ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<am2315_crc16>:
  endbr64
  mov    %rdi,%rcx
  test   %sil,%sil
  je     L7
  movzbl %sil,%esi
  mov    $0xffffffff,%r8d
  lea    (%rdi,%rsi,1),%rdi
  xchg   %ax,%ax
L10:
  movzbl (%rcx),%edx
  add    $0x1,%rcx
  xor    %r8d,%edx
  mov    %edx,%eax
  shr    %ax
  mov    %eax,%esi
  xor    $0xa001,%si
  and    $0x1,%edx
  cmovne %esi,%eax
  mov    %eax,%edx
  shr    %dx
  mov    %edx,%esi
  xor    $0xa001,%si
  test   $0x1,%al
  cmovne %esi,%edx
  mov    %edx,%eax
  shr    %ax
  mov    %eax,%esi
  xor    $0xa001,%si
  and    $0x1,%edx
  cmovne %esi,%eax
  mov    %eax,%edx
  shr    %dx
  mov    %edx,%esi
  xor    $0xa001,%si
  test   $0x1,%al
  cmovne %esi,%edx
  mov    %edx,%eax
  shr    %ax
  mov    %eax,%esi
  xor    $0xa001,%si
  and    $0x1,%edx
  cmovne %esi,%eax
  mov    %eax,%edx
  shr    %dx
  mov    %edx,%esi
  xor    $0xa001,%si
  test   $0x1,%al
  cmovne %esi,%edx
  mov    %edx,%eax
  shr    %ax
  mov    %eax,%esi
  xor    $0xa001,%si
  and    $0x1,%edx
  cmovne %esi,%eax
  mov    %eax,%r8d
  shr    %r8w
  mov    %r8d,%edx
  xor    $0xa001,%dx
  test   $0x1,%al
  cmovne %edx,%r8d
  cmp    %rdi,%rcx
  jne    L10
  mov    %r8d,%eax
  ret
L7:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
uint16_t am2315_crc16(unsigned char *ptr, unsigned char len) {
  unsigned short crc = 0xFFFF;
  unsigned char i;
  while (len--) {
    crc ^= *ptr++;
    for (i = 0; i < 8; i++) {
      if (crc & 0x01) {
        crc >>= 1;
        crc ^= 0xA001;
      } else {
        crc >>= 1;
      }
    }
  }
  return crc;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<utils_safe_read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
L6:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L13
  mov    D16(%rip),%rax
  mov    (%rax),%rdx
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L6
L13:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
ssize_t utils_safe_read(int fd, void *buf, size_t count) {
  ssize_t n;
  do {
    n = read(fd, buf, count);
  } while (n < 0 && errno == EINTR);
  return n;
}
```
user: What is the c source code of the assembly code below?

```asm
<utils_safe_read>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%rbp
  mov    %rdx,%rbx
L17:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  jns    L2
  mov    D4(%rip),%rcx
  mov    D7(%rip),%rdx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L17
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t utils_safe_read(int fd, void *buf, size_t count) {
  ssize_t n;
  do {
    n = read(fd, buf, count);
  } while (n < 0 && errno == EINTR);
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<utils_safe_read>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  jmp    L22
  nopl   0x0(%rax,%rax,1)
L5:
  mov    D17(%rip),%rdx
  mov    D14(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L1
L22:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  js     L5
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t utils_safe_read(int fd, void *buf, size_t count) {
  ssize_t n;
  do {
    n = read(fd, buf, count);
  } while (n < 0 && errno == EINTR);
  return n;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<utils_safe_read>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L14:
  mov    D8(%rip),%rdx
  mov    D5(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L13
L15:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %rax,%rax
  js     L14
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
ssize_t utils_safe_read(int fd, void *buf, size_t count) {
  ssize_t n;
  do {
    n = read(fd, buf, count);
  } while (n < 0 && errno == EINTR);
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ends_with>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L9
  mov    $0x0,%eax
  jmp    L15
L9:
  mov    -0xc(%rbp),%eax
  cltq
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  sub    %rdx,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcasecmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L15:
  leave
  ret
```

assistant: ```c
int ends_with(const char *haystack, const char *needle) {
  const char *end;
  int nlen = strlen(needle);
  int hlen = strlen(haystack);
  if (nlen > hlen)
    return 0;
  end = haystack + hlen - nlen;
  return (strcasecmp(end, needle) ? 0 : 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ends_with>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    $0x0,%edx
  cmp    %eax,%r12d
  jle    L7
L12:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  cltq
  movslq %r12d,%r12
  sub    %r12,%rax
  lea    (%rbx,%rax,1),%rdi
  mov    %rbp,%rsi
  call   <strcasecmp@plt>
  test   %rax,%rax
  sete   %dl
  movzbl %dl,%edx
  jmp    L12
```

assistant: ```c
int ends_with(const char *haystack, const char *needle) {
  const char *end;
  int nlen = strlen(needle);
  int hlen = strlen(haystack);
  if (nlen > hlen)
    return 0;
  end = haystack + hlen - nlen;
  return (strcasecmp(end, needle) ? 0 : 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ends_with>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rbx,%rdi
  movslq %eax,%rbp
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %eax,%ebp
  jg     L4
  cltq
  mov    %r12,%rsi
  sub    %rbp,%rax
  lea    (%rbx,%rax,1),%rdi
  call   <strcasecmp@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  sete   %r8b
L4:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int ends_with(const char *haystack, const char *needle) {
  const char *end;
  int nlen = strlen(needle);
  int hlen = strlen(haystack);
  if (nlen > hlen)
    return 0;
  end = haystack + hlen - nlen;
  return (strcasecmp(end, needle) ? 0 : 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ends_with>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rbx,%rdi
  movslq %eax,%rbp
  call   <strlen@plt>
  xor    %r8d,%r8d
  cmp    %eax,%ebp
  jg     L3
  cltq
  mov    %r12,%rsi
  sub    %rbp,%rax
  lea    (%rbx,%rax,1),%rdi
  call   <strcasecmp@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  sete   %r8b
L3:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int ends_with(const char *haystack, const char *needle) {
  const char *end;
  int nlen = strlen(needle);
  int hlen = strlen(haystack);
  if (nlen > hlen)
    return 0;
  end = haystack + hlen - nlen;
  return (strcasecmp(end, needle) ? 0 : 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x40(%rbp)
  movl   $0x2,-0x3c(%rbp)
  movl   $0x3,-0x38(%rbp)
  movl   $0x4,-0x34(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %eax,-0x4c(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x50(%rbp)
  jmp    L16
L25:
  lea    -0x40(%rbp),%rdx
  mov    -0x50(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    -0x50(%rbp),%eax
  cltq
  mov    %rdx,-0x30(%rbp,%rax,8)
  mov    -0x50(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x50(%rbp)
L16:
  cmpl   $0x3,-0x50(%rbp)
  jle    L25
  mov    -0x48(%rbp),%rax
  lea    0x10(%rax),%rcx
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdx
  mov    %rcx,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  add    $0x8,%rax
  lea    -0x30(%rbp),%rdx
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x48(%rbp),%rax
  mov    0x8(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
data_type: string
name: unpack_data
arguments: label: D0
data_type: string
name: unpack_data
arguments: label: D22
data_type: string
name: unpack_data
arguments: label: D37
data_type: string
name: unpack_data
arguments: label: D46
data_type: string

tool: D8: Array c:

tool: D0: %-3d 

tool: D22: 
B++ Move %d %d bytes


tool: D37: 
A++ Move %d %d bytes


tool: D46: 
*a[1] = %d
(*b)[2]=%d

assistant: ```c
void main() {
  int c[4] = {1, 2, 3, 4};
  int i;
  int *a[4];
  int(*b)[4];
  int m = c;
  b = &c;
  printf("Array c:\n");
  for (i = 0; i < 4; i++) {
    a[i] = &c[i];
    printf("%-3d ", *a[i]);
  }
  printf("\nB++ Move %d %d bytes\n", b + 1, b);
  printf("\nA++ Move %d %d bytes\n", a + 1, a);
  printf("\n*a[1] = %d\n(*b)[2]=%d", *a[1], (*b)[2]);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x10(%rsp),%rbp
  mov    %rbp,%r13
  lea    D21(%rip),%r12
L5:
  mov    %rbx,0x0(%rbp)
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  add    $0x8,%rbp
  cmp    %r13,%rbx
  jne    L5
  mov    %rsp,%rdx
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdx
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%edx
  mov    0x18(%rsp),%rax
  mov    (%rax),%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D35
type_guessed: string
name: read_data
arguments: data_tag: D21
type_guessed: string
name: read_data
arguments: data_tag: D17
type_guessed: string
name: read_data
arguments: data_tag: D28
type_guessed: string
name: read_data
arguments: data_tag: D4
type_guessed: string

tool: D35: Array c:

tool: D21: %-3d 

tool: D17: 
B++ Move %d %d bytes


tool: D28: 
A++ Move %d %d bytes


tool: D4: 
*a[1] = %d
(*b)[2]=%d

assistant: ```c
void main() {
  int c[4] = {1, 2, 3, 4};
  int i;
  int *a[4];
  int(*b)[4];
  int m = c;
  b = &c;
  printf("Array c:\n");
  for (i = 0; i < 4; i++) {
    a[i] = &c[i];
    printf("%-3d ", *a[i]);
  }
  printf("\nB++ Move %d %d bytes\n", b + 1, b);
  printf("\nA++ Move %d %d bytes\n", a + 1, a);
  printf("\n*a[1] = %d\n(*b)[2]=%d", *a[1], (*b)[2]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D38(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  lea    D32(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x200000001,%rax
  lea    0x10(%rsp),%rbp
  mov    %rsp,%r15
  mov    %r15,%rbx
  mov    %rbp,%r14
  mov    %rbp,%r13
  mov    %rax,(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x8(%rsp)
  call   <puts@plt>
L9:
  mov    (%rbx),%esi
  mov    %rbx,0x0(%rbp)
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  add    $0x8,%rbp
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L9
  mov    %r15,%rdx
  mov    %r14,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %r14,%rdx
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  mov    0x8(%rsp),%edx
  lea    D45(%rip),%rdi
  mov    (%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D38
type_infered: string
name: disassemble_data
arguments: label: D32
type_infered: string
name: disassemble_data
arguments: label: D14
type_infered: string
name: disassemble_data
arguments: label: D22
type_infered: string
name: disassemble_data
arguments: label: D45
type_infered: string

tool: D38: Array c:

tool: D32: %-3d 

tool: D14: 
B++ Move %d %d bytes


tool: D22: 
A++ Move %d %d bytes


tool: D45: 
*a[1] = %d
(*b)[2]=%d

assistant: ```c
void main() {
  int c[4] = {1, 2, 3, 4};
  int i;
  int *a[4];
  int(*b)[4];
  int m = c;
  b = &c;
  printf("Array c:\n");
  for (i = 0; i < 4; i++) {
    a[i] = &c[i];
    printf("%-3d ", *a[i]);
  }
  printf("\nB++ Move %d %d bytes\n", b + 1, b);
  printf("\nA++ Move %d %d bytes\n", a + 1, a);
  printf("\n*a[1] = %d\n(*b)[2]=%d", *a[1], (*b)[2]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D26(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  lea    D58(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  movdqa D7(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rsp,%r15
  movaps %xmm0,(%rsp)
  mov    %r15,%rbx
  mov    %rbp,%r14
  mov    %rbp,%r13
  call   <puts@plt>
L17:
  mov    (%rbx),%esi
  mov    %rbx,0x0(%rbp)
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  add    $0x8,%rbp
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L17
  mov    %r15,%rdx
  mov    %r14,%rsi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x18(%rsp),%rsi
  mov    %r14,%rdx
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  mov    0x8(%rsp),%edx
  lea    D10(%rip),%rdi
  mov    (%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D26
datatype: string
name: unpack_data
arguments: tag: D58
datatype: string
name: unpack_data
arguments: tag: D7
datatype: byte[16]
name: unpack_data
arguments: tag: D0
datatype: string
name: unpack_data
arguments: tag: D17
datatype: string
name: unpack_data
arguments: tag: D10
datatype: string

tool: D26: Array c:

tool: D58: %-3d 

tool: {"D7": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D0: 
B++ Move %d %d bytes


tool: D17: 
A++ Move %d %d bytes


tool: D10: 
*a[1] = %d
(*b)[2]=%d

assistant: ```c
void main() {
  int c[4] = {1, 2, 3, 4};
  int i;
  int *a[4];
  int(*b)[4];
  int m = c;
  b = &c;
  printf("Array c:\n");
  for (i = 0; i < 4; i++) {
    a[i] = &c[i];
    printf("%-3d ", *a[i]);
  }
  printf("\nB++ Move %d %d bytes\n", b + 1, b);
  printf("\nA++ Move %d %d bytes\n", a + 1, a);
  printf("\n*a[1] = %d\n(*b)[2]=%d", *a[1], (*b)[2]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sendAndRecv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <send@plt>
  cmp    $0xffffffff,%eax
  jne    L33
  mov    $0x0,%edi
  call   <exit@plt>
  pxor   %xmm0,%xmm0
  jmp    L35
L33:
  mov    $0x0,%eax
  call   <getWallTime@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <recvUntil@plt>
  test   %eax,%eax
  jne    L19
  mov    $0x0,%edi
  call   <exit@plt>
  pxor   %xmm0,%xmm0
  jmp    L35
L19:
  mov    $0x0,%eax
  call   <getWallTime@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
L35:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double sendAndRecv(int sockfd, char *text, char *response) {
  double rtt;
  if (send(sockfd, text, strlen(text), 0) == -1) {
    exit(0);
    return 0;
  }
  rtt = getWallTime();
  if (!recvUntil(sockfd, response, '\n')) {
    exit(0);
    return 0;
  }
  rtt = getWallTime() - rtt;
  return rtt;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sendAndRecv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%ecx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <send@plt>
  cmp    $0xffffffff,%eax
  je     L28
  mov    $0x0,%eax
  call   <getWallTime@plt>
  movq   %xmm0,%rbx
  mov    $0xa,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <recvUntil@plt>
  test   %eax,%eax
  je     L19
  mov    $0x0,%eax
  call   <getWallTime@plt>
  movq   %rbx,%xmm1
  subsd  %xmm1,%xmm0
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  mov    $0x0,%edi
  call   <exit@plt>
  pxor   %xmm0,%xmm0
  jmp    L39
L19:
  mov    $0x0,%edi
  call   <exit@plt>
  pxor   %xmm0,%xmm0
  jmp    L39
```

assistant: ```c
double sendAndRecv(int sockfd, char *text, char *response) {
  double rtt;
  if (send(sockfd, text, strlen(text), 0) == -1) {
    exit(0);
    return 0;
  }
  rtt = getWallTime();
  if (!recvUntil(sockfd, response, '\n')) {
    exit(0);
    return 0;
  }
  rtt = getWallTime() - rtt;
  return rtt;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sendAndRecv>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x10,%rsp
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <send@plt>
  cmp    $0xffffffff,%eax
  je     L6
  xor    %eax,%eax
  call   <getWallTime@plt>
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %r12d,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <recvUntil@plt>
  test   %eax,%eax
  je     L6
  xor    %eax,%eax
  call   <getWallTime@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x10,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double sendAndRecv(int sockfd, char *text, char *response) {
  double rtt;
  if (send(sockfd, text, strlen(text), 0) == -1) {
    exit(0);
    return 0;
  }
  rtt = getWallTime();
  if (!recvUntil(sockfd, response, '\n')) {
    exit(0);
    return 0;
  }
  rtt = getWallTime() - rtt;
  return rtt;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sendAndRecv>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x10,%rsp
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <send@plt>
  cmp    $0xffffffff,%eax
  je     L19
  xor    %eax,%eax
  call   <getWallTime@plt>
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %r12d,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <recvUntil@plt>
  test   %eax,%eax
  je     L19
  xor    %eax,%eax
  call   <getWallTime@plt>
  subsd  0x8(%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x10,%rsp
  pxor   %xmm0,%xmm0
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double sendAndRecv(int sockfd, char *text, char *response) {
  double rtt;
  if (send(sockfd, text, strlen(text), 0) == -1) {
    exit(0);
    return 0;
  }
  rtt = getWallTime();
  if (!recvUntil(sockfd, response, '\n')) {
    exit(0);
    return 0;
  }
  rtt = getWallTime() - rtt;
  return rtt;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
L15:
  mov    $0x0,%eax
  call   <foo@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L12:
  mov    $0x0,%eax
  call   <jump@plt>
  test   %rax,%rax
  je     L39
  jmp    L15
L39:
  mov    $0x0,%eax
  call   <foo@plt>
  addl   $0x1,-0x4(%rbp)
L34:
  cmpl   $0x9,-0x4(%rbp)
  jle    L12
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
start:
  foo();
  for (int i = 0; i < 10; ++i) {
    if (jump())
      goto start;
    foo();
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbx
L11:
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0xa,%ebx
L8:
  mov    $0x0,%eax
  call   <jump@plt>
  test   %rax,%rax
  jne    L11
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L8
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
start:
  foo();
  for (int i = 0; i < 10; ++i) {
    if (jump())
      goto start;
    foo();
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
L19:
  xor    %eax,%eax
  mov    $0xa,%ebx
  call   <foo@plt>
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  call   <jump@plt>
  test   %rax,%rax
  jne    L19
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L17
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
start:
  foo();
  for (int i = 0; i < 10; ++i) {
    if (jump())
      goto start;
    foo();
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
L8:
  xor    %eax,%eax
  mov    $0xa,%ebx
  call   <foo@plt>
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  call   <jump@plt>
  test   %rax,%rax
  jne    L8
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L0
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(void) {
start:
  foo();
  for (int i = 0; i < 10; ++i) {
    if (jump())
      goto start;
    foo();
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<xioctl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
L10:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <ioctl@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L18
  mov    D8(%rip),%rax
  mov    (%rax),%rdx
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L10
L18:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int xioctl(int fh, int request, void *arg) {
  int r;
  do {
    r = ioctl(fh, request, arg);
  } while (-1 == r && EINTR == errno);
  return r;
}
```