ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dtype_kind>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbx,%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <dtype_normalize@plt>
  movzbl 0x1(%rbx),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char dtype_kind(const char *dtype) {
  char ndtype[8];
  dtype_normalize(ndtype, dtype);
  return dtype[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_best_eci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x3,-0x2c(%rbp)
  mov    -0x3c(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L16:
  cmp    %rdx,%rsp
  je     L12
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L16
L12:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L15
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L15:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
L40:
  lea    -0x3c(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x38(%rbp),%rsi
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <utf_to_eci@plt>
  test   %rax,%rax
  jne    L27
  mov    -0x2c(%rbp),%eax
  jmp    L53
L27:
  addl   $0x1,-0x2c(%rbp)
  cmpl   $0x18,-0x2c(%rbp)
  jle    L40
  mov    $0x1a,%eax
L53:
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int get_best_eci(unsigned char source[], int length) {
  int eci = 3;
  unsigned char local_source[length + 1];
  do {
    if (utf_to_eci(eci, source, local_source, &length) == 0) {
      return eci;
    }
    eci++;
  } while (eci < 25);
  return 26;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_best_eci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%r12
  mov    %esi,%eax
  mov    %esi,-0x34(%rbp)
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x28(%rbp)
  xor    %edx,%edx
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L21:
  cmp    %rdx,%rsp
  je     L59
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L21
L59:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L30
  orq    $0x0,-0x8(%rsp,%rax,1)
L30:
  mov    %rsp,%r14
  mov    $0x3,%ebx
  lea    -0x34(%rbp),%r13
L57:
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <utf_to_eci@plt>
  test   %rax,%rax
  je     L53
  add    $0x1,%ebx
  cmp    $0x19,%ebx
  jne    L57
  mov    $0x1a,%ebx
L53:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    %ebx,%eax
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_best_eci(unsigned char source[], int length) {
  int eci = 3;
  unsigned char local_source[length + 1];
  do {
    if (utf_to_eci(eci, source, local_source, &length) == 0) {
      return eci;
    }
    eci++;
  } while (eci < 25);
  return 26;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_best_eci>:
  endbr64
  push   %rbp
  mov    %esi,%eax
  add    $0x1,%eax
  cltq
  mov    %rsp,%rbp
  push   %r14
  add    $0xf,%rax
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %esi,-0x34(%rbp)
  mov    %rsp,%rcx
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x28(%rbp)
  xor    %edx,%edx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L44
L62:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L62
L44:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L55
L58:
  mov    %rsp,%r14
  mov    $0x3,%r12d
  lea    -0x34(%rbp),%r13
  jmp    L9
  nopl   (%rax)
L23:
  add    $0x1,%r12d
  cmp    $0x19,%r12d
  je     L40
L9:
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <utf_to_eci@plt>
  test   %rax,%rax
  jne    L23
L4:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L74
  lea    -0x20(%rbp),%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nop
L40:
  mov    $0x1a,%r12d
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L55:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L58
L74:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_best_eci(unsigned char source[], int length) {
  int eci = 3;
  unsigned char local_source[length + 1];
  do {
    if (utf_to_eci(eci, source, local_source, &length) == 0) {
      return eci;
    }
    eci++;
  } while (eci < 25);
  return 26;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_best_eci>:
  endbr64
  push   %rbp
  mov    %esi,%eax
  add    $0x1,%eax
  cltq
  mov    %rsp,%rbp
  push   %r14
  add    $0xf,%rax
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %esi,-0x34(%rbp)
  mov    %rsp,%rcx
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x28(%rbp)
  xor    %edx,%edx
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L8
L38:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L38
L8:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L28
L26:
  mov    %rsp,%r14
  mov    $0x3,%r12d
  lea    -0x34(%rbp),%r13
  jmp    L56
  nopl   (%rax)
L49:
  add    $0x1,%r12d
  cmp    $0x19,%r12d
  je     L87
L56:
  mov    %r13,%rcx
  mov    %r14,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <utf_to_eci@plt>
  test   %rax,%rax
  jne    L49
L47:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  lea    -0x20(%rbp),%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nop
L87:
  mov    $0x1a,%r12d
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L28:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L26
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_best_eci(unsigned char source[], int length) {
  int eci = 3;
  unsigned char local_source[length + 1];
  do {
    if (utf_to_eci(eci, source, local_source, &length) == 0) {
      return eci;
    }
    eci++;
  } while (eci < 25);
  return 26;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ximage_convert_bgr233>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L23:
  movl   $0x0,-0x8(%rbp)
  jmp    L24
L0:
  mov    -0x8(%rbp),%eax
  shl    $0x2,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xb(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xa(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x2,%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  and    $0xffffffc0,%eax
  mov    %eax,%edx
  movzbl -0xa(%rbp),%eax
  shr    $0x2,%al
  and    $0x38,%eax
  or     %eax,%edx
  movzbl -0xb(%rbp),%eax
  shr    $0x5,%al
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L0
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  add    %rax,-0x28(%rbp)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L23
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ximage_convert_bgr233(const unsigned char *__restrict src, int srcstride, unsigned char *__restrict dst, int dststride, int w, int h) {
  unsigned char r, g, b;
  int x, y;
  for (y = 0; y < h; y++) {
    for (x = 0; x < w; x++) {
      r = src[4 * x + 0];
      g = src[4 * x + 1];
      b = src[4 * x + 2];
      dst[x] = (b & 0xC0) | ((g >> 2) & 0x38) | ((r >> 5) & 0x7);
    }
    src += srcstride;
    dst += dststride;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<ximage_convert_bgr233>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %esi,%eax
  mov    %rdx,%rsi
  mov    %r8d,%ebx
  mov    %r9d,%ebp
  mov    $0x0,%r11d
  mov    %r8d,%r10d
  movslq %eax,%r9
  movslq %ecx,%r8
  test   %ebp,%ebp
  jg     L26
L17:
  pop    %rbx
  pop    %rbp
  ret
L1:
  movzbl 0x1(%rdi,%rdx,4),%eax
  shr    $0x2,%al
  and    $0x38,%eax
  movzbl 0x2(%rdi,%rdx,4),%ecx
  and    $0xffffffc0,%ecx
  or     %ecx,%eax
  movzbl (%rdi,%rdx,4),%ecx
  shr    $0x5,%cl
  or     %ecx,%eax
  mov    %al,(%rsi,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdx,%r10
  jne    L1
L21:
  add    %r9,%rdi
  add    %r8,%rsi
  add    $0x1,%r11d
  cmp    %r11d,%ebp
  je     L17
L26:
  mov    $0x0,%edx
  test   %ebx,%ebx
  jg     L1
  jmp    L21
```

assistant: ```c
void ximage_convert_bgr233(const unsigned char *__restrict src, int srcstride, unsigned char *__restrict dst, int dststride, int w, int h) {
  unsigned char r, g, b;
  int x, y;
  for (y = 0; y < h; y++) {
    for (x = 0; x < w; x++) {
      r = src[4 * x + 0];
      g = src[4 * x + 1];
      b = src[4 * x + 2];
      dst[x] = (b & 0xC0) | ((g >> 2) & 0x38) | ((r >> 5) & 0x7);
    }
    src += srcstride;
    dst += dststride;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ximage_convert_bgr233>:
  endbr64
  test   %r9d,%r9d
  jle    L32
  push   %rbp
  mov    %rdx,%r10
  mov    %r9d,%ebp
  xor    %r11d,%r11d
  push   %rbx
  mov    %r8d,%ebx
  movslq %esi,%r9
  movslq %ecx,%r8
  movslq %ebx,%rsi
L7:
  xor    %edx,%edx
  test   %ebx,%ebx
  jle    L14
  cs nopw 0x0(%rax,%rax,1)
L21:
  movzbl 0x1(%rdi,%rdx,4),%eax
  movzbl 0x2(%rdi,%rdx,4),%ecx
  shr    $0x2,%al
  and    $0xffffffc0,%ecx
  and    $0x38,%eax
  or     %ecx,%eax
  movzbl (%rdi,%rdx,4),%ecx
  shr    $0x5,%cl
  or     %ecx,%eax
  mov    %al,(%r10,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L21
L14:
  add    $0x1,%r11d
  add    %r9,%rdi
  add    %r8,%r10
  cmp    %r11d,%ebp
  jne    L7
  pop    %rbx
  pop    %rbp
  ret
L32:
  ret
```

assistant: ```c
void ximage_convert_bgr233(const unsigned char *__restrict src, int srcstride, unsigned char *__restrict dst, int dststride, int w, int h) {
  unsigned char r, g, b;
  int x, y;
  for (y = 0; y < h; y++) {
    for (x = 0; x < w; x++) {
      r = src[4 * x + 0];
      g = src[4 * x + 1];
      b = src[4 * x + 2];
      dst[x] = (b & 0xC0) | ((g >> 2) & 0x38) | ((r >> 5) & 0x7);
    }
    src += srcstride;
    dst += dststride;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<is_odd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L2
  mov    $0xffffffff,%eax
  jmp    L19
L2:
  mov    $0x0,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int is_odd(int *val) {
  if (*val % 2 == 0)
    return -1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_odd>:
  endbr64
  mov    (%rdi),%eax
  and    $0x1,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int is_odd(int *val) {
  if (*val % 2 == 0)
    return -1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_odd>:
  endbr64
  mov    (%rdi),%eax
  and    $0x1,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int is_odd(int *val) {
  if (*val % 2 == 0)
    return -1;
  else
    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
<is_odd>:
  endbr64
  mov    (%rdi),%eax
  and    $0x1,%eax
  sub    $0x1,%eax
  ret
```

assistant: ```c
int is_odd(int *val) {
  if (*val % 2 == 0)
    return -1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<lcm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L17
L1:
  mov    -0x14(%rbp),%eax
  add    %eax,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L1
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int lcm(int a, int b) {
  int m = a;
  while (m % b) {
    m += a;
  }
  return m;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lcm>:
  endbr64
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L20
  mov    %edi,%ecx
L0:
  add    %edi,%ecx
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L0
L16:
  mov    %ecx,%eax
  ret
L20:
  mov    %edi,%ecx
  jmp    L16
```

assistant: ```c
int lcm(int a, int b) {
  int m = a;
  while (m % b) {
    m += a;
  }
  return m;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lcm>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%r8d
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L17
L3:
  add    %edi,%r8d
  mov    %r8d,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L3
L17:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int lcm(int a, int b) {
  int m = a;
  while (m % b) {
    m += a;
  }
  return m;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lcm>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%r8d
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L15
L16:
  add    %edi,%r8d
  mov    %r8d,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L16
L15:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int lcm(int a, int b) {
  int m = a;
  while (m % b) {
    m += a;
  }
  return m;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x2c,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *get_name(char **line) {
  char *name = *line;
  char *p = strchr(name, ',');
  *p = '\0';
  *line = p + 1;
  return name;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_name>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    (%rdi),%rbp
  mov    $0x2c,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  movb   $0x0,(%rax)
  add    $0x1,%rax
  mov    %rax,(%rbx)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *get_name(char **line) {
  char *name = *line;
  char *p = strchr(name, ',');
  *p = '\0';
  *line = p + 1;
  return name;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_name>:
  endbr64
  push   %r12
  mov    $0x2c,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  mov    %r12,%rdi
  call   <strchr@plt>
  movb   $0x0,(%rax)
  add    $0x1,%rax
  mov    %rax,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_name(char **line) {
  char *name = *line;
  char *p = strchr(name, ',');
  *p = '\0';
  *line = p + 1;
  return name;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_name>:
  endbr64
  push   %r12
  mov    $0x2c,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  mov    %r12,%rdi
  call   <strchr@plt>
  movb   $0x0,(%rax)
  add    $0x1,%rax
  mov    %rax,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_name(char **line) {
  char *name = *line;
  char *p = strchr(name, ',');
  *p = '\0';
  *line = p + 1;
  return name;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0xc(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  jbe    L6
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L26
L6:
  mov    -0xc(%rbp),%eax
  mov    $0x2,%edi
  movd   %eax,%xmm0
  call   <pow@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L26:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_infered: string
name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D27
type_infered: string
name: retrieve_data
arguments: data_tag: D4
type_infered: string

tool: D2: Digite um numero: 

tool: D7: %f

tool: D27: Raiz quadrada = %.2f

tool: D4: Quadrado = %.2f

assistant: ```c
int main() {
  float num;
  printf("Digite um numero: ");
  scanf("%f", &num);
  if (num > 0) {
    printf("Raiz quadrada = %.2f", sqrt(num));
  } else {
    printf("Quadrado = %.2f", pow(num, 2));
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  comiss D8(%rip),%xmm0
  jbe    L14
  call   <sqrt@plt>
  lea    D49(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L29:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L14:
  mov    $0x2,%edi
  call   <pow@plt>
  lea    D6(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L29
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D21
data_type: string
name: extract_symbolic_data
arguments: tag: D30
data_type: string
name: extract_symbolic_data
arguments: tag: D8
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D49
data_type: string
name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D21: Digite um numero: 

tool: D30: %f

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

tool: D49: Raiz quadrada = %.2f

tool: D6: Quadrado = %.2f

assistant: ```c
int main() {
  float num;
  printf("Digite um numero: ");
  scanf("%f", &num);
  if (num > 0) {
    printf("Raiz quadrada = %.2f", sqrt(num));
  } else {
    printf("Quadrado = %.2f", pow(num, 2));
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D46(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  comiss D33(%rip),%xmm0
  jbe    L2
  call   <sqrt@plt>
  lea    D32(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  mov    $0x2,%edi
  call   <pow@plt>
  lea    D45(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L8
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D46
type_guessed: string
name: unpack_data
arguments: tag: D13
type_guessed: string
name: unpack_data
arguments: tag: D33
type_guessed: byte[4]
name: unpack_data
arguments: tag: D32
type_guessed: string
name: unpack_data
arguments: tag: D45
type_guessed: string

tool: D46: Digite um numero: 

tool: D13: %f

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

tool: D32: Raiz quadrada = %.2f

tool: D45: Quadrado = %.2f

assistant: ```c
int main() {
  float num;
  printf("Digite um numero: ");
  scanf("%f", &num);
  if (num > 0) {
    printf("Raiz quadrada = %.2f", sqrt(num));
  } else {
    printf("Quadrado = %.2f", pow(num, 2));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  comiss D43(%rip),%xmm0
  jbe    L13
  call   <sqrt@plt>
  lea    D4(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L13:
  mov    $0x2,%edi
  call   <pow@plt>
  lea    D48(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  jmp    L23
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: string
name: unpack_data
arguments: data_tag: D30
type: string
name: unpack_data
arguments: data_tag: D43
type: byte[4]
name: unpack_data
arguments: data_tag: D4
type: string
name: unpack_data
arguments: data_tag: D48
type: string

tool: D5: Digite um numero: 

tool: D30: %f

tool: {"D43": [0x00,0x00,0x00,0x00]}

tool: D4: Raiz quadrada = %.2f

tool: D48: Quadrado = %.2f

assistant: ```c
int main() {
  float num;
  printf("Digite um numero: ");
  scanf("%f", &num);
  if (num > 0) {
    printf("Raiz quadrada = %.2f", sqrt(num));
  } else {
    printf("Quadrado = %.2f", pow(num, 2));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<string_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L6
L13:
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long string_length(const char *s) {
  unsigned long i = 0;
  for (; *s != '\0'; ++i) {
    ;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L15
L3:
  jmp    L3
L15:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned long string_length(const char *s) {
  unsigned long i = 0;
  for (; *s != '\0'; ++i) {
    ;
  }
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L4
L11:
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long string_length(const char *s) {
  unsigned long i = 0;
  for (; *s != '\0'; ++i) {
    ;
  }
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<string_length>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L1
L5:
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long string_length(const char *s) {
  unsigned long i = 0;
  for (; *s != '\0'; ++i) {
    ;
  }
  return i;
}
```