ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<iterateB>:
  endbr64
  test   %edi,%edi
  jle    L5
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D9(%rip),%r12
L9:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string

tool: D9: iterateB k = %d


assistant: ```c
void iterateB(int n) {
  int k = 0;
  for (k = 0; k < n; k++) {
    printf("iterateB k = %d\n", k);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<iterateB>:
  endbr64
  test   %edi,%edi
  jle    L14
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  ret
```

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

tool: D0: iterateB k = %d


assistant: ```c
void iterateB(int n) {
  int k = 0;
  for (k = 0; k < n; k++) {
    printf("iterateB k = %d\n", k);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<iterateB>:
  endbr64
  test   %edi,%edi
  jle    L1
  push   %r12
  lea    D1(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  ret
```

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

tool: D1: iterateB k = %d


assistant: ```c
void iterateB(int n) {
  int k = 0;
  for (k = 0; k < n; k++) {
    printf("iterateB k = %d\n", k);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<utf8_encode_nt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <utf8_encode@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L13
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L14
L13:
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
L14:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int utf8_encode_nt(unsigned int unicode_value, char *dest, size_t size) {
  int len = utf8_encode(unicode_value, dest, size - 1);
  if (len > 0)
    dest[len] = 0;
  else
    dest[0] = 0;
  return len;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<utf8_encode_nt>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x1,%rdx
  call   <utf8_encode@plt>
  test   %eax,%eax
  jle    L18
  movslq %eax,%rdx
  movb   $0x0,(%rbx,%rdx,1)
L12:
  pop    %rbx
  ret
L18:
  movb   $0x0,(%rbx)
  jmp    L12
```

assistant: ```c
int utf8_encode_nt(unsigned int unicode_value, char *dest, size_t size) {
  int len = utf8_encode(unicode_value, dest, size - 1);
  if (len > 0)
    dest[len] = 0;
  else
    dest[0] = 0;
  return len;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<utf8_encode_nt>:
  endbr64
  push   %rbx
  sub    $0x1,%rdx
  mov    %rsi,%rbx
  call   <utf8_encode@plt>
  test   %eax,%eax
  jle    L9
  movslq %eax,%rdx
  movb   $0x0,(%rbx,%rdx,1)
  pop    %rbx
  ret
  xchg   %ax,%ax
L9:
  movb   $0x0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int utf8_encode_nt(unsigned int unicode_value, char *dest, size_t size) {
  int len = utf8_encode(unicode_value, dest, size - 1);
  if (len > 0)
    dest[len] = 0;
  else
    dest[0] = 0;
  return len;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<utf8_encode_nt>:
  endbr64
  push   %rbx
  sub    $0x1,%rdx
  mov    %rsi,%rbx
  call   <utf8_encode@plt>
  test   %eax,%eax
  jle    L2
  movslq %eax,%rdx
  movb   $0x0,(%rbx,%rdx,1)
  pop    %rbx
  ret
  xchg   %ax,%ax
L2:
  movb   $0x0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int utf8_encode_nt(unsigned int unicode_value, char *dest, size_t size) {
  int len = utf8_encode(unicode_value, dest, size - 1);
  if (len > 0)
    dest[len] = 0;
  else
    dest[0] = 0;
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<valid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  mov    %rdi,-0xf8(%rbp)
  mov    %rsi,-0x100(%rbp)
  mov    %edx,-0x104(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xe0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0xd,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0x70(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0xd,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0xe8(%rbp)
  jmp    L60
L0:
  mov    -0xe8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0xf8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    -0xe0(%rbp,%rax,4),%eax
  mov    -0xe8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0xf8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  lea    -0x61(%rdx),%ecx
  lea    0x1(%rax),%edx
  movslq %ecx,%rax
  mov    %edx,-0xe0(%rbp,%rax,4)
  mov    -0xe8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x100(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x61,%eax
  cltq
  mov    -0x70(%rbp,%rax,4),%eax
  mov    -0xe8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x100(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movsbl %dl,%edx
  lea    -0x61(%rdx),%ecx
  lea    0x1(%rax),%edx
  movslq %ecx,%rax
  mov    %edx,-0x70(%rbp,%rax,4)
  addl   $0x1,-0xe8(%rbp)
L60:
  mov    -0xe8(%rbp),%eax
  cmp    -0x104(%rbp),%eax
  jl     L0
  movl   $0x0,-0xe4(%rbp)
  jmp    L34
L51:
  mov    -0xe4(%rbp),%eax
  cltq
  mov    -0xe0(%rbp,%rax,4),%edx
  mov    -0xe4(%rbp),%eax
  cltq
  mov    -0x70(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  je     L6
  mov    $0x0,%eax
  jmp    L68
L6:
  addl   $0x1,-0xe4(%rbp)
L34:
  cmpl   $0x19,-0xe4(%rbp)
  jle    L51
  mov    $0x1,%eax
L68:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int a[26] = {0}, b[26] = {0};
  for (int i = 0; i < n; i++) {
    a[s1[i] - 'a'] += 1;
    b[s2[i] - 'a'] += 1;
  }
  for (int i = 0; i < 26; i++)
    if (a[i] != b[i])
      return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<valid>:
  endbr64
  sub    $0xe8,%rsp
  mov    %rdi,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0xd,%ecx
  rep stos %rax,%es:(%rdi)
  lea    0x70(%rsp),%rdi
  mov    $0xd,%ecx
  rep stos %rax,%es:(%rdi)
  test   %edx,%edx
  jle    L37
  mov    %edx,%edx
L0:
  movsbl (%r8,%rcx,1),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  movsbl (%rsi,%rcx,1),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x70(%rsp,%rax,4)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L0
L37:
  mov    $0x0,%eax
  lea    0x70(%rsp),%rdx
L22:
  mov    (%rdx,%rax,1),%edi
  cmp    %edi,(%rsp,%rax,1)
  jne    L28
  add    $0x4,%rax
  cmp    $0x68,%rax
  jne    L22
  mov    $0x1,%eax
  jmp    L36
L28:
  mov    $0x0,%eax
L36:
  mov    0xd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0xe8,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int a[26] = {0}, b[26] = {0};
  for (int i = 0; i < n; i++) {
    a[s1[i] - 'a'] += 1;
    b[s2[i] - 'a'] += 1;
  }
  for (int i = 0; i < 26; i++)
    if (a[i] != b[i])
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<valid>:
  endbr64
  sub    $0xe8,%rsp
  mov    %rdi,%r9
  mov    %rsi,%r10
  mov    $0xd,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r8
  lea    0x70(%rsp),%rsi
  mov    %r8,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0xd,%ecx
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  test   %edx,%edx
  jle    L30
  movslq %edx,%rdx
  nopl   (%rax)
L36:
  movsbl (%r9,%rcx,1),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  movsbl (%r10,%rcx,1),%eax
  add    $0x1,%rcx
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x70(%rsp,%rax,4)
  cmp    %rcx,%rdx
  jne    L36
  mov    (%rsp),%edx
  mov    0x70(%rsp),%ecx
L26:
  xor    %eax,%eax
  jmp    L63
  nopl   0x0(%rax)
L54:
  add    $0x4,%rax
  cmp    $0x68,%rax
  je     L47
  mov    (%r8,%rax,1),%edx
  mov    (%rsi,%rax,1),%ecx
L63:
  cmp    %ecx,%edx
  je     L54
  xor    %eax,%eax
L65:
  mov    0xd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0xe8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  mov    $0x1,%eax
  jmp    L65
L30:
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L26
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int a[26] = {0}, b[26] = {0};
  for (int i = 0; i < n; i++) {
    a[s1[i] - 'a'] += 1;
    b[s2[i] - 'a'] += 1;
  }
  for (int i = 0; i < 26; i++)
    if (a[i] != b[i])
      return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<valid>:
  endbr64
  sub    $0xe8,%rsp
  mov    %rdi,%r9
  mov    %rsi,%r10
  mov    $0xd,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r8
  lea    0x70(%rsp),%rsi
  mov    %r8,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0xd,%ecx
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  test   %edx,%edx
  jle    L35
  movslq %edx,%rdx
  nopl   0x0(%rax)
L14:
  movsbl (%r9,%rcx,1),%eax
  sub    $0x61,%eax
  cltq
  addl   $0x1,(%rsp,%rax,4)
  movsbl (%r10,%rcx,1),%eax
  add    $0x1,%rcx
  sub    $0x61,%eax
  cltq
  addl   $0x1,0x70(%rsp,%rax,4)
  cmp    %rcx,%rdx
  jne    L14
L35:
  xor    %eax,%eax
  jmp    L43
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x4,%rax
  cmp    $0x68,%rax
  je     L63
L43:
  mov    (%rsi,%rax,1),%edi
  cmp    %edi,(%r8,%rax,1)
  je     L21
  xor    %eax,%eax
L29:
  mov    0xd8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0xe8,%rsp
  ret
L63:
  mov    $0x1,%eax
  jmp    L29
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int valid(char *s1, char *s2, int n) {
  int a[26] = {0}, b[26] = {0};
  for (int i = 0; i < n; i++) {
    a[s1[i] - 'a'] += 1;
    b[s2[i] - 'a'] += 1;
  }
  for (int i = 0; i < 26; i++)
    if (a[i] != b[i])
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L39
L23:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L22
  mov    -0x4(%rbp),%eax
  jmp    L26
L22:
  addl   $0x1,-0x4(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L23
  mov    $0xffffffff,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
int pos(const char *s, const int c) {
  int i;
  for (i = 0; s[i] != '\0'; i++) {
    if (s[i] == c) {
      return i;
    }
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<pos>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  mov    $0x0,%edx
L38:
  mov    %edx,%ecx
  movsbl %al,%eax
  cmp    %esi,%eax
  je     L39
  add    $0x1,%rdx
  movzbl (%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L38
  mov    $0xffffffff,%ecx
L39:
  mov    %ecx,%eax
  ret
L31:
  mov    $0xffffffff,%ecx
  jmp    L39
```

assistant: ```c
int pos(const char *s, const int c) {
  int i;
  for (i = 0; s[i] != '\0'; i++) {
    if (s[i] == c) {
      return i;
    }
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<pos>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L17
  xor    %edx,%edx
  jmp    L24
  nop
L32:
  add    $0x1,%rdx
  movsbl (%rdi,%rdx,1),%eax
  test   %al,%al
  je     L17
L24:
  mov    %edx,%r8d
  cmp    %esi,%eax
  jne    L32
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int pos(const char *s, const int c) {
  int i;
  for (i = 0; s[i] != '\0'; i++) {
    if (s[i] == c) {
      return i;
    }
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pos>:
  endbr64
  movsbl (%rdi),%eax
  test   %al,%al
  je     L9
  xor    %edx,%edx
  jmp    L4
  nop
L31:
  add    $0x1,%rdx
  movsbl (%rdi,%rdx,1),%eax
  test   %al,%al
  je     L9
L4:
  mov    %edx,%r8d
  cmp    %esi,%eax
  jne    L31
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int pos(const char *s, const int c) {
  int i;
  for (i = 0; s[i] != '\0'; i++) {
    if (s[i] == c) {
      return i;
    }
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L3
  cmpl   $0x0,-0x18(%rbp)
  jne    L29
L3:
  mov    $0x1,%eax
  jmp    L9
L29:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <f>
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <f>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
L9:
  leave
  ret
```

assistant: ```c
int f(int dx, int dy) {
  int a, b;
  if ((dx == 0) || (dy == 0)) {
    return 1;
  }
  a = f(dx - 1, dy);
  b = f(dx, dy - 1);
  return (a + b);
}
```
user: What is the c source code of the assembly code below:

```asm
<f>:
  endbr64
  test   %edi,%edi
  je     L18
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    $0x1,%eax
  test   %esi,%esi
  je     L5
  lea    -0x1(%rdi),%edi
  call   <f>
  mov    %eax,%r12d
  lea    -0x1(%rbx),%esi
  mov    %ebp,%edi
  call   <f>
  add    %r12d,%eax
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int f(int dx, int dy) {
  int a, b;
  if ((dx == 0) || (dy == 0)) {
    return 1;
  }
  a = f(dx - 1, dy);
  b = f(dx, dy - 1);
  return (a + b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f>:
  endbr64
  test   %edi,%edi
  je     L111
  mov    %esi,%r11d
  test   %esi,%esi
  je     L111
  push   %r15
  mov    %edi,%r10d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  movl   $0x0,0x8(%rsp)
L131:
  mov    $0x1,%eax
  cmp    $0x1,%r10d
  je     L43
  mov    %r11d,0xc(%rsp)
  mov    %r10d,%ebp
  movl   $0x0,0x10(%rsp)
  mov    %r11d,0x1c(%rsp)
L103:
  mov    $0x1,%eax
  cmp    $0x2,%ebp
  je     L130
  mov    0xc(%rsp),%eax
  movl   $0x0,0x18(%rsp)
  mov    %eax,0x14(%rsp)
L91:
  mov    $0x1,%r10d
  cmp    $0x3,%ebp
  je     L53
  mov    0x14(%rsp),%r11d
  xor    %r10d,%r10d
L182:
  mov    $0x1,%edx
  cmp    $0x4,%ebp
  je     L78
  mov    %r11d,%r12d
  xor    %edx,%edx
L11:
  mov    $0x1,%eax
  cmp    $0x5,%ebp
  je     L42
  mov    %r12d,%ecx
  xor    %ebx,%ebx
L96:
  mov    $0x1,%r15d
  cmp    $0x6,%ebp
  je     L145
  mov    %ecx,%r8d
  xor    %r15d,%r15d
L183:
  mov    $0x1,%r13d
  cmp    $0x7,%ebp
  je     L165
  mov    %r8d,%r14d
  xor    %r13d,%r13d
L159:
  mov    $0x1,%eax
  cmp    $0x8,%ebp
  je     L151
  lea    -0x9(%rbp),%edi
  mov    %r14d,%esi
  xor    %r9d,%r9d
L99:
  mov    %r9d,0x3c(%rsp)
  mov    %r8d,0x38(%rsp)
  mov    %ecx,0x34(%rsp)
  mov    %edx,0x30(%rsp)
  mov    %r10d,0x2c(%rsp)
  mov    %r11d,0x28(%rsp)
  mov    %esi,0x24(%rsp)
  mov    %edi,0x20(%rsp)
  call   <f>
  mov    0x3c(%rsp),%r9d
  mov    0x24(%rsp),%esi
  mov    0x20(%rsp),%edi
  mov    0x28(%rsp),%r11d
  add    %eax,%r9d
  sub    $0x1,%esi
  mov    0x2c(%rsp),%r10d
  mov    0x30(%rsp),%edx
  mov    0x34(%rsp),%ecx
  mov    0x38(%rsp),%r8d
  jne    L99
  lea    0x1(%r9),%eax
L151:
  add    %eax,%r13d
  sub    $0x1,%r14d
  jne    L159
  add    $0x1,%r13d
L165:
  add    %r13d,%r15d
  sub    $0x1,%r8d
  jne    L183
  add    $0x1,%r15d
L145:
  add    %r15d,%ebx
  sub    $0x1,%ecx
  jne    L96
  lea    0x1(%rbx),%eax
L42:
  add    %eax,%edx
  sub    $0x1,%r12d
  jne    L11
  add    $0x1,%edx
L78:
  add    %edx,%r10d
  sub    $0x1,%r11d
  jne    L182
  add    $0x1,%r10d
L53:
  add    %r10d,0x18(%rsp)
  subl   $0x1,0x14(%rsp)
  jne    L91
  mov    0x18(%rsp),%eax
  add    $0x1,%eax
L130:
  add    %eax,0x10(%rsp)
  subl   $0x1,0xc(%rsp)
  jne    L103
  mov    0x10(%rsp),%eax
  mov    0x1c(%rsp),%r11d
  mov    %ebp,%r10d
  add    $0x1,%eax
L43:
  add    %eax,0x8(%rsp)
  sub    $0x1,%r11d
  jne    L131
  mov    0x8(%rsp),%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  add    $0x1,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L111:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int f(int dx, int dy) {
  int a, b;
  if ((dx == 0) || (dy == 0)) {
    return 1;
  }
  a = f(dx - 1, dy);
  b = f(dx, dy - 1);
  return (a + b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<f>:
  endbr64
  test   %edi,%edi
  je     L8
  mov    %esi,%r11d
  test   %esi,%esi
  je     L8
  push   %r15
  mov    %edi,%r10d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  movl   $0x0,0x8(%rsp)
L61:
  mov    $0x1,%eax
  cmp    $0x1,%r10d
  je     L0
  mov    %r11d,0xc(%rsp)
  mov    %r10d,%ebp
  movl   $0x0,0x10(%rsp)
  mov    %r11d,0x1c(%rsp)
L113:
  mov    $0x1,%eax
  cmp    $0x2,%ebp
  je     L18
  mov    0xc(%rsp),%eax
  movl   $0x0,0x18(%rsp)
  mov    %eax,0x14(%rsp)
L163:
  mov    $0x1,%r10d
  cmp    $0x3,%ebp
  je     L45
  mov    0x14(%rsp),%r11d
  xor    %r10d,%r10d
L130:
  mov    $0x1,%edx
  cmp    $0x4,%ebp
  je     L174
  mov    %r11d,%r12d
  xor    %edx,%edx
L94:
  mov    $0x1,%eax
  cmp    $0x5,%ebp
  je     L148
  mov    %r12d,%ecx
  xor    %ebx,%ebx
L140:
  mov    $0x1,%r15d
  cmp    $0x6,%ebp
  je     L75
  mov    %ecx,%r8d
  xor    %r15d,%r15d
L153:
  mov    $0x1,%r13d
  cmp    $0x7,%ebp
  je     L165
  mov    %r8d,%r14d
  xor    %r13d,%r13d
L19:
  mov    $0x1,%eax
  cmp    $0x8,%ebp
  je     L143
  lea    -0x9(%rbp),%edi
  mov    %r14d,%esi
  xor    %r9d,%r9d
L55:
  mov    %r9d,0x3c(%rsp)
  mov    %r8d,0x38(%rsp)
  mov    %ecx,0x34(%rsp)
  mov    %edx,0x30(%rsp)
  mov    %r10d,0x2c(%rsp)
  mov    %r11d,0x28(%rsp)
  mov    %esi,0x24(%rsp)
  mov    %edi,0x20(%rsp)
  call   <f>
  mov    0x3c(%rsp),%r9d
  mov    0x24(%rsp),%esi
  mov    0x20(%rsp),%edi
  mov    0x28(%rsp),%r11d
  add    %eax,%r9d
  sub    $0x1,%esi
  mov    0x2c(%rsp),%r10d
  mov    0x30(%rsp),%edx
  mov    0x34(%rsp),%ecx
  mov    0x38(%rsp),%r8d
  jne    L55
  lea    0x1(%r9),%eax
L143:
  add    %eax,%r13d
  sub    $0x1,%r14d
  jne    L19
  add    $0x1,%r13d
L165:
  add    %r13d,%r15d
  sub    $0x1,%r8d
  jne    L153
  add    $0x1,%r15d
L75:
  add    %r15d,%ebx
  sub    $0x1,%ecx
  jne    L140
  lea    0x1(%rbx),%eax
L148:
  add    %eax,%edx
  sub    $0x1,%r12d
  jne    L94
  add    $0x1,%edx
L174:
  add    %edx,%r10d
  sub    $0x1,%r11d
  jne    L130
  add    $0x1,%r10d
L45:
  add    %r10d,0x18(%rsp)
  subl   $0x1,0x14(%rsp)
  jne    L163
  mov    0x18(%rsp),%eax
  add    $0x1,%eax
L18:
  add    %eax,0x10(%rsp)
  subl   $0x1,0xc(%rsp)
  jne    L113
  mov    0x10(%rsp),%eax
  mov    0x1c(%rsp),%r11d
  mov    %ebp,%r10d
  add    $0x1,%eax
L0:
  add    %eax,0x8(%rsp)
  sub    $0x1,%r11d
  jne    L61
  mov    0x8(%rsp),%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  add    $0x1,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int f(int dx, int dy) {
  int a, b;
  if ((dx == 0) || (dy == 0)) {
    return 1;
  }
  a = f(dx - 1, dy);
  b = f(dx, dy - 1);
  return (a + b);
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L7
  cmpl   $0x9,-0x14(%rbp)
  ja     L7
L29:
  subl   $0x1,-0x14(%rbp)
  shll   -0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L29
L7:
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(unsigned int n) {
  int i, r = 1;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
      } while (n > 0);
    }
  }
  return r + n;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L26
  cmp    $0x9,%edi
  ja     L26
L23:
  add    %eax,%eax
  sub    $0x1,%edi
  jne    L23
L26:
  add    %edi,%eax
  ret
```

assistant: ```c
int foo(unsigned int n) {
  int i, r = 1;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
      } while (n > 0);
    }
  }
  return r + n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L18
  cmp    $0x9,%edi
  ja     L19
  nopw   0x0(%rax,%rax,1)
L22:
  add    %eax,%eax
  sub    $0x1,%edi
  jne    L22
L18:
  ret
L19:
  lea    0x1(%rdi),%eax
  ret
```

assistant: ```c
int foo(unsigned int n) {
  int i, r = 1;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
      } while (n > 0);
    }
  }
  return r + n;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L83
  cmp    $0x9,%edi
  ja     L88
  cmp    $0x1,%edi
  je     L41
  mov    %edi,%eax
  shr    %eax
  cmp    $0x1,%eax
  je     L87
  cmp    $0x2,%eax
  je     L72
  cmp    $0x4,%eax
  jne    L40
  movq   D11(%rip),%xmm0
L7:
  movd   %xmm0,%eax
  pshufd $0xe5,%xmm0,%xmm1
  movd   %xmm1,%edx
  imul   %edx,%eax
  and    $0x1,%edi
  je     L83
L41:
  add    %eax,%eax
L83:
  ret
  nopw   0x0(%rax,%rax,1)
L88:
  lea    0x1(%rdi),%eax
  ret
  nopl   0x0(%rax)
L87:
  movq   D2(%rip),%xmm0
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L72:
  movq   D33(%rip),%xmm0
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L40:
  movq   D5(%rip),%xmm0
  jmp    L7
```

tool_calls: name: unpack_data
arguments: data_label: D11
datatype: byte[8]
name: unpack_data
arguments: data_label: D2
datatype: byte[8]
name: unpack_data
arguments: data_label: D33
datatype: byte[8]
name: unpack_data
arguments: data_label: D5
datatype: byte[8]

tool: {"D11": [0x10,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

tool: {"D2": [0x02,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: {"D33": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D5": [0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

assistant: ```c
int foo(unsigned int n) {
  int i, r = 1;
  if (n > 0) {
    asm("");
    if (n < 10) {
      asm("");
      do {
        --n;
        r *= 2;
      } while (n > 0);
    }
  }
  return r + n;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<GetNextDigit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <IsDigit@plt>
  cmp    $0x1,%eax
  jne    L9
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
L9:
  nop
  leave
  ret
```

assistant: ```c
void GetNextDigit(int *dest, char *src) {
  if (IsDigit(src[0]) == 1) {
    *dest *= 10;
    *dest += src[0] - '0';
  }
}
```