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

```asm
<atomic>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movb   $0x1,0xe(%rsp)
  movb   $0x0,0xf(%rsp)
  movl   $0x1,0x14(%rsp)
  lock addl $0x1,0x10(%rsp)
  lock subb $0x2,0xe(%rsp)
  mov    $0x3,%esi
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <__sync_fetch_and_min@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <__sync_fetch_and_max@plt>
  lea    0x14(%rsp),%rbp
  mov    $0x5,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <__sync_fetch_and_umin@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <__sync_fetch_and_umax@plt>
  mov    $0x7,%eax
  xchg   %eax,0x10(%rsp)
  mov    $0x4,%edx
  mov    $0x7b8,%ecx
  mov    %edx,%eax
  lock cmpxchg %ecx,0x10(%rsp)
  mov    %edx,%eax
  lock cmpxchg %ecx,0x10(%rsp)
  lock andl $0x9,0x10(%rsp)
  lock orl $0xa,0x10(%rsp)
  lock xorl $0xb,0x10(%rsp)
  lock addl $0x1,0x10(%rsp)
  lock subl $0x2,0x10(%rsp)
  lock andb $0x3,0xe(%rsp)
  lock orb $0x4,0xe(%rsp)
  movzbl 0xe(%rsp),%eax
L9:
  mov    %eax,%edx
  xor    $0x5,%edx
  lock cmpxchg %dl,0xe(%rsp)
  jne    L9
  xor    %ecx,%ecx
  mov    %rcx,%rax
  lock cmpxchg %rcx,0x0
  xor    %eax,%eax
  mov    $0x1,%esi
  lock cmpxchg %sil,0xf(%rsp)
  test   %al,%al
  movsbl %dl,%edx
  mov    $0x2a,%eax
  cmove  %edx,%eax
  mov    %eax,%r8d
  mov    %rcx,%rax
  lock cmpxchg %rcx,0x0
  movl   $0x0,0x10(%rsp)
  lock orq $0x0,(%rsp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  mov    %r8d,%eax
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int atomic(void) {
  int old;
  int val = 1;
  char valc = 1;
  _Bool valb = 0;
  unsigned int uval = 1;
  int cmp = 0;
  old = __sync_fetch_and_add(&val, 1);
  old = __sync_fetch_and_sub(&valc, 2);
  old = __sync_fetch_and_min(&val, 3);
  old = __sync_fetch_and_max(&val, 4);
  old = __sync_fetch_and_umin(&uval, 5u);
  old = __sync_fetch_and_umax(&uval, 6u);
  old = __sync_lock_test_and_set(&val, 7);
  old = __sync_val_compare_and_swap(&val, 4, 1976);
  old = __sync_bool_compare_and_swap(&val, 4, 1976);
  old = __sync_fetch_and_and(&val, 0x9);
  old = __sync_fetch_and_or(&val, 0xa);
  old = __sync_fetch_and_xor(&val, 0xb);
  old = __sync_add_and_fetch(&val, 1);
  old = __sync_sub_and_fetch(&val, 2);
  old = __sync_and_and_fetch(&valc, 3);
  old = __sync_or_and_fetch(&valc, 4);
  old = __sync_xor_and_fetch(&valc, 5);
  __sync_val_compare_and_swap((void **)0, (void *)0, (void *)0);
  if (__sync_val_compare_and_swap(&valb, 0, 1)) {
    old = 42;
  }
  __sync_bool_compare_and_swap((void **)0, (void *)0, (void *)0);
  __sync_lock_release(&val);
  __sync_synchronize();
  return old;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atomic>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  movb   $0x1,0xe(%rsp)
  movb   $0x0,0xf(%rsp)
  movl   $0x1,0x14(%rsp)
  lock addl $0x1,0x10(%rsp)
  lock subb $0x2,0xe(%rsp)
  mov    $0x3,%esi
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <__sync_fetch_and_min@plt>
  mov    %rbp,%rdi
  mov    $0x4,%esi
  xor    %eax,%eax
  call   <__sync_fetch_and_max@plt>
  lea    0x14(%rsp),%rbp
  mov    $0x5,%esi
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <__sync_fetch_and_umin@plt>
  mov    $0x6,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <__sync_fetch_and_umax@plt>
  mov    $0x7,%eax
  xchg   %eax,0x10(%rsp)
  mov    $0x4,%edx
  mov    $0x7b8,%ecx
  mov    %edx,%eax
  lock cmpxchg %ecx,0x10(%rsp)
  mov    %edx,%eax
  lock cmpxchg %ecx,0x10(%rsp)
  lock andl $0x9,0x10(%rsp)
  lock orl $0xa,0x10(%rsp)
  lock xorl $0xb,0x10(%rsp)
  lock addl $0x1,0x10(%rsp)
  lock subl $0x2,0x10(%rsp)
  lock andb $0x3,0xe(%rsp)
  lock orb $0x4,0xe(%rsp)
  movzbl 0xe(%rsp),%eax
L14:
  mov    %eax,%edx
  xor    $0x5,%edx
  lock cmpxchg %dl,0xe(%rsp)
  jne    L14
  xor    %ecx,%ecx
  mov    %rcx,%rax
  lock cmpxchg %rcx,0x0
  xor    %eax,%eax
  mov    $0x1,%esi
  lock cmpxchg %sil,0xf(%rsp)
  test   %al,%al
  movsbl %dl,%edx
  mov    $0x2a,%eax
  cmove  %edx,%eax
  mov    %eax,%r8d
  mov    %rcx,%rax
  lock cmpxchg %rcx,0x0
  movl   $0x0,0x10(%rsp)
  lock orq $0x0,(%rsp)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  mov    %r8d,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int atomic(void) {
  int old;
  int val = 1;
  char valc = 1;
  _Bool valb = 0;
  unsigned int uval = 1;
  int cmp = 0;
  old = __sync_fetch_and_add(&val, 1);
  old = __sync_fetch_and_sub(&valc, 2);
  old = __sync_fetch_and_min(&val, 3);
  old = __sync_fetch_and_max(&val, 4);
  old = __sync_fetch_and_umin(&uval, 5u);
  old = __sync_fetch_and_umax(&uval, 6u);
  old = __sync_lock_test_and_set(&val, 7);
  old = __sync_val_compare_and_swap(&val, 4, 1976);
  old = __sync_bool_compare_and_swap(&val, 4, 1976);
  old = __sync_fetch_and_and(&val, 0x9);
  old = __sync_fetch_and_or(&val, 0xa);
  old = __sync_fetch_and_xor(&val, 0xb);
  old = __sync_add_and_fetch(&val, 1);
  old = __sync_sub_and_fetch(&val, 2);
  old = __sync_and_and_fetch(&valc, 3);
  old = __sync_or_and_fetch(&valc, 4);
  old = __sync_xor_and_fetch(&valc, 5);
  __sync_val_compare_and_swap((void **)0, (void *)0, (void *)0);
  if (__sync_val_compare_and_swap(&valb, 0, 1)) {
    old = 42;
  }
  __sync_bool_compare_and_swap((void **)0, (void *)0, (void *)0);
  __sync_lock_release(&val);
  __sync_synchronize();
  return old;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gcd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
  mov    -0x8(%rbp),%eax
  jmp    L4
L2:
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <gcd>
L4:
  leave
  ret
```

assistant: ```c
int gcd(int p, int q) {
  if (p == 0)
    return q;
  gcd(q, p % q);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edi
  test   %eax,%eax
  jne    L1
  mov    %esi,%eax
  ret
L1:
  sub    $0x8,%rsp
  cltd
  idiv   %esi
  mov    %edx,%esi
  call   <gcd>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gcd(int p, int q) {
  if (p == 0)
    return q;
  gcd(q, p % q);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<gcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %edi,%edi
  je     L9
  nopl   0x0(%rax)
L19:
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  mov    %ecx,%eax
  test   %ecx,%ecx
  jne    L19
L9:
  mov    %edx,%eax
  ret
```

assistant: ```c
int gcd(int p, int q) {
  if (p == 0)
    return q;
  gcd(q, p % q);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<gcd>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edx
  test   %edi,%edi
  je     L2
  nopl   0x0(%rax)
L16:
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  mov    %ecx,%eax
  test   %ecx,%ecx
  jne    L16
L2:
  mov    %edx,%eax
  ret
```

assistant: ```c
int gcd(int p, int q) {
  if (p == 0)
    return q;
  gcd(q, p % q);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  je     L44
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L44:
  cmpl   $0x1,-0x14(%rbp)
  je     L29
  cmpl   $0x2,-0x14(%rbp)
  je     L2
  jmp    L38
L29:
  mov    -0x8(%rbp),%rax
  add    $0x5,%rax
  movb   $0x7,(%rax)
  jmp    L20
L2:
  mov    -0x8(%rbp),%rax
  add    $0xb,%rax
  movb   $0xf,(%rax)
  jmp    L20
L38:
  movq   $0x0,-0x8(%rbp)
  nop
L20:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *foo(int bar, char *baz) {
  char *tmp;
  if (bar & 3)
    tmp = baz;
  switch (bar) {
  case 1:
    tmp[5] = 7;
    break;
  case 2:
    tmp[11] = 15;
    break;
  default:
    tmp = 0;
    break;
  }
  return tmp;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  test   $0x3,%dil
  mov    $0x0,%eax
  cmovne %rsi,%rax
  cmp    $0x1,%edi
  je     L9
  cmp    $0x2,%edi
  je     L13
  mov    $0x0,%eax
  ret
L9:
  movb   $0x7,0x5(%rax)
  ret
L13:
  movb   $0xf,0xb(%rax)
  ret
```

assistant: ```c
char *foo(int bar, char *baz) {
  char *tmp;
  if (bar & 3)
    tmp = baz;
  switch (bar) {
  case 1:
    tmp[5] = 7;
    break;
  case 2:
    tmp[11] = 15;
    break;
  default:
    tmp = 0;
    break;
  }
  return tmp;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  xor    %eax,%eax
  test   $0x3,%dil
  cmovne %rsi,%rax
  cmp    $0x1,%edi
  je     L5
  cmp    $0x2,%edi
  jne    L9
  movb   $0xf,0xb(%rax)
  ret
  nopl   (%rax)
L9:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  movb   $0x7,0x5(%rax)
  ret
```

assistant: ```c
char *foo(int bar, char *baz) {
  char *tmp;
  if (bar & 3)
    tmp = baz;
  switch (bar) {
  case 1:
    tmp[5] = 7;
    break;
  case 2:
    tmp[11] = 15;
    break;
  default:
    tmp = 0;
    break;
  }
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  xor    %eax,%eax
  test   $0x3,%dil
  cmovne %rsi,%rax
  cmp    $0x1,%edi
  je     L9
  cmp    $0x2,%edi
  jne    L15
  movb   $0xf,0xb(%rax)
  ret
  nopl   (%rax)
L15:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  movb   $0x7,0x5(%rax)
  ret
```

assistant: ```c
char *foo(int bar, char *baz) {
  char *tmp;
  if (bar & 3)
    tmp = baz;
  switch (bar) {
  case 1:
    tmp[5] = 7;
    break;
  case 2:
    tmp[11] = 15;
    break;
  default:
    tmp = 0;
    break;
  }
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  D20(%rip),%xmm0
  pxor   %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  mov    $0x0,%ecx
  cltd
  idiv   %ecx
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    $0x0,%ecx
  cltd
  idiv   %ecx
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D20
datatype: f64
name: retrieve_data
arguments: tag: D7
datatype: string

tool: D20: 1

tool: D7: %lf


assistant: ```c
int main() {
  printf("%lf\n", 1.0 / 0.0);
  printf("%lf\n", 0.0 / 0.0);
  printf("%lf\n", 1 / 0);
  printf("%lf\n", 0 / 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  movsd  D17(%rip),%xmm0
  pxor   %xmm1,%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  lea    D14(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  cvttsd2si %xmm0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  mov    $0x0,%ecx
  cltd
  idiv   %ecx
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_label: D17
datatype: f64
name: parse_data
arguments: data_label: D14
datatype: string

tool: D17: 1

tool: D14: %lf


assistant: ```c
int main() {
  printf("%lf\n", 1.0 / 0.0);
  printf("%lf\n", 0.0 / 0.0);
  printf("%lf\n", 1 / 0);
  printf("%lf\n", 0 / 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  movsd  D9(%rip),%xmm0
  pxor   %xmm1,%xmm1
  push   %rbp
  lea    D11(%rip),%rbp
  mov    %rbp,%rdi
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  divsd  %xmm0,%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  ud2
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D9
type: f64
name: extract_symbolic_data
arguments: data_tag: D11
type: string

tool: D9: 1

tool: D11: %lf


assistant: ```c
int main() {
  printf("%lf\n", 1.0 / 0.0);
  printf("%lf\n", 0.0 / 0.0);
  printf("%lf\n", 1 / 0);
  printf("%lf\n", 0 / 0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  movsd  D9(%rip),%xmm0
  pxor   %xmm1,%xmm1
  push   %rbp
  lea    D19(%rip),%rbp
  mov    %rbp,%rdi
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  divsd  %xmm0,%xmm0
  cvttsd2si %xmm0,%esi
  call   <printf@plt>
  ud2
```

tool_calls: name: parse_data
arguments: data_tag: D9
type: f64
name: parse_data
arguments: data_tag: D19
type: string

tool: D9: 1

tool: D19: %lf


assistant: ```c
int main() {
  printf("%lf\n", 1.0 / 0.0);
  printf("%lf\n", 0.0 / 0.0);
  printf("%lf\n", 1 / 0);
  printf("%lf\n", 0 / 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<putc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <scif_putc@plt>
  cmpl   $0xa,-0x4(%rbp)
  jne    L5
  mov    $0xd,%edi
  call   <scif_putc@plt>
L5:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int putc(int c) {
  scif_putc(c);
  if (c == '\n') {
    scif_putc('\r');
  }
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<putc>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  movsbl %dil,%edi
  call   <scif_putc@plt>
  cmp    $0xa,%ebx
  je     L4
L10:
  mov    %ebx,%eax
  pop    %rbx
  ret
L4:
  mov    $0xd,%edi
  call   <scif_putc@plt>
  jmp    L10
```

assistant: ```c
int putc(int c) {
  scif_putc(c);
  if (c == '\n') {
    scif_putc('\r');
  }
  return c;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<putc>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  movsbl %dil,%edi
  call   <scif_putc@plt>
  cmp    $0xa,%r12d
  je     L1
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L1:
  mov    $0xd,%edi
  call   <scif_putc@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int putc(int c) {
  scif_putc(c);
  if (c == '\n') {
    scif_putc('\r');
  }
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<putc>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  movsbl %dil,%edi
  call   <scif_putc@plt>
  cmp    $0xa,%r12d
  je     L8
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L8:
  mov    $0xd,%edi
  call   <scif_putc@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int putc(int c) {
  scif_putc(c);
  if (c == '\n') {
    scif_putc('\r');
  }
  return c;
}
```
user: Translate the below assembly code into C programming language:

```asm
<progress>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movb   $0x42,-0x1(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  je     L23
  cmpl   $0x2,-0x14(%rbp)
  jg     L19
  cmpl   $0x0,-0x14(%rbp)
  je     L72
  cmpl   $0x1,-0x14(%rbp)
  je     L37
  jmp    L19
L72:
  movb   $0x2e,-0x1(%rbp)
  jmp    L71
L37:
  movb   $0x2b,-0x1(%rbp)
  jmp    L71
L23:
  movb   $0x2a,-0x1(%rbp)
  jmp    L71
L19:
  movb   $0xa,-0x1(%rbp)
  nop
L71:
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  movsbl -0x1(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <fputc@plt>
  nop
  leave
  ret
```

assistant: ```c
void progress(int p, int n, void *arg) {
  char c = 'B';
  switch (p) {
  case 0:
    c = '.';
    break;
  case 1:
    c = '+';
    break;
  case 2:
    c = '*';
    break;
  default:
    c = '\n';
    break;
  }
  fputc(c, stderr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<progress>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2b,%eax
  cmp    $0x1,%edi
  je     L5
  mov    $0x2a,%eax
  cmp    $0x2,%edi
  je     L5
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0x24,%eax
  add    $0xa,%eax
L5:
  movsbl %al,%edi
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  call   <fputc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void progress(int p, int n, void *arg) {
  char c = 'B';
  switch (p) {
  case 0:
    c = '.';
    break;
  case 1:
    c = '+';
    break;
  case 2:
    c = '*';
    break;
  default:
    c = '\n';
    break;
  }
  fputc(c, stderr);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  je     L7
  mov    $0x0,%eax
  call   <abort@plt>
L7:
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void ok(int i) {
  if (i != 1)
    abort();
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ok>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jne    L17
L12:
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
L17:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L12
```

assistant: ```c
void ok(int i) {
  if (i != 1)
    abort();
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ok>:
  endbr64
  cmp    $0x1,%edi
  jne    L2
  xor    %edi,%edi
  jmp    1070 <exit@plt>
L2:
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <abort@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void ok(int i) {
  if (i != 1)
    abort();
  exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ok>:
  endbr64
  cmp    $0x1,%edi
  jne    L6
  xor    %edi,%edi
  jmp    1070 <exit@plt>
L6:
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <abort@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void ok(int i) {
  if (i != 1)
    abort();
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<DrawTile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,%eax
  mov    %al,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <move@plt>
  movsbl -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <addch@plt>
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  lea    0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <move@plt>
  movsbl -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <addch@plt>
  nop
  leave
  ret
```

assistant: ```c
void DrawTile(int y, int x, char tile) {
  move((int)y, (int)(x * 2));
  addch(tile);
  move((int)y, (int)(x * 2) + 1);
  addch(tile);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<DrawTile>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %edx,%ebx
  lea    (%rsi,%rsi,1),%ebp
  mov    %ebp,%esi
  call   <move@plt>
  movsbl %bl,%ebx
  mov    %ebx,%edi
  call   <addch@plt>
  lea    0x1(%rbp),%esi
  mov    %r12d,%edi
  call   <move@plt>
  mov    %ebx,%edi
  call   <addch@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void DrawTile(int y, int x, char tile) {
  move((int)y, (int)(x * 2));
  addch(tile);
  move((int)y, (int)(x * 2) + 1);
  addch(tile);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<DrawTile>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  lea    (%rsi,%rsi,1),%ebx
  movsbl %bpl,%ebp
  mov    %ebx,%esi
  call   <move@plt>
  mov    %ebp,%edi
  call   <addch@plt>
  mov    %r12d,%edi
  lea    0x1(%rbx),%esi
  call   <move@plt>
  pop    %rbx
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <addch@plt>
```

assistant: ```c
void DrawTile(int y, int x, char tile) {
  move((int)y, (int)(x * 2));
  addch(tile);
  move((int)y, (int)(x * 2) + 1);
  addch(tile);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<DrawTile>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  lea    (%rsi,%rsi,1),%ebx
  movsbl %bpl,%ebp
  mov    %ebx,%esi
  call   <move@plt>
  mov    %ebp,%edi
  call   <addch@plt>
  mov    %r12d,%edi
  lea    0x1(%rbx),%esi
  call   <move@plt>
  pop    %rbx
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <addch@plt>
```

assistant: ```c
void DrawTile(int y, int x, char tile) {
  move((int)y, (int)(x * 2));
  addch(tile);
  move((int)y, (int)(x * 2) + 1);
  addch(tile);
}
```
user: What is the c source code of the assembly code below?

```asm
<blur_matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %edx,%r15d
  mov    %ecx,%ebx
  cmp    %rsi,%rdi
  je     L1
  mov    %edx,%r13d
  imul   %ecx,%r13d
  shl    $0x3,%r13d
  mov    %r13d,%edx
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  call   <memcpy@plt>
  sub    $0x1,%r15d
  mov    %r12,0x8(%rsp)
  cmp    $0x1,%r15d
  jle    L71
L32:
  mov    %rbp,%r8
  mov    0x8(%rsp),%rax
  lea    0x8(%rax),%r10
  lea    -0x3(%rbx),%eax
  lea    0x10(,%rax,8),%r11
  mov    $0x1,%r14d
  movsd  D15(%rip),%xmm6
  movsd  D19(%rip),%xmm5
  movsd  D24(%rip),%xmm4
  jmp    L45
L1:
  mov    %ecx,%esi
  mov    %edx,%edi
  call   <new_matrix2@plt>
  mov    %rax,0x8(%rsp)
  mov    %r15d,%r13d
  imul   %ebx,%r13d
  shl    $0x3,%r13d
  mov    %r13d,%edx
  mov    0x0(%rbp),%rsi
  mov    (%rax),%rdi
  call   <memcpy@plt>
  sub    $0x1,%r15d
  cmp    $0x1,%r15d
  jg     L32
L56:
  mov    %r13d,%edx
  mov    0x8(%rsp),%rbx
  mov    (%rbx),%rsi
  mov    (%r12),%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdi
  call   <free_matrix2@plt>
  jmp    L71
L78:
  mov    0x8(%r8),%rcx
  movapd %xmm6,%xmm3
  mulsd  (%rcx,%rax,1),%xmm3
  mov    0x10(%r8),%r9
  mov    (%r8),%rdi
  movsd  (%r9,%rax,1),%xmm1
  addsd  (%rdi,%rax,1),%xmm1
  mov    %rax,%rdx
  add    $0x8,%rax
  mov    (%r10),%rsi
  movsd  (%r9,%rax,1),%xmm0
  addsd  -0x8(%r9,%rdx,1),%xmm0
  addsd  (%rdi,%rax,1),%xmm0
  addsd  -0x8(%rdi,%rdx,1),%xmm0
  mulsd  %xmm5,%xmm0
  addsd  (%rcx,%rax,1),%xmm1
  addsd  -0x8(%rcx,%rdx,1),%xmm1
  mulsd  %xmm4,%xmm1
  addsd  %xmm3,%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi,%rdx,1)
  cmp    %rax,%r11
  jne    L78
L69:
  add    $0x1,%r14d
  add    $0x8,%r8
  add    $0x8,%r10
  cmp    %r15d,%r14d
  jge    L70
L45:
  mov    $0x8,%eax
  cmp    $0x2,%ebx
  jg     L78
  jmp    L69
L70:
  cmp    %rbp,%r12
  je     L56
L71:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D15
type: byte[8]
name: parse_data
arguments: data_tag: D19
type: byte[8]
name: parse_data
arguments: data_tag: D24
type: byte[8]

tool: {"D15": [0xa8,0x35,0xcd,0x3b,0x4e,0xd1,0xe3,0x3f]}

tool: {"D19": [0x4d,0xf3,0x8e,0x53,0x74,0x24,0x87,0x3f]}

tool: {"D24": [0xf9,0xa0,0x67,0xb3,0xea,0x73,0xb5,0x3f]}

assistant: ```c
void blur_matrix(double **dst, double **src, int n, int m) {
  double G[3] = {.6193, .0838, .0113};
  double **I = (dst == src ? new_matrix2(n, m) : dst);
  memcpy(I[0], src[0], n * m * sizeof(double));
  int i, j;
  for (i = 1; i < n - 1; i++)
    for (j = 1; j < m - 1; j++)
      I[i][j] = G[0] * src[i][j] + G[1] * (src[i + 1][j] + src[i - 1][j] + src[i][j + 1] + src[i][j - 1]) + G[2] * (src[i + 1][j + 1] + src[i + 1][j - 1] + src[i - 1][j + 1] + src[i - 1][j - 1]);
  if (dst == src) {
    memcpy(dst[0], I[0], n * m * sizeof(double));
    free_matrix2(I);
  }
}
```