ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D21(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getch@plt>
  mov    $0x32,%eax
  pxor   %xmm0,%xmm0
  movss  D19(%rip),%xmm4
  movss  D8(%rip),%xmm5
  movaps %xmm4,%xmm2
  movaps %xmm4,%xmm1
  nopw   0x0(%rax,%rax,1)
L5:
  movaps %xmm1,%xmm3
  addss  %xmm5,%xmm1
  divss  %xmm2,%xmm3
  addss  %xmm4,%xmm2
  addss  %xmm3,%xmm0
  sub    $0x1,%eax
  jne    L5
  lea    D37(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D21
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: byte[4]
name: extract_symbolic_data
arguments: label: D8
type_infered: byte[4]
name: extract_symbolic_data
arguments: label: D37
type_infered: string

tool: D21: Vamos calcular uma sequencia legal!

(Sequencia do Tipo: 1/1 + 3/2 + 5/3 + 7/4 + ... + 99/50)


tool: {"D19": [0x00,0x00,0x80,0x3f]}

tool: {"D8": [0x00,0x00,0x00,0x40]}

tool: D37: A soma e: %.4f


assistant: ```c
int main() {
  float dividendo, divisor, soma;
  printf("Vamos calcular uma sequencia legal!\n\n(Sequencia do Tipo: 1/1 + 3/2 + 5/3 + 7/4 + ... + 99/50)\n\n");
  soma = 0.0;
  getch();
  dividendo = 1;
  for (divisor = 1; divisor <= 50; divisor++) {
    soma = soma + (dividendo / divisor);
    dividendo += 2;
  }
  printf("A soma e: %.4f\n", soma);
  getch();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getch@plt>
  mov    $0x32,%eax
  pxor   %xmm0,%xmm0
  movss  D39(%rip),%xmm4
  movss  D13(%rip),%xmm5
  movaps %xmm4,%xmm2
  movaps %xmm4,%xmm1
  nopw   0x0(%rax,%rax,1)
L8:
  movaps %xmm1,%xmm3
  addss  %xmm5,%xmm1
  divss  %xmm2,%xmm3
  addss  %xmm4,%xmm2
  addss  %xmm3,%xmm0
  sub    $0x1,%eax
  jne    L8
  lea    D19(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <getch@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D20
datatype: string
name: unpack_data
arguments: data_label: D39
datatype: byte[4]
name: unpack_data
arguments: data_label: D13
datatype: byte[4]
name: unpack_data
arguments: data_label: D19
datatype: string

tool: D20: Vamos calcular uma sequencia legal!

(Sequencia do Tipo: 1/1 + 3/2 + 5/3 + 7/4 + ... + 99/50)


tool: {"D39": [0x00,0x00,0x80,0x3f]}

tool: {"D13": [0x00,0x00,0x00,0x40]}

tool: D19: A soma e: %.4f


assistant: ```c
int main() {
  float dividendo, divisor, soma;
  printf("Vamos calcular uma sequencia legal!\n\n(Sequencia do Tipo: 1/1 + 3/2 + 5/3 + 7/4 + ... + 99/50)\n\n");
  soma = 0.0;
  getch();
  dividendo = 1;
  for (divisor = 1; divisor <= 50; divisor++) {
    soma = soma + (dividendo / divisor);
    dividendo += 2;
  }
  printf("A soma e: %.4f\n", soma);
  getch();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L18
L27:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <fact@plt>
  add    %eax,-0x10(%rbp)
L18:
  cmpl   $0x0,-0x14(%rbp)
  jg     L27
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jne    L15
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L31
L15:
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L31:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type_guessed: string
name: disassemble_data
arguments: data_label: D7
type_guessed: string
name: disassemble_data
arguments: data_label: D34
type_guessed: string
name: disassemble_data
arguments: data_label: D26
type_guessed: string

tool: D0: Enter the number: 

tool: D7: %d

tool: D34: 
This is a strong number

tool: D26: 
This is not a strong number

assistant: ```c
int main(void) {
  int n, t, s = 0, r;
  printf("Enter the number: ");
  scanf("%d", &n);
  t = n;
  while (t > 0) {
    r = t % 10;
    t = t / 10;
    s = s + fact(r);
  }
  (s == n) ? printf("\nThis is a strong number") : printf("\nThis is not a strong number");
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D37(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L58
  mov    $0x0,%ebp
L12:
  movslq %ebx,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  mov    %edi,%edx
  lea    (%rdi,%rdi,4),%eax
  add    %eax,%eax
  mov    %ebx,%edi
  sub    %eax,%edi
  mov    %ebx,%r12d
  mov    %edx,%ebx
  call   <fact@plt>
  add    %eax,%ebp
  cmp    $0x9,%r12d
  jg     L12
L55:
  cmp    %ebp,0x4(%rsp)
  je     L28
  lea    D1(%rip),%rdi
  call   <printf@plt>
L49:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L58:
  mov    $0x0,%ebp
  jmp    L55
L28:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L49
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
type: string
name: parse_data
arguments: label: D37
type: string
name: parse_data
arguments: label: D1
type: string
name: parse_data
arguments: label: D13
type: string

tool: D3: Enter the number: 

tool: D37: %d

tool: D1: 
This is not a strong number

tool: D13: 
This is a strong number

assistant: ```c
int main(void) {
  int n, t, s = 0, r;
  printf("Enter the number: ");
  scanf("%d", &n);
  t = n;
  while (t > 0) {
    r = t % 10;
    t = t / 10;
    s = s + fact(r);
  }
  (s == n) ? printf("\nThis is a strong number") : printf("\nThis is not a strong number");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  lea    D28(%rip),%rdi
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L31
  mov    $0xcccccccd,%r12d
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %ebx,%edi
  mov    %ebx,%r13d
  imul   %r12,%rdi
  shr    $0x23,%rdi
  lea    (%rdi,%rdi,4),%edx
  mov    %edi,%eax
  mov    %ebx,%edi
  add    %edx,%edx
  mov    %eax,%ebx
  sub    %edx,%edi
  call   <fact@plt>
  add    %eax,%ebp
  cmp    $0x9,%r13d
  jg     L27
L31:
  cmp    %ebp,0x4(%rsp)
  je     L9
  lea    D20(%rip),%rdi
  call   <printf@plt>
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  lea    D19(%rip),%rdi
  call   <printf@plt>
  jmp    L0
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D28
type: string
name: parse_data
arguments: label: D4
type: string
name: parse_data
arguments: label: D20
type: string
name: parse_data
arguments: label: D19
type: string

tool: D28: Enter the number: 

tool: D4: %d

tool: D20: 
This is not a strong number

tool: D19: 
This is a strong number

assistant: ```c
int main(void) {
  int n, t, s = 0, r;
  printf("Enter the number: ");
  scanf("%d", &n);
  t = n;
  while (t > 0) {
    r = t % 10;
    t = t / 10;
    s = s + fact(r);
  }
  (s == n) ? printf("\nThis is a strong number") : printf("\nThis is not a strong number");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D3(%rip),%rdi
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  jle    L23
  mov    $0xcccccccd,%r12d
  nopl   0x0(%rax,%rax,1)
L29:
  mov    %ebx,%edi
  mov    %ebx,%r13d
  imul   %r12,%rdi
  shr    $0x23,%rdi
  lea    (%rdi,%rdi,4),%edx
  mov    %edi,%eax
  mov    %ebx,%edi
  add    %edx,%edx
  mov    %eax,%ebx
  sub    %edx,%edi
  call   <fact@plt>
  add    %eax,%ebp
  cmp    $0x9,%r13d
  jg     L29
L23:
  cmp    %ebp,0x4(%rsp)
  je     L19
  lea    D2(%rip),%rdi
  call   <printf@plt>
L15:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L15
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type_guessed: string
name: extract_symbolic_data
arguments: label: D31
type_guessed: string
name: extract_symbolic_data
arguments: label: D2
type_guessed: string
name: extract_symbolic_data
arguments: label: D4
type_guessed: string

tool: D3: Enter the number: 

tool: D31: %d

tool: D2: 
This is not a strong number

tool: D4: 
This is a strong number

assistant: ```c
int main(void) {
  int n, t, s = 0, r;
  printf("Enter the number: ");
  scanf("%d", &n);
  t = n;
  while (t > 0) {
    r = t % 10;
    t = t / 10;
    s = s + fact(r);
  }
  (s == n) ? printf("\nThis is a strong number") : printf("\nThis is not a strong number");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_dirname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L5
L22:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L33
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L16
L33:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L18
L16:
  subl   $0x1,-0x8(%rbp)
L5:
  cmpl   $0x0,-0x8(%rbp)
  jns    L22
  nop
L18:
  nop
  leave
  ret
```

assistant: ```c
void my_dirname(char *path) {
  int n;
  int len = strlen(path);
  for (n = len - 1; n >= 0; n--) {
    if (path[n] == '/' || path[n] == '\\') {
      path[n] = 0;
      break;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<my_dirname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edx
  sub    $0x1,%edx
  js     L44
  movslq %edx,%rdx
  add    %rbx,%rdx
  movslq %eax,%rcx
  lea    -0x2(%rbx,%rcx,1),%rcx
  lea    -0x1(%rax),%eax
  sub    %rax,%rcx
L9:
  mov    %rdx,%rsi
  movzbl (%rdx),%eax
  cmp    $0x2f,%al
  je     L4
  cmp    $0x5c,%al
  je     L4
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L9
  jmp    L44
L4:
  movb   $0x0,(%rsi)
L44:
  pop    %rbx
  ret
```

assistant: ```c
void my_dirname(char *path) {
  int n;
  int len = strlen(path);
  for (n = len - 1; n >= 0; n--) {
    if (path[n] == '/' || path[n] == '\\') {
      path[n] = 0;
      break;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<my_dirname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%esi
  sub    $0x1,%esi
  js     L3
  cltq
  movslq %esi,%rdx
  mov    %esi,%esi
  lea    -0x2(%rbx,%rax,1),%rcx
  add    %rbx,%rdx
  sub    %rsi,%rcx
  cs nopw 0x0(%rax,%rax,1)
L18:
  movzbl (%rdx),%eax
  cmp    $0x2f,%al
  je     L24
  cmp    $0x5c,%al
  je     L24
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L18
L3:
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L24:
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
```

assistant: ```c
void my_dirname(char *path) {
  int n;
  int len = strlen(path);
  for (n = len - 1; n >= 0; n--) {
    if (path[n] == '/' || path[n] == '\\') {
      path[n] = 0;
      break;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<my_dirname>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%esi
  sub    $0x1,%esi
  js     L39
  cltq
  movslq %esi,%rdx
  mov    %esi,%esi
  lea    -0x2(%rbx,%rax,1),%rcx
  add    %rbx,%rdx
  sub    %rsi,%rcx
  cs nopw 0x0(%rax,%rax,1)
L32:
  movzbl (%rdx),%eax
  cmp    $0x2f,%al
  je     L8
  cmp    $0x5c,%al
  je     L8
  sub    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L32
L39:
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  movb   $0x0,(%rdx)
  pop    %rbx
  ret
```

assistant: ```c
void my_dirname(char *path) {
  int n;
  int len = strlen(path);
  for (n = len - 1; n >= 0; n--) {
    if (path[n] == '/' || path[n] == '\\') {
      path[n] = 0;
      break;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<r_hash_fn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x1505,-0x8(%rbp)
  jmp    L9
L7:
  mov    -0x8(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L9:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L7
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long r_hash_fn(char *key) {
  unsigned long hash = 5381;
  int c;
  while (c = *key++)
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<r_hash_fn>:
  endbr64
  lea    0x1(%rdi),%rcx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L16
  mov    $0x1505,%eax
L18:
  mov    %rax,%rsi
  shl    $0x5,%rsi
  add    %rsi,%rax
  movsbq %dl,%rdx
  add    %rdx,%rax
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L18
  ret
L16:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned long r_hash_fn(char *key) {
  unsigned long hash = 5381;
  int c;
  while (c = *key++)
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<r_hash_fn>:
  endbr64
  movsbq (%rdi),%rdx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L18
  nopl   (%rax)
L1:
  mov    %rax,%rsi
  add    $0x1,%rcx
  shl    $0x5,%rsi
  add    %rsi,%rax
  add    %rdx,%rax
  movsbq -0x1(%rcx),%rdx
  test   %dl,%dl
  jne    L1
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  ret
```

assistant: ```c
unsigned long r_hash_fn(char *key) {
  unsigned long hash = 5381;
  int c;
  while (c = *key++)
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: Translate the below assembly code into C programming language:

```asm
<r_hash_fn>:
  endbr64
  movsbq (%rdi),%rdx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L5
  nopl   (%rax)
L7:
  mov    %rax,%rsi
  add    $0x1,%rcx
  shl    $0x5,%rsi
  add    %rsi,%rax
  add    %rdx,%rax
  movsbq -0x1(%rcx),%rdx
  test   %dl,%dl
  jne    L7
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
unsigned long r_hash_fn(char *key) {
  unsigned long hash = 5381;
  int c;
  while (c = *key++)
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<thread_function>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
L0:
  mov    $0x0,%eax
  call   <foo@plt>
  jmp    L0
```

assistant: ```c
void *thread_function(void *arg) {
  while (1)
    foo();
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<thread_function>:
  endbr64
  sub    $0x8,%rsp
L7:
  mov    $0x0,%eax
  call   <foo@plt>
  jmp    L7
```

assistant: ```c
void *thread_function(void *arg) {
  while (1)
    foo();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<thread_function>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <foo@plt>
  jmp    L4
```

assistant: ```c
void *thread_function(void *arg) {
  while (1)
    foo();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<thread_function>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %eax,%eax
  call   <foo@plt>
  jmp    L0
```

assistant: ```c
void *thread_function(void *arg) {
  while (1)
    foo();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ansieeol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  call   <ttputc@plt>
  mov    $0x5b,%edi
  call   <ttputc@plt>
  mov    $0x4b,%edi
  call   <ttputc@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int ansieeol() {
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ttputc('K');
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ansieeol>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  call   <ttputc@plt>
  mov    $0x5b,%edi
  call   <ttputc@plt>
  mov    $0x4b,%edi
  call   <ttputc@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ansieeol() {
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ttputc('K');
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ansieeol>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  call   <ttputc@plt>
  mov    $0x5b,%edi
  call   <ttputc@plt>
  mov    $0x4b,%edi
  call   <ttputc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ansieeol() {
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ttputc('K');
  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
<ansieeol>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ttflush@plt>
  mov    $0x1b,%edi
  call   <ttputc@plt>
  mov    $0x5b,%edi
  call   <ttputc@plt>
  mov    $0x4b,%edi
  call   <ttputc@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ansieeol() {
  ttflush();
  ttputc(0x1B);
  ttputc('[');
  ttputc('K');
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<my_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <mem_alloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  leave
  ret
```

assistant: ```c
char *my_strdup(const char *str) {
  size_t len = strlen(str) + 1;
  char *dup = mem_alloc(len);
  return memcpy(dup, str, len);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<my_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdi
  call   <mem_alloc@plt>
  mov    %rax,%rdi
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *my_strdup(const char *str) {
  size_t len = strlen(str) + 1;
  char *dup = mem_alloc(len);
  return memcpy(dup, str, len);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <mem_alloc@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <memcpy@plt>
```

assistant: ```c
char *my_strdup(const char *str) {
  size_t len = strlen(str) + 1;
  char *dup = mem_alloc(len);
  return memcpy(dup, str, len);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<my_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdi
  call   <mem_alloc@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <memcpy@plt>
```

assistant: ```c
char *my_strdup(const char *str) {
  size_t len = strlen(str) + 1;
  char *dup = mem_alloc(len);
  return memcpy(dup, str, len);
}
```
user: Translate the below assembly code into C programming language:

```asm
<iposix_path_splitext>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L59
L37:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L62
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L85
L62:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L82
  subl   $0x1,-0x10(%rbp)
L59:
  cmpl   $0x0,-0x10(%rbp)
  jns    L37
  jmp    L85
L82:
  nop
L85:
  cmpq   $0x0,-0x20(%rbp)
  je     L67
  mov    -0x24(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jle    L41
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L41:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L67
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L67:
  mov    -0x4(%rbp),%eax
  sub    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L113
  movl   $0x0,-0x8(%rbp)
L113:
  mov    -0x8(%rbp),%edx
  mov    -0x28(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x8(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  je     L26
  cmpl   $0x0,-0x8(%rbp)
  jle    L39
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L39:
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jge    L26
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L26:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int iposix_path_splitext(const char *path, char *p1, int l1, char *p2, int l2) {
  int length, i, k, size;
  length = (int)strlen(path);
  for (i = length - 1, k = length; i >= 0; i--) {
    if (path[i] == '.') {
      k = i;
      break;
    } else if ('/' == '/') {
      if (path[i] == '/')
        break;
    } else {
      if (path[i] == '/' || path[i] == '\\')
        break;
    }
  }
  if (p1) {
    size = k < l1 ? k : l1;
    if (size > 0)
      memcpy(p1, path, size);
    if (size < l1)
      p1[size] = 0;
  }
  size = length - k - 1;
  if (size < 0)
    size = 0;
  size = size < l2 ? size : l2;
  if (p2) {
    if (size > 0)
      memcpy(p2, path + k + 1, size);
    if (size < l2)
      p2[size] = 0;
  }
  return 0;
}
```