
  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
<Bar3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
datatype: string

tool: D8:  Hello from Bar3: one = %d, two = %d!
 

assistant: ```c
extern long Bar3(long one, long two) {
  printf(" Hello from Bar3: one = %d, two = %d!\n ", one, two);
  return one + two;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Bar3>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdx
  mov    %rdi,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type: string

tool: D4:  Hello from Bar3: one = %d, two = %d!
 

assistant: ```c
extern long Bar3(long one, long two) {
  printf(" Hello from Bar3: one = %d, two = %d!\n ", one, two);
  return one + two;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Bar3>:
  endbr64
  push   %rbp
  mov    %rsi,%rdx
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  lea    D4(%rip),%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%rax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type_guessed: string

tool: D4:  Hello from Bar3: one = %d, two = %d!
 

assistant: ```c
extern long Bar3(long one, long two) {
  printf(" Hello from Bar3: one = %d, two = %d!\n ", one, two);
  return one + two;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Bar3>:
  endbr64
  push   %rbp
  mov    %rsi,%rdx
  mov    %rsi,%rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  lea    D8(%rip),%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%rax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
data_type: string

tool: D8:  Hello from Bar3: one = %d, two = %d!
 

assistant: ```c
extern long Bar3(long one, long two) {
  printf(" Hello from Bar3: one = %d, two = %d!\n ", one, two);
  return one + two;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<google_earth_output_closing>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type: string

tool: D5: </Folder>
</Document>
</kml>


assistant: ```c
void google_earth_output_closing(FILE *output_file) {
  fprintf(output_file, "</Folder>\n"
                       "</Document>\n"
                       "</kml>\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<google_earth_output_closing>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  call   <fprintf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
type: string

tool: D0: </Folder>
</Document>
</kml>


assistant: ```c
void google_earth_output_closing(FILE *output_file) {
  fprintf(output_file, "</Folder>\n"
                       "</Document>\n"
                       "</kml>\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<google_earth_output_closing>:
  endbr64
  lea    D0(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D0
data_type: string

tool: D0: </Folder>
</Document>
</kml>


assistant: ```c
void google_earth_output_closing(FILE *output_file) {
  fprintf(output_file, "</Folder>\n"
                       "</Document>\n"
                       "</kml>\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<google_earth_output_closing>:
  endbr64
  lea    D4(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
datatype: string

tool: D4: </Folder>
</Document>
</kml>


assistant: ```c
void google_earth_output_closing(FILE *output_file) {
  fprintf(output_file, "</Folder>\n"
                       "</Document>\n"
                       "</kml>\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  mov    %rsi,%rcx
  mov    (%rdi),%rax
  sub    $0x99f4bb,%rax
  mov    $0x0,%edx
  jne    L2
  mov    %rax,%rsi
  shr    $0x20,%rsi
  mov    %rax,%rdx
  shr    $0x30,%rdx
  shl    $0x20,%rdx
  movabs $0xffff0000ffffffff,%rdi
  and    %rdi,%rax
  or     %rdx,%rax
  mov    %rax,%rdx
  mov    %si,%dx
  shl    $0x30,%rax
  movabs $0xffffffffffff,%rsi
  and    %rsi,%rdx
  or     %rdx,%rax
  shl    $0x8,%edx
  sar    $0x18,%edx
  mov    %rax,%rsi
  shr    $0x20,%rsi
  and    $0xff0000,%esi
  and    $0xffffffffff00ffff,%rax
  or     %rsi,%rax
  movzbl %dl,%edx
  shl    $0x30,%rdx
  movabs $0xff00ffffffffffff,%rsi
  and    %rsi,%rax
  or     %rax,%rdx
L2:
  add    $0x5adf563,%rdx
  mov    %rdx,(%rcx)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy12;
  char copy13;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          state[local1] -= state[0UL];
          state[local1] = state[0UL] + state[0UL];
        } else {
          state[local1] += state[local1];
        }
      } else {
        copy12 = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = *((unsigned short *)(&state[local1]) + 2);
        *((unsigned short *)(&state[local1]) + 2) = copy12;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        copy13 = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 6);
        *((char *)(&state[0UL]) + 6) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 95286627UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  mov    $0x5adf563,%edx
  sub    $0x99f4bb,%rax
  jne    L5
  and    $0xff000000,%eax
  lea    0x5adf563(%rax),%rdx
L5:
  mov    %rdx,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  unsigned short copy12;
  char copy13;
  {
    state[0UL] = input[0UL] + 0xffffffffff660b45UL;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        if (state[0UL] > local1) {
          state[local1] -= state[0UL];
          state[local1] = state[0UL] + state[0UL];
        } else {
          state[local1] += state[local1];
        }
      } else {
        copy12 = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = *((unsigned short *)(&state[local1]) + 2);
        *((unsigned short *)(&state[local1]) + 2) = copy12;
        copy12 = *((unsigned short *)(&state[local1]) + 0);
        *((unsigned short *)(&state[local1]) + 0) = *((unsigned short *)(&state[local1]) + 3);
        *((unsigned short *)(&state[local1]) + 3) = copy12;
        copy13 = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 6);
        *((char *)(&state[0UL]) + 6) = copy13;
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + 95286627UL;
  }
}
```