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

```asm
<write_offlinepage>:
  endbr64
  push   %r14
  push   %r13
  mov    %esi,%r13d
  mov    $0xc8,%esi
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <send_http_header@plt>
  cmp    $0xffffffff,%r13d
  je     L6
  mov    %r13d,%ecx
  lea    D1(%rip),%rdx
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    $0x13,%esi
  call   <snprintf@plt>
L19:
  mov    $0x64,%edi
  call   <getText@plt>
  mov    %r12d,%edi
  mov    %r14,%rdx
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <write_st@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  lea    D3(%rip),%rdx
  mov    $0x13,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <snprintf@plt>
  jmp    L19
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
type: string
name: read_data
arguments: tag: D3
type: string

tool: D1: WR-%d

tool: D3: Alle

assistant: ```c
void write_offlinepage(int fd, int idx) {
  send_http_header(fd, 200);
  char text[20];
  if (idx == -1)
    snprintf(text, 19, "Alle");
  else
    snprintf(text, 19, "WR-%d", idx);
  char *txt = getText(100);
  write_st(fd, txt, text);
  free(txt);
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  add    $0x2,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <THROW@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string

tool: D1: foo(): throw %d!


assistant: ```c
void foo() {
  int i = rand() % 2 + 2;
  printf("foo(): throw %d!\n", i);
  THROW(i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rax,%rdx,1),%ebx
  and    $0x1,%ebx
  sub    %edx,%ebx
  add    $0x2,%ebx
  mov    %ebx,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <THROW@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type_infered: string

tool: D9: foo(): throw %d!


assistant: ```c
void foo() {
  int i = rand() % 2 + 2;
  printf("foo(): throw %d!\n", i);
  THROW(i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <rand@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rax,%rdx,1),%ebp
  and    $0x1,%ebp
  sub    %edx,%ebp
  add    $0x2,%ebp
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <THROW@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D4
datatype: string

tool: D4: foo(): throw %d!


assistant: ```c
void foo() {
  int i = rand() % 2 + 2;
  printf("foo(): throw %d!\n", i);
  THROW(i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <rand@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rax,%rdx,1),%ebp
  and    $0x1,%ebp
  sub    %edx,%ebp
  add    $0x2,%ebp
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <THROW@plt>
```

tool_calls: name: parse_data
arguments: data_label: D9
datatype: string

tool: D9: foo(): throw %d!


assistant: ```c
void foo() {
  int i = rand() % 2 + 2;
  printf("foo(): throw %d!\n", i);
  THROW(i);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<SetDisplayOffset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0xd3,-0xa(%rbp)
  movzbl -0x14(%rbp),%eax
  and    $0x3f,%eax
  mov    %al,-0x9(%rbp)
  lea    -0xa(%rbp),%rax
  mov    $0x2,%esi
  mov    %rax,%rdi
  call   <oled_SendCommand@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void SetDisplayOffset(char a) {
  unsigned char d[] = {0xD3, a & 0x3F};
  oled_SendCommand(d, sizeof(d));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<SetDisplayOffset>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0xd3,0x6(%rsp)
  and    $0x3f,%edi
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  mov    $0x2,%esi
  call   <oled_SendCommand@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void SetDisplayOffset(char a) {
  unsigned char d[] = {0xD3, a & 0x3F};
  oled_SendCommand(d, sizeof(d));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SetDisplayOffset>:
  endbr64
  sub    $0x18,%rsp
  and    $0x3f,%edi
  mov    $0x2,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  movb   $0xd3,0x6(%rsp)
  call   <oled_SendCommand@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void SetDisplayOffset(char a) {
  unsigned char d[] = {0xD3, a & 0x3F};
  oled_SendCommand(d, sizeof(d));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<SetDisplayOffset>:
  endbr64
  sub    $0x18,%rsp
  and    $0x3f,%edi
  mov    $0x2,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %dil,0x7(%rsp)
  lea    0x6(%rsp),%rdi
  movb   $0xd3,0x6(%rsp)
  call   <oled_SendCommand@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void SetDisplayOffset(char a) {
  unsigned char d[] = {0xD3, a & 0x3F};
  oled_SendCommand(d, sizeof(d));
}
```
user: What is the c source code of the assembly code below:

```asm
<uart_flush>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L6:
  mov    $0x20215054,%edi
  call   <GET32@plt>
  and    $0x100,%eax
  test   %eax,%eax
  je     L11
  jmp    L6
L11:
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void uart_flush(void) {
  while (1) {
    if ((GET32(0x20215054) & 0x100) == 0)
      break;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uart_flush>:
  endbr64
  sub    $0x8,%rsp
L5:
  mov    $0x20215054,%edi
  call   <GET32@plt>
  test   $0x1,%ah
  jne    L5
  add    $0x8,%rsp
  ret
```

assistant: ```c
void uart_flush(void) {
  while (1) {
    if ((GET32(0x20215054) & 0x100) == 0)
      break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<uart_flush>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  mov    $0x20215054,%edi
  call   <GET32@plt>
  test   $0x1,%ah
  jne    L1
  add    $0x8,%rsp
  ret
```

assistant: ```c
void uart_flush(void) {
  while (1) {
    if ((GET32(0x20215054) & 0x100) == 0)
      break;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<uart_flush>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x20215054,%edi
  call   <GET32@plt>
  test   $0x1,%ah
  jne    L9
  add    $0x8,%rsp
  ret
```

assistant: ```c
void uart_flush(void) {
  while (1) {
    if ((GET32(0x20215054) & 0x100) == 0)
      break;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jg     L3
  call   <abort@plt>
L3:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(int i) {
  if (i < 5)
    __builtin_abort();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  cmp    $0x4,%edi
  jle    L3
  mov    $0x0,%eax
  ret
L3:
  sub    $0x8,%rsp
  call   <abort@plt>
```

assistant: ```c
int __attribute__((no, noclone)) foo(int i) {
  if (i < 5)
    __builtin_abort();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  cmp    $0x4,%edi
  jle    1060 <foo.cold>
  xor    %eax,%eax
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(int i) {
  if (i < 5)
    __builtin_abort();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<foo>:
  endbr64
  cmp    $0x4,%edi
  jle    1060 <foo.cold>
  xor    %eax,%eax
  ret
```

assistant: ```c
int __attribute__((no, noclone)) foo(int i) {
  if (i < 5)
    __builtin_abort();
  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
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D23
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D65
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D10
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D51
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D76
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D44
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D53
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D7
type_infered: string

tool: D23: Usage: "%s". This is a two state discrete exponential time simulator.


tool: D65: Please supply 4 or 5 arguments:

tool: D10: 	1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.

tool: D51: 	2) Substitution rate of second state, same format as above.

tool: D76: 	3) Number of sites, i.e. number of separate replications.

tool: D44: 	4) Number of minimum extent units. I.e duration of each replication, it's calculated by

tool: D53: 		The smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.

tool: D7: 	5) specified random number seed (optional, if not supplied, a random seed will be generated).

assistant: ```c
void usage(char *progname) {
  printf("Usage: \"%s\". This is a two state discrete exponential time simulator.\n", progname);
  printf("Please supply 4 or 5 arguments:\n");
  printf("\t1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.\n");
  printf("\t2) Substitution rate of second state, same format as above.\n");
  printf("\t3) Number of sites, i.e. number of separate replications.\n");
  printf("\t4) Number of minimum extent units. I.e duration of each replication, it's calculated by");
  printf("\t\tThe smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.\n");
  printf("\t5) specified random number seed (optional, if not supplied, a random seed will be generated).\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D71(%rip),%rdi
  call   <puts@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
  lea    D79(%rip),%rdi
  call   <puts@plt>
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  lea    D66(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D51
datatype: string
name: unpack_data
arguments: label: D71
datatype: string
name: unpack_data
arguments: label: D6
datatype: string
name: unpack_data
arguments: label: D15
datatype: string
name: unpack_data
arguments: label: D79
datatype: string
name: unpack_data
arguments: label: D38
datatype: string
name: unpack_data
arguments: label: D4
datatype: string
name: unpack_data
arguments: label: D66
datatype: string

tool: D51: Usage: "%s". This is a two state discrete exponential time simulator.


tool: D71: Please supply 4 or 5 arguments:

tool: D6: 	1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.

tool: D15: 	2) Substitution rate of second state, same format as above.

tool: D79: 	3) Number of sites, i.e. number of separate replications.

tool: D38: 	4) Number of minimum extent units. I.e duration of each replication, it's calculated by

tool: D4: 		The smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.

tool: D66: 	5) specified random number seed (optional, if not supplied, a random seed will be generated).

assistant: ```c
void usage(char *progname) {
  printf("Usage: \"%s\". This is a two state discrete exponential time simulator.\n", progname);
  printf("Please supply 4 or 5 arguments:\n");
  printf("\t1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.\n");
  printf("\t2) Substitution rate of second state, same format as above.\n");
  printf("\t3) Number of sites, i.e. number of separate replications.\n");
  printf("\t4) Number of minimum extent units. I.e duration of each replication, it's calculated by");
  printf("\t\tThe smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.\n");
  printf("\t5) specified random number seed (optional, if not supplied, a random seed will be generated).\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  lea    D0(%rip),%rdi
  call   <puts@plt>
  lea    D66(%rip),%rdi
  call   <puts@plt>
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D53(%rip),%rdi
  call   <puts@plt>
  lea    D77(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D23
data_type: string
name: unpack_data
arguments: data_tag: D0
data_type: string
name: unpack_data
arguments: data_tag: D66
data_type: string
name: unpack_data
arguments: data_tag: D63
data_type: string
name: unpack_data
arguments: data_tag: D53
data_type: string
name: unpack_data
arguments: data_tag: D77
data_type: string

tool: D6: Usage: "%s". This is a two state discrete exponential time simulator.


tool: D16: Please supply 4 or 5 arguments:

tool: D23: 	1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.

tool: D0: 	2) Substitution rate of second state, same format as above.

tool: D66: 	3) Number of sites, i.e. number of separate replications.

tool: D63: 	4) Number of minimum extent units. I.e duration of each replication, it's calculated by

tool: D53: 		The smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.

tool: D77: 	5) specified random number seed (optional, if not supplied, a random seed will be generated).

assistant: ```c
void usage(char *progname) {
  printf("Usage: \"%s\". This is a two state discrete exponential time simulator.\n", progname);
  printf("Please supply 4 or 5 arguments:\n");
  printf("\t1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.\n");
  printf("\t2) Substitution rate of second state, same format as above.\n");
  printf("\t3) Number of sites, i.e. number of separate replications.\n");
  printf("\t4) Number of minimum extent units. I.e duration of each replication, it's calculated by");
  printf("\t\tThe smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.\n");
  printf("\t5) specified random number seed (optional, if not supplied, a random seed will be generated).\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
  lea    D74(%rip),%rdi
  call   <puts@plt>
  lea    D69(%rip),%rdi
  call   <puts@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
  lea    D46(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  lea    D71(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: parse_data
arguments: label: D11
type: string
name: parse_data
arguments: label: D17
type: string
name: parse_data
arguments: label: D74
type: string
name: parse_data
arguments: label: D69
type: string
name: parse_data
arguments: label: D2
type: string
name: parse_data
arguments: label: D46
type: string
name: parse_data
arguments: label: D10
type: string
name: parse_data
arguments: label: D71
type: string

tool: D11: Usage: "%s". This is a two state discrete exponential time simulator.


tool: D17: Please supply 4 or 5 arguments:

tool: D74: 	1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.

tool: D69: 	2) Substitution rate of second state, same format as above.

tool: D2: 	3) Number of sites, i.e. number of separate replications.

tool: D46: 	4) Number of minimum extent units. I.e duration of each replication, it's calculated by

tool: D10: 		The smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.

tool: D71: 	5) specified random number seed (optional, if not supplied, a random seed will be generated).

assistant: ```c
void usage(char *progname) {
  printf("Usage: \"%s\". This is a two state discrete exponential time simulator.\n", progname);
  printf("Please supply 4 or 5 arguments:\n");
  printf("\t1) Substitution rate of first state, either as fraction or float. Eg. 1/20 or .05 would be a rate of 1 in 20.\n");
  printf("\t2) Substitution rate of second state, same format as above.\n");
  printf("\t3) Number of sites, i.e. number of separate replications.\n");
  printf("\t4) Number of minimum extent units. I.e duration of each replication, it's calculated by");
  printf("\t\tThe smallest substrate by average will occur once per 1 extent unit,which is an int, not a float.\n");
  printf("\t5) specified random number seed (optional, if not supplied, a random seed will be generated).\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<out2in_written_socks_reply>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <expect_amqp_header@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int out2in_written_socks_reply(struct pollfd *in_fd, struct pollfd *out_fd, struct _session *sess) {
  expect_amqp_header(in_fd, out_fd, sess);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<out2in_written_socks_reply>:
  endbr64
  sub    $0x8,%rsp
  call   <expect_amqp_header@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int out2in_written_socks_reply(struct pollfd *in_fd, struct pollfd *out_fd, struct _session *sess) {
  expect_amqp_header(in_fd, out_fd, sess);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<out2in_written_socks_reply>:
  endbr64
  sub    $0x8,%rsp
  call   <expect_amqp_header@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int out2in_written_socks_reply(struct pollfd *in_fd, struct pollfd *out_fd, struct _session *sess) {
  expect_amqp_header(in_fd, out_fd, sess);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<out2in_written_socks_reply>:
  endbr64
  sub    $0x8,%rsp
  call   <expect_amqp_header@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int out2in_written_socks_reply(struct pollfd *in_fd, struct pollfd *out_fd, struct _session *sess) {
  expect_amqp_header(in_fd, out_fd, sess);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    %r8d,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  addl   $0x1,-0x2c(%rbp)
  subl   $0x1,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L23
L5:
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edi
  mov    -0x14(%rbp),%eax
  cltd
  idiv   %edi
  cltq
  mov    -0x10(%rbp,%rax,4),%ecx
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x14(%rbp),%eax
  cltd
  idiv   %esi
  mov    %eax,%esi
  mov    -0x38(%rbp),%eax
  lea    (%rcx,%rax,1),%edx
  movslq %esi,%rax
  mov    %edx,-0x10(%rbp,%rax,4)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x5,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  movb   $0x3,(%rdx,%rax,1)
  addl   $0x1,-0x14(%rbp)
L23:
  mov    -0x34(%rbp),%eax
  add    %eax,%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L5
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void print(char m[][66], int h, int k, int n, int f) {
  int b[2], y;
  h = h + 1;
  k = k - 1;
  b[0] = h - 1;
  b[1] = k;
  for (y = 0; y < 2 * n - 1; y++) {
    b[y / (n + 1)] = b[y / (n + 1)] + f;
    m[b[0]][b[1]] = 3;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<print>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %esi,(%rsp)
  sub    $0x1,%edx
  mov    %edx,0x4(%rsp)
  lea    (%rcx,%rcx,1),%r9d
  cmp    $0x1,%r9d
  jle    L14
  sub    $0x1,%r9d
  mov    $0x0,%esi
  add    $0x1,%ecx
L13:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  cltq
  add    %r8d,(%rsp,%rax,4)
  movslq (%rsp),%rdx
  mov    %rdx,%rax
  shl    $0x5,%rax
  add    %rdx,%rax
  lea    (%rdi,%rax,2),%rdx
  movslq 0x4(%rsp),%rax
  movb   $0x3,(%rdx,%rax,1)
  add    $0x1,%esi
  cmp    %esi,%r9d
  jne    L13
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void print(char m[][66], int h, int k, int n, int f) {
  int b[2], y;
  h = h + 1;
  k = k - 1;
  b[0] = h - 1;
  b[1] = k;
  for (y = 0; y < 2 * n - 1; y++) {
    b[y / (n + 1)] = b[y / (n + 1)] + f;
    m[b[0]][b[1]] = 3;
  }
}
```