ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<s9_Decompression>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsi,(%rsp)
  test   %edx,%edx
  jle    L8
  mov    %rdi,%r12
  mov    %edx,%ebp
  mov    %rdi,%rbx
  mov    %rsp,%r13
  nopl   0x0(%rax)
L12:
  mov    %rbx,%rdi
  mov    %r13,%rsi
  add    $0x4,%rbx
  call   <s9_decode@plt>
  sub    %eax,%ebp
  test   %ebp,%ebp
  jg     L12
  mov    %rbx,%rax
  sub    %r12,%rax
  sar    $0x2,%rax
L6:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %eax,%eax
  jmp    L6
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int s9_Decompression(unsigned int *input, unsigned int *output, int size) {
  unsigned int *_p = output;
  int num;
  unsigned *tmp = input;
  int left = size;
  while (left > 0) {
    num = s9_decode(tmp, &_p);
    tmp++;
    left -= num;
  }
  return tmp - input;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<verifica_login>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe8,%rsp
  mov    %edi,%r14d
  mov    %esi,0x8(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%rsi
  lea    D12(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%r12
  mov    $0x0,%ebp
  movl   $0x0,0x4(%rsp)
  mov    $0x0,%r15d
  lea    0x10(%rsp),%r13
  test   %rax,%rax
  je     L47
L122:
  mov    %r12,%rdi
  call   <feof@plt>
  or     %ebp,%eax
  mov    %eax,%ebx
  jne    L54
  mov    %r12,%rdx
  mov    $0xc8,%esi
  mov    %r13,%rdi
  call   <fgets@plt>
  mov    $0x1,%esi
  mov    %r13,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    %r14d,%eax
  jne    L122
  mov    $0x2,%esi
  mov    %r13,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    %ebx,%ebp
  cmp    0x8(%rsp),%eax
  jne    L122
  mov    $0x4,%esi
  mov    %r13,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    $0x2,%eax
  jne    L81
L4:
  mov    %ebx,%ebp
  jmp    L122
L47:
  mov    $0x4,%ebp
  jmp    L29
L81:
  mov    $0x3,%esi
  mov    %r13,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    0xc(%rsp),%eax
  mov    $0x1,%eax
  cmove  %eax,%r15d
  mov    $0x4,%esi
  mov    %r13,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    %eax,%ebx
  cmp    $0x1,%eax
  je     L26
  mov    $0x1,%ebx
  jmp    L4
L26:
  mov    %eax,0x4(%rsp)
  jmp    L4
L54:
  mov    %r12,%rdi
  call   <fclose@plt>
  test   %ebp,%ebp
  je     L29
  cmpl   $0x1,0x4(%rsp)
  je     L34
  test   %r15d,%r15d
  mov    $0x2,%ebp
  cmovne %r15d,%ebp
L29:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L93
  mov    %ebp,%eax
  add    $0xe8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  mov    $0x3,%ebp
  jmp    L29
L93:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D18
datatype: string
name: unpack_data
arguments: data_tag: D12
datatype: string

tool: D18: r

tool: D12: usuarios.txt

assistant: ```c
int verifica_login(int agencia, int cc, int senha) {
  int correta = 0, bloqueada = 0, encontrou = 0;
  FILE *arq_usuarios;
  char buffer[200];
  arq_usuarios = fopen("usuarios.txt", "r");
  if (arq_usuarios != NULL) {
    while ((!feof(arq_usuarios)) && (encontrou == 0)) {
      fgets(buffer, 200, arq_usuarios);
      if ((atoi(captura(buffer, 1)) == agencia) && (atoi(captura(buffer, 2)) == cc) && (atoi(captura(buffer, 4)) != 2)) {
        encontrou = 1;
        if (atoi(captura(buffer, 3)) == senha)
          correta = 1;
        if (atoi(captura(buffer, 4)) == 1)
          bloqueada = 1;
      }
    }
    fclose(arq_usuarios);
    if (encontrou == 0)
      return 0;
    else if (bloqueada == 1)
      return 3;
    else if (correta == 0)
      return 2;
    else
      return 1;
  } else
    return 4;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<verifica_login>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  lea    D10(%rip),%rdi
  push   %rbx
  sub    $0xe8,%rsp
  mov    %esi,0x8(%rsp)
  lea    D19(%rip),%rsi
  lea    0x10(%rsp),%rbx
  mov    %edx,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%r13
  test   %rax,%rax
  jne    L104
  jmp    L51
L129:
  test   %r12d,%r12d
  jne    L21
  mov    %r13,%rdx
  mov    $0xc8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %rbx,%rdi
  mov    $0x1,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    %ebp,%eax
  je     L132
L104:
  mov    %r13,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L129
  mov    %r13,%rdi
  call   <fclose@plt>
  test   %r12d,%r12d
  jne    L25
L1:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0xe8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  mov    %r13,%rdi
  call   <fclose@plt>
L25:
  mov    $0x2,%r12d
  mov    $0x3,%eax
  sub    %r14d,%r12d
  cmp    $0x1,%r15d
  cmove  %eax,%r12d
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L132:
  mov    %rbx,%rdi
  mov    $0x2,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    0x8(%rsp),%eax
  jne    L104
  mov    %rbx,%rdi
  mov    $0x4,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    $0x2,%eax
  je     L104
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    $0x4,%esi
  cmp    0xc(%rsp),%eax
  mov    %rbx,%rdi
  mov    $0x1,%eax
  cmove  %eax,%r14d
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    %eax,%r12d
  cmp    $0x1,%eax
  je     L57
  mov    $0x1,%r12d
  jmp    L104
L51:
  mov    $0x4,%r12d
  jmp    L1
L57:
  mov    $0x1,%r15d
  jmp    L104
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
data_type: string
name: extract_symbolic_data
arguments: tag: D19
data_type: string

tool: D10: usuarios.txt

tool: D19: r

assistant: ```c
int verifica_login(int agencia, int cc, int senha) {
  int correta = 0, bloqueada = 0, encontrou = 0;
  FILE *arq_usuarios;
  char buffer[200];
  arq_usuarios = fopen("usuarios.txt", "r");
  if (arq_usuarios != NULL) {
    while ((!feof(arq_usuarios)) && (encontrou == 0)) {
      fgets(buffer, 200, arq_usuarios);
      if ((atoi(captura(buffer, 1)) == agencia) && (atoi(captura(buffer, 2)) == cc) && (atoi(captura(buffer, 4)) != 2)) {
        encontrou = 1;
        if (atoi(captura(buffer, 3)) == senha)
          correta = 1;
        if (atoi(captura(buffer, 4)) == 1)
          bloqueada = 1;
      }
    }
    fclose(arq_usuarios);
    if (encontrou == 0)
      return 0;
    else if (bloqueada == 1)
      return 3;
    else if (correta == 0)
      return 2;
    else
      return 1;
  } else
    return 4;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<verifica_login>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  lea    D5(%rip),%rdi
  push   %rbx
  sub    $0xe8,%rsp
  mov    %esi,0x8(%rsp)
  lea    D17(%rip),%rsi
  lea    0x10(%rsp),%rbx
  mov    %edx,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%r13
  test   %rax,%rax
  jne    L137
  jmp    L85
L86:
  test   %r12d,%r12d
  jne    L17
  mov    %r13,%rdx
  mov    $0xc8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %rbx,%rdi
  mov    $0x1,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    %ebp,%eax
  je     L4
L137:
  mov    %r13,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L86
  mov    %r13,%rdi
  call   <fclose@plt>
  test   %r12d,%r12d
  jne    L54
L93:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L94
  add    $0xe8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %r13,%rdi
  call   <fclose@plt>
L54:
  mov    $0x2,%r12d
  mov    $0x3,%eax
  sub    %r14d,%r12d
  cmp    $0x1,%r15d
  cmove  %eax,%r12d
  jmp    L93
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %rbx,%rdi
  mov    $0x2,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    0x8(%rsp),%eax
  jne    L137
  mov    %rbx,%rdi
  mov    $0x4,%esi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  cmp    $0x2,%eax
  je     L137
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    $0x4,%esi
  cmp    0xc(%rsp),%eax
  mov    %rbx,%rdi
  mov    $0x1,%eax
  cmove  %eax,%r14d
  call   <captura@plt>
  mov    %eax,%edi
  call   <atoi@plt>
  mov    %eax,%r12d
  cmp    $0x1,%eax
  je     L42
  mov    $0x1,%r12d
  jmp    L137
L85:
  mov    $0x4,%r12d
  jmp    L93
L42:
  mov    $0x1,%r15d
  jmp    L137
L94:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type: string
name: disassemble_data
arguments: data_tag: D17
type: string

tool: D5: usuarios.txt

tool: D17: r

assistant: ```c
int verifica_login(int agencia, int cc, int senha) {
  int correta = 0, bloqueada = 0, encontrou = 0;
  FILE *arq_usuarios;
  char buffer[200];
  arq_usuarios = fopen("usuarios.txt", "r");
  if (arq_usuarios != NULL) {
    while ((!feof(arq_usuarios)) && (encontrou == 0)) {
      fgets(buffer, 200, arq_usuarios);
      if ((atoi(captura(buffer, 1)) == agencia) && (atoi(captura(buffer, 2)) == cc) && (atoi(captura(buffer, 4)) != 2)) {
        encontrou = 1;
        if (atoi(captura(buffer, 3)) == senha)
          correta = 1;
        if (atoi(captura(buffer, 4)) == 1)
          bloqueada = 1;
      }
    }
    fclose(arq_usuarios);
    if (encontrou == 0)
      return 0;
    else if (bloqueada == 1)
      return 3;
    else if (correta == 0)
      return 2;
    else
      return 1;
  } else
    return 4;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<syscon_total_degree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x1c,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int syscon_total_degree(int *d) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(28, d, b, c, 0);
  return fail;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<syscon_total_degree>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x1c,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int syscon_total_degree(int *d) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(28, d, b, c, 0);
  return fail;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<syscon_total_degree>:
  endbr64
  mov    %rdi,%rsi
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x1c,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int syscon_total_degree(int *d) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(28, d, b, c, 0);
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<syscon_total_degree>:
  endbr64
  mov    %rdi,%rsi
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x1c,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int syscon_total_degree(int *d) {
  int *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(28, d, b, c, 0);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<DigitsCount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L1
L7:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x9,-0x14(%rbp)
  jg     L7
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int DigitsCount(int number) {
  int count;
  count = 1;
  while (number >= 10) {
    number /= 10;
    count++;
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<DigitsCount>:
  endbr64
  cmp    $0x9,%edi
  jle    L18
  mov    $0x1,%edx
L13:
  mov    %edi,%ecx
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edi
  add    $0x1,%edx
  cmp    $0x63,%ecx
  jg     L13
L29:
  mov    %edx,%eax
  ret
L18:
  mov    $0x1,%edx
  jmp    L29
```

assistant: ```c
int DigitsCount(int number) {
  int count;
  count = 1;
  while (number >= 10) {
    number /= 10;
    count++;
  }
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<DigitsCount>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x9,%edi
  jle    L2
  mov    $0xcccccccd,%edx
  nopl   0x0(%rax)
L13:
  mov    %edi,%edi
  add    $0x1,%r8d
  mov    %rdi,%rax
  imul   %rdx,%rdi
  shr    $0x23,%rdi
  cmp    $0x63,%eax
  jg     L13
L2:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int DigitsCount(int number) {
  int count;
  count = 1;
  while (number >= 10) {
    number /= 10;
    count++;
  }
  return count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<DigitsCount>:
  endbr64
  mov    $0x1,%r8d
  cmp    $0x9,%edi
  jle    L17
  mov    $0xcccccccd,%edx
  nopl   0x0(%rax)
L4:
  mov    %edi,%edi
  add    $0x1,%r8d
  mov    %rdi,%rax
  imul   %rdx,%rdi
  shr    $0x23,%rdi
  cmp    $0x63,%eax
  jg     L4
L17:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int DigitsCount(int number) {
  int count;
  count = 1;
  while (number >= 10) {
    number /= 10;
    count++;
  }
  return count;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ui_draw_hline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <nc_set_yx@plt>
  jmp    L12
L3:
  mov    $0x2d,%edi
  call   <nc_addch@plt>
L12:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jg     L3
  nop
  nop
  leave
  ret
```

assistant: ```c
void ui_draw_hline(int y, int x, int w) {
  nc_set_yx(y, x);
  while (w-- > 0)
    nc_addch('-');
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ui_draw_hline>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebp
  call   <nc_set_yx@plt>
  lea    -0x1(%rbp),%ebx
  test   %ebp,%ebp
  jle    L12
L15:
  mov    $0x2d,%edi
  call   <nc_addch@plt>
  sub    $0x1,%ebx
  cmp    $0xffffffff,%ebx
  jne    L15
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ui_draw_hline(int y, int x, int w) {
  nc_set_yx(y, x);
  while (w-- > 0)
    nc_addch('-');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ui_draw_hline>:
  endbr64
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <nc_set_yx@plt>
  test   %ebp,%ebp
  jle    L15
  lea    -0x1(%rbp),%ebx
  nopl   0x0(%rax,%rax,1)
L0:
  mov    $0x2d,%edi
  call   <nc_addch@plt>
  sub    $0x1,%ebx
  jae    L0
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ui_draw_hline(int y, int x, int w) {
  nc_set_yx(y, x);
  while (w-- > 0)
    nc_addch('-');
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ui_draw_hline>:
  endbr64
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <nc_set_yx@plt>
  test   %ebp,%ebp
  jle    L8
  lea    -0x1(%rbp),%ebx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0x2d,%edi
  call   <nc_addch@plt>
  sub    $0x1,%ebx
  jae    L3
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ui_draw_hline(int y, int x, int w) {
  nc_set_yx(y, x);
  while (w-- > 0)
    nc_addch('-');
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gob_encode_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <gob_encode_uint64@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L21
  mov    -0x4(%rbp),%eax
  mov    -0x30(%rbp),%rdx
  sub    %edx,%eax
  jmp    L10
L21:
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  sub    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L20
  mov    -0x30(%rbp),%rax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  neg    %eax
  jmp    L10
L20:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x30(%rbp),%rax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
L10:
  leave
  ret
```

assistant: ```c
int gob_encode_bytes(char *buf, size_t buf_size, void *s, size_t len) {
  int nbytes = gob_encode_uint64(buf, buf_size, len);
  if (nbytes < 0) {
    return nbytes - len;
  }
  buf += nbytes;
  buf_size -= nbytes;
  if (len >= buf_size) {
    return -(nbytes + len);
  }
  memcpy(buf, s, len);
  return nbytes + len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gob_encode_bytes>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%r14
  mov    %rcx,%r12
  mov    %rcx,%rdx
  call   <gob_encode_uint64@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L13
  cltq
  sub    %rax,%rbp
  cmp    %rbp,%r12
  jb     L33
  add    %r12d,%ebx
  mov    %ebx,%eax
  neg    %eax
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  sub    %r12d,%eax
  jmp    L31
L33:
  lea    0x0(%r13,%rax,1),%rdi
  mov    %r12,%rdx
  mov    %r14,%rsi
  call   <memcpy@plt>
  lea    (%rbx,%r12,1),%eax
  jmp    L31
```

assistant: ```c
int gob_encode_bytes(char *buf, size_t buf_size, void *s, size_t len) {
  int nbytes = gob_encode_uint64(buf, buf_size, len);
  if (nbytes < 0) {
    return nbytes - len;
  }
  buf += nbytes;
  buf_size -= nbytes;
  if (len >= buf_size) {
    return -(nbytes + len);
  }
  memcpy(buf, s, len);
  return nbytes + len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gob_encode_bytes>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  mov    %rcx,%rdx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <gob_encode_uint64@plt>
  test   %eax,%eax
  js     L15
  movslq %eax,%rdx
  lea    (%rax,%r13,1),%ebx
  sub    %rdx,%rbp
  cmp    %rbp,%r13
  jb     L2
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  neg    %eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  lea    (%r14,%rdx,1),%rdi
  mov    %r12,%rsi
  mov    %r13,%rdx
  call   <memcpy@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  pop    %rbx
  sub    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int gob_encode_bytes(char *buf, size_t buf_size, void *s, size_t len) {
  int nbytes = gob_encode_uint64(buf, buf_size, len);
  if (nbytes < 0) {
    return nbytes - len;
  }
  buf += nbytes;
  buf_size -= nbytes;
  if (len >= buf_size) {
    return -(nbytes + len);
  }
  memcpy(buf, s, len);
  return nbytes + len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gob_encode_bytes>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  mov    %rcx,%rdx
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <gob_encode_uint64@plt>
  test   %eax,%eax
  js     L14
  movslq %eax,%rdx
  lea    (%rax,%r13,1),%ebx
  sub    %rdx,%rbp
  cmp    %rbp,%r13
  jb     L16
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  neg    %eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  lea    (%r14,%rdx,1),%rdi
  mov    %r12,%rsi
  mov    %r13,%rdx
  call   <memcpy@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  pop    %rbx
  sub    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int gob_encode_bytes(char *buf, size_t buf_size, void *s, size_t len) {
  int nbytes = gob_encode_uint64(buf, buf_size, len);
  if (nbytes < 0) {
    return nbytes - len;
  }
  buf += nbytes;
  buf_size -= nbytes;
  if (len >= buf_size) {
    return -(nbytes + len);
  }
  memcpy(buf, s, len);
  return nbytes + len;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D54(%rip),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  D8(%rip),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D13(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D36(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  D30(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D6(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x6,-0x44(%rbp)
  mov    -0x44(%rbp),%edx
  lea    -0x40(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <median@plt>
  mov    %eax,%esi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D54
type_guessed: f64
name: unpack_data
arguments: data_tag: D8
type_guessed: f64
name: unpack_data
arguments: data_tag: D13
type_guessed: f64
name: unpack_data
arguments: data_tag: D36
type_guessed: f64
name: unpack_data
arguments: data_tag: D30
type_guessed: f64
name: unpack_data
arguments: data_tag: D6
type_guessed: f64
name: unpack_data
arguments: data_tag: D43
type_guessed: string

tool: D54: 12.2

tool: D8: 10

tool: D13: 13.4

tool: D36: 14.5

tool: D30: 56.2

tool: D6: 23.3

tool: D43: Median is %lf.


assistant: ```c
int main(int argc, const char *argv[]) {
  double array[] = {12.2, 10.0, 13.4, 14.5, 56.2, 23.3};
  int nArray = sizeof(array) / sizeof(array[0]);
  printf("Median is %lf.\n", median(array, nArray));
  return 0;
}
```