ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%r12
L45:
  mov    %rsp,%rbp
L38:
  mov    $0x200,%edx
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L0
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %ebx,%eax
  je     L38
  mov    %r12,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L45
L0:
  cmp    $0xffffffff,%eax
  je     L21
L57:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  lea    D10(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L57
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D4
type: string
name: unpack_data
arguments: data_label: D10
type: string

tool: D4: write

tool: D10: read

assistant: ```c
main(int argc, char *argv[]) {
  int n;
  char buf[512];
  while ((n = read(0, buf, 512)) > 0)
    if (write(1, buf, n) != n) {
      perror("write");
      exit(1);
    };
  ;
  if (n == -1) {
    perror("read");
    exit(1);
  };
  ;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L50
  xchg   %ax,%ax
L16:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %ebx,%eax
  jne    L40
L50:
  mov    $0x200,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L16
  cmp    $0xffffffff,%eax
  je     L5
L31:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L62
  add    $0x210,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L40:
  mov    %r12,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L50
L5:
  lea    D3(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L31
L62:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string
name: read_data
arguments: data_tag: D3
data_type: string

tool: D4: write

tool: D3: read

assistant: ```c
main(int argc, char *argv[]) {
  int n;
  char buf[512];
  while ((n = read(0, buf, 512)) > 0)
    if (write(1, buf, n) != n) {
      perror("write");
      exit(1);
    };
  ;
  if (n == -1) {
    perror("read");
    exit(1);
  };
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L59
  xchg   %ax,%ax
L27:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %ebx,%eax
  jne    L58
L59:
  mov    $0x200,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L27
  cmp    $0xffffffff,%eax
  je     L8
L39:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x210,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L58:
  mov    %r12,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L59
L8:
  lea    D16(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L39
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
type: string
name: parse_data
arguments: tag: D16
type: string

tool: D10: write

tool: D16: read

assistant: ```c
main(int argc, char *argv[]) {
  int n;
  char buf[512];
  while ((n = read(0, buf, 512)) > 0)
    if (write(1, buf, n) != n) {
      perror("write");
      exit(1);
    };
  ;
  if (n == -1) {
    perror("read");
    exit(1);
  };
  ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_27_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  xor    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  and    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  or     -0x14(%rbp),%eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  not    %eax
  and    -0x10(%rbp),%eax
  and    -0x18(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_27_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T ^ S;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_27_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  shl    $0x8,%esi
  or     %eax,%esi
  mov    %edi,%edx
  xor    %esi,%edx
  movzbl %dl,%edx
  xor    %edi,%edx
  mov    %esi,%eax
  not    %eax
  or     %edi,%eax
  not    %edx
  and    %edx,%eax
  not    %edi
  and    %esi,%edi
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_27_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T ^ S;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xrop3_27_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%edx
  shl    $0x8,%esi
  or     %eax,%esi
  xor    %esi,%edx
  mov    %esi,%eax
  movzbl %dl,%edx
  not    %eax
  xor    %edi,%edx
  or     %edi,%eax
  not    %edi
  not    %edx
  and    %esi,%edi
  and    %edx,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_27_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T ^ S;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xrop3_27_0_1>:
  endbr64
  movzbl %dil,%eax
  mov    %eax,%edi
  shl    $0x8,%edi
  or     %eax,%edi
  movzbl %sil,%eax
  mov    %eax,%esi
  mov    %edi,%edx
  shl    $0x8,%esi
  or     %eax,%esi
  xor    %esi,%edx
  mov    %esi,%eax
  movzbl %dl,%edx
  not    %eax
  xor    %edi,%edx
  or     %edi,%eax
  not    %edi
  not    %edx
  and    %esi,%edi
  and    %edx,%eax
  movzbl %dil,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
unsigned xrop3_27_0_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T ^ S;
  stk2 = D & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & S) | (stk1 & (~T)) | (T & (~S) & D);
}
```
user: What is the c source code of the assembly code below?

```asm
<i2c_smbus_write_byte_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x18(%rbp)
  mov    %al,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movzbl -0x1c(%rbp),%eax
  mov    %al,-0x9(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%ecx
  movzbl -0x18(%rbp),%edx
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  lea    -0x9(%rbp),%rdi
  mov    -0x14(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <i2c_smbus_access@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
long i2c_smbus_write_byte_data(int file, unsigned char command, unsigned char value) {
  union i2c_smbus_data data;
  data.byte = value;
  return i2c_smbus_access(file, I2C_SMBUS_WRITE, command, I2C_SMBUS_BYTE_DATA, &data);
}
```
user: Translate the below assembly code into C programming language:

```asm
<i2c_smbus_write_byte_data>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %dl,0x7(%rsp)
  movzbl %sil,%edx
  lea    0x7(%rsp),%r8
  mov    D13(%rip),%rax
  mov    (%rax),%ecx
  mov    D19(%rip),%rax
  mov    (%rax),%esi
  call   <i2c_smbus_access@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long i2c_smbus_write_byte_data(int file, unsigned char command, unsigned char value) {
  union i2c_smbus_data data;
  data.byte = value;
  return i2c_smbus_access(file, I2C_SMBUS_WRITE, command, I2C_SMBUS_BYTE_DATA, &data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<i2c_smbus_write_byte_data>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  mov    %dl,0x7(%rsp)
  movzbl %sil,%edx
  lea    0x7(%rsp),%r8
  mov    (%rax),%ecx
  mov    D11(%rip),%rax
  mov    (%rax),%esi
  call   <i2c_smbus_access@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long i2c_smbus_write_byte_data(int file, unsigned char command, unsigned char value) {
  union i2c_smbus_data data;
  data.byte = value;
  return i2c_smbus_access(file, I2C_SMBUS_WRITE, command, I2C_SMBUS_BYTE_DATA, &data);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<i2c_smbus_write_byte_data>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D16(%rip),%rax
  mov    %dl,0x7(%rsp)
  movzbl %sil,%edx
  lea    0x7(%rsp),%r8
  mov    (%rax),%ecx
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  call   <i2c_smbus_access@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long i2c_smbus_write_byte_data(int file, unsigned char command, unsigned char value) {
  union i2c_smbus_data data;
  data.byte = value;
  return i2c_smbus_access(file, I2C_SMBUS_WRITE, command, I2C_SMBUS_BYTE_DATA, &data);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<br_httpd_get_port>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jne    L11
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L8
L11:
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <ntohs@plt>
  leave
  ret
```

assistant: ```c
uint16_t br_httpd_get_port(int af, const struct sockaddr *addr) {
  in_port_t port = 0;
  if (af == AF_INET6) {
    struct sockaddr_in6 *a = (struct sockaddr_in6 *)addr;
    port = a->sin6_port;
  } else {
    struct sockaddr_in *a = (struct sockaddr_in *)addr;
    port = a->sin_port;
  }
  return ntohs(port);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<br_httpd_get_port>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%edi
  call   <ntohs@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
uint16_t br_httpd_get_port(int af, const struct sockaddr *addr) {
  in_port_t port = 0;
  if (af == AF_INET6) {
    struct sockaddr_in6 *a = (struct sockaddr_in6 *)addr;
    port = a->sin6_port;
  } else {
    struct sockaddr_in *a = (struct sockaddr_in *)addr;
    port = a->sin_port;
  }
  return ntohs(port);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<br_httpd_get_port>:
  endbr64
  mov    (%rsi),%edi
  jmp    1050 <ntohs@plt>
```

assistant: ```c
uint16_t br_httpd_get_port(int af, const struct sockaddr *addr) {
  in_port_t port = 0;
  if (af == AF_INET6) {
    struct sockaddr_in6 *a = (struct sockaddr_in6 *)addr;
    port = a->sin6_port;
  } else {
    struct sockaddr_in *a = (struct sockaddr_in *)addr;
    port = a->sin_port;
  }
  return ntohs(port);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<br_httpd_get_port>:
  endbr64
  mov    (%rsi),%edi
  jmp    1050 <ntohs@plt>
```

assistant: ```c
uint16_t br_httpd_get_port(int af, const struct sockaddr *addr) {
  in_port_t port = 0;
  if (af == AF_INET6) {
    struct sockaddr_in6 *a = (struct sockaddr_in6 *)addr;
    port = a->sin6_port;
  } else {
    struct sockaddr_in *a = (struct sockaddr_in *)addr;
    port = a->sin_port;
  }
  return ntohs(port);
}
```
user: What is the c source code of the assembly code below:

```asm
<printFileReverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L26
L38:
  movl   $0x0,-0x4(%rbp)
  jmp    L39
L34:
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x7,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L39:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L34
  mov    $0xa,%edi
  call   <putchar@plt>
  subl   $0x1,-0x8(%rbp)
L26:
  cmpl   $0x0,-0x8(%rbp)
  jns    L38
  nop
  nop
  leave
  ret
```

assistant: ```c
void printFileReverse(char content[][128], int *line, int charNum[]) {
  printf("\n");
  for (int i = *line - 1; i >= 0; i--) {
    for (int j = 0; j < charNum[i]; j++) {
      printf("%c", content[i][j]);
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printFileReverse>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x0(%rbp),%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  js     L50
  cltq
  lea    0x0(,%rax,4),%rcx
  lea    (%rbx,%rcx,1),%r13
  shl    $0x7,%rax
  add    %rax,%r14
  lea    -0x4(%rbx,%rcx,1),%r15
  lea    -0x1(%rdx),%eax
  shl    $0x2,%rax
  sub    %rax,%r15
  jmp    L42
L53:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <putchar@plt>
  add    $0x1,%rbx
  cmp    %ebx,(%r12)
  jg     L53
L26:
  mov    $0xa,%edi
  call   <putchar@plt>
  sub    $0x4,%r13
  add    $0xffffffffffffff80,%r14
  cmp    %r15,%r13
  je     L50
L42:
  mov    %r13,%r12
  mov    %r14,%rbp
  mov    $0x0,%ebx
  cmpl   $0x0,0x0(%r13)
  jg     L53
  jmp    L26
L50:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void printFileReverse(char content[][128], int *line, int charNum[]) {
  printf("\n");
  for (int i = *line - 1; i >= 0; i--) {
    for (int j = 0; j < charNum[i]; j++) {
      printf("%c", content[i][j]);
    }
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printFileReverse>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    $0xa,%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  mov    0x0(%rbp),%eax
  sub    $0x1,%eax
  js     L17
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    0x0(,%rdx,4),%rcx
  shl    $0x2,%rax
  shl    $0x7,%rdx
  lea    -0x4(%rbx,%rcx,1),%r13
  lea    (%rbx,%rcx,1),%rbp
  add    %rdx,%r12
  sub    %rax,%r13
  nopl   (%rax)
L35:
  mov    0x0(%rbp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L36
  nopl   0x0(%rax)
L10:
  movsbl (%r12,%rbx,1),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %ebx,0x0(%rbp)
  jg     L10
L36:
  mov    $0xa,%edi
  sub    $0x4,%rbp
  add    $0xffffffffffffff80,%r12
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L35
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void printFileReverse(char content[][128], int *line, int charNum[]) {
  printf("\n");
  for (int i = *line - 1; i >= 0; i--) {
    for (int j = 0; j < charNum[i]; j++) {
      printf("%c", content[i][j]);
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printFileReverse>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    $0xa,%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  mov    0x0(%rbp),%eax
  sub    $0x1,%eax
  js     L11
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    0x0(,%rdx,4),%rcx
  shl    $0x2,%rax
  shl    $0x7,%rdx
  lea    -0x4(%rbx,%rcx,1),%r13
  lea    (%rbx,%rcx,1),%rbp
  add    %rdx,%r12
  sub    %rax,%r13
  nopl   (%rax)
L23:
  mov    0x0(%rbp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L7
  nopl   0x0(%rax)
L3:
  movsbl (%r12,%rbx,1),%edi
  add    $0x1,%rbx
  call   <putchar@plt>
  cmp    %ebx,0x0(%rbp)
  jg     L3
L7:
  mov    $0xa,%edi
  sub    $0x4,%rbp
  add    $0xffffffffffffff80,%r12
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L23
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void printFileReverse(char content[][128], int *line, int charNum[]) {
  printf("\n");
  for (int i = *line - 1; i >= 0; i--) {
    for (int j = 0; j < charNum[i]; j++) {
      printf("%c", content[i][j]);
    }
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  sub    -0x10(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
L8:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x11(%rbp),%eax
  mov    %al,(%rdx)
  cmpb   $0x0,-0x11(%rbp)
  jne    L8
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  char c;
  char *s = (char *)src;
  const ptrdiff_t off = dest - s - 1;
  do {
    c = *s++;
    s[off] = c;
  } while (c != 0);
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  mov    %rdi,%rcx
  sub    %rsi,%rcx
L7:
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%edx
  mov    %dl,-0x1(%rsi,%rcx,1)
  test   %dl,%dl
  jne    L7
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  char c;
  char *s = (char *)src;
  const ptrdiff_t off = dest - s - 1;
  do {
    c = *s++;
    s[off] = c;
  } while (c != 0);
  return dest;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rcx
  mov    %rdi,%rax
  sub    %rsi,%rcx
  nopl   (%rax)
L4:
  movzbl (%rsi),%edx
  add    $0x1,%rsi
  mov    %dl,-0x1(%rsi,%rcx,1)
  test   %dl,%dl
  jne    L4
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  char c;
  char *s = (char *)src;
  const ptrdiff_t off = dest - s - 1;
  do {
    c = *s++;
    s[off] = c;
  } while (c != 0);
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rcx
  mov    %rdi,%rax
  sub    %rsi,%rcx
  nopl   (%rax)
L3:
  movzbl (%rsi),%edx
  add    $0x1,%rsi
  mov    %dl,-0x1(%rsi,%rcx,1)
  test   %dl,%dl
  jne    L3
  ret
```

assistant: ```c
char *strcpy(char *dest, const char *src) {
  char c;
  char *s = (char *)src;
  const ptrdiff_t off = dest - s - 1;
  do {
    c = *s++;
    s[off] = c;
  } while (c != 0);
  return dest;
}
```
user: What is the c source code of the assembly code below?

```asm
<print_hex8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x8,-0x4(%rbp)
  jmp    L13
L18:
  mov    -0x14(%rbp),%eax
  shr    $0x1c,%eax
  mov    %eax,%edi
  call   <print_hex1@plt>
  shll   $0x4,-0x14(%rbp)
  subl   $0x1,-0x4(%rbp)
L13:
  cmpl   $0x0,-0x4(%rbp)
  jne    L18
  nop
  nop
  leave
  ret
```

assistant: ```c
void print_hex8(unsigned v) {
  int i;
  for (i = 8; i; --i) {
    print_hex1((v >> 28) & 0xF);
    v <<= 4;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_hex8>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x8,%ebp
L1:
  mov    %ebx,%edi
  shr    $0x1c,%edi
  call   <print_hex1@plt>
  shl    $0x4,%ebx
  sub    $0x1,%ebp
  jne    L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void print_hex8(unsigned v) {
  int i;
  for (i = 8; i; --i) {
    print_hex1((v >> 28) & 0xF);
    v <<= 4;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<print_hex8>:
  endbr64
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L6:
  mov    %ebx,%edi
  shl    $0x4,%ebx
  shr    $0x1c,%edi
  call   <print_hex1@plt>
  sub    $0x1,%ebp
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void print_hex8(unsigned v) {
  int i;
  for (i = 8; i; --i) {
    print_hex1((v >> 28) & 0xF);
    v <<= 4;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_hex8>:
  endbr64
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L6:
  mov    %ebx,%edi
  shl    $0x4,%ebx
  shr    $0x1c,%edi
  call   <print_hex1@plt>
  sub    $0x1,%ebp
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void print_hex8(unsigned v) {
  int i;
  for (i = 8; i; --i) {
    print_hex1((v >> 28) & 0xF);
    v <<= 4;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<nondet_assume>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L0
  mov    $0x0,%eax
  call   <nondet_cancel@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void nondet_assume(int expr) {
  if (!expr) {
    nondet_cancel();
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<nondet_assume>:
  endbr64
  test   %edi,%edi
  je     L7
  ret
L7:
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <nondet_cancel@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void nondet_assume(int expr) {
  if (!expr) {
    nondet_cancel();
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nondet_assume>:
  endbr64
  test   %edi,%edi
  je     L1
  ret
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  jmp    1050 <nondet_cancel@plt>
```

assistant: ```c
void nondet_assume(int expr) {
  if (!expr) {
    nondet_cancel();
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nondet_assume>:
  endbr64
  test   %edi,%edi
  je     L5
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  jmp    1050 <nondet_cancel@plt>
```

assistant: ```c
void nondet_assume(int expr) {
  if (!expr) {
    nondet_cancel();
  }
}
```