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

```asm
<get_ps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r12
  push   %rbx
  sub    $0x50,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%r12
  lea    D17(%rip),%rax
  mov    %rax,-0x48(%rbp)
  lea    D12(%rip),%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %eax,%eax
  add    %ebx,%eax
  mov    %eax,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L18:
  cmp    %rdx,%rsp
  je     L26
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L18
L26:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L14
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L14:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x48(%rbp),%rdi
  mov    -0x48(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  mov    -0x4c(%rbp),%esi
  mov    -0x30(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x54(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <exec_and_send@plt>
  mov    %r12,%rsp
  nop
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D17
type_guessed: string
name: parse_data
arguments: label: D12
type_guessed: string

tool: D17: ps -ao pid,cmd

tool: D12: %s | grep -v "%s"

assistant: ```c
void get_ps(int clientfd) {
  char *cmd = "ps -ao pid,cmd";
  char *fmt = "%s | grep -v \"%s\"";
  int size = strlen(fmt) + 2 * strlen(cmd);
  char buffer[size];
  snprintf(buffer, size, fmt, cmd, cmd);
  exec_and_send(clientfd, buffer);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<get_ps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D38(%rip),%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  lea    D36(%rip),%rdi
  call   <strlen@plt>
  lea    (%rbx,%rax,2),%esi
  movslq %esi,%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L20:
  cmp    %rdx,%rsp
  je     L17
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L20
L17:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L14
  orq    $0x0,-0x8(%rsp,%rax,1)
L14:
  lea    D36(%rip),%r8
  mov    %r8,%rcx
  lea    D38(%rip),%rdx
  mov    %rsp,%rdi
  call   <snprintf@plt>
  mov    %rsp,%rsi
  mov    %r12d,%edi
  call   <exec_and_send@plt>
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D38
type: string
name: unpack_data
arguments: tag: D36
type: string

tool: D38: %s | grep -v "%s"

tool: D36: ps -ao pid,cmd

assistant: ```c
void get_ps(int clientfd) {
  char *cmd = "ps -ao pid,cmd";
  char *fmt = "%s | grep -v \"%s\"";
  int size = strlen(fmt) + 2 * strlen(cmd);
  char buffer[size];
  snprintf(buffer, size, fmt, cmd, cmd);
  exec_and_send(clientfd, buffer);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_ps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  lea    D19(%rip),%r14
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  mov    %r13,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    (%rbx,%rax,2),%esi
  movslq %esi,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L12
L9:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L9
L12:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L37
L40:
  mov    %rsp,%rdi
  lea    D19(%rip),%r8
  mov    %r14,%rcx
  mov    %r13,%rdx
  call   <snprintf@plt>
  mov    %rsp,%rsi
  mov    %r12d,%edi
  call   <exec_and_send@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L37:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L40
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D19
datatype: string
name: retrieve_data
arguments: data_label: D18
datatype: string

tool: D19: ps -ao pid,cmd

tool: D18: %s | grep -v "%s"

assistant: ```c
void get_ps(int clientfd) {
  char *cmd = "ps -ao pid,cmd";
  char *fmt = "%s | grep -v \"%s\"";
  int size = strlen(fmt) + 2 * strlen(cmd);
  char buffer[size];
  snprintf(buffer, size, fmt, cmd, cmd);
  exec_and_send(clientfd, buffer);
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_ps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  lea    D29(%rip),%r14
  push   %r13
  lea    D21(%rip),%r13
  push   %r12
  mov    %edi,%r12d
  mov    %r13,%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    (%rbx,%rax,2),%esi
  movslq %esi,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L36
L34:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L34
L36:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L46
L27:
  mov    %rsp,%rdi
  lea    D29(%rip),%r8
  mov    %r14,%rcx
  mov    %r13,%rdx
  call   <snprintf@plt>
  mov    %rsp,%rsi
  mov    %r12d,%edi
  call   <exec_and_send@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L46:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L27
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D29
type_guessed: string
name: retrieve_data
arguments: data_label: D21
type_guessed: string

tool: D29: ps -ao pid,cmd

tool: D21: %s | grep -v "%s"

assistant: ```c
void get_ps(int clientfd) {
  char *cmd = "ps -ao pid,cmd";
  char *fmt = "%s | grep -v \"%s\"";
  int size = strlen(fmt) + 2 * strlen(cmd);
  char buffer[size];
  snprintf(buffer, size, fmt, cmd, cmd);
  exec_and_send(clientfd, buffer);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<assert_null_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L6
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <assert_fail@plt>
L6:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
datatype: string

tool: D5: Expected <null> but was not

assistant: ```c
void assert_null_(const char *file, int line, void *pointer) {
  if (pointer != NULL) {
    assert_fail(file, line, "Expected <null> but was not");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<assert_null_>:
  endbr64
  test   %rdx,%rdx
  je     L1
  sub    $0x8,%rsp
  lea    D0(%rip),%rdx
  call   <assert_fail@plt>
  add    $0x8,%rsp
  ret
L1:
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
type: string

tool: D0: Expected <null> but was not

assistant: ```c
void assert_null_(const char *file, int line, void *pointer) {
  if (pointer != NULL) {
    assert_fail(file, line, "Expected <null> but was not");
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<assert_null_>:
  endbr64
  test   %rdx,%rdx
  je     L2
  lea    D7(%rip),%rdx
  jmp    1050 <assert_fail@plt>
  nopl   (%rax)
L2:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
data_type: string

tool: D7: Expected <null> but was not

assistant: ```c
void assert_null_(const char *file, int line, void *pointer) {
  if (pointer != NULL) {
    assert_fail(file, line, "Expected <null> but was not");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<assert_null_>:
  endbr64
  test   %rdx,%rdx
  je     L3
  lea    D6(%rip),%rdx
  jmp    1050 <assert_fail@plt>
  nopl   (%rax)
L3:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D6: Expected <null> but was not

assistant: ```c
void assert_null_(const char *file, int line, void *pointer) {
  if (pointer != NULL) {
    assert_fail(file, line, "Expected <null> but was not");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<untrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  add    %eax,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L12:
  mov    -0xc(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  add    %edx,%edx
  add    $0x1,%edx
  mov    %edx,%ecx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jb     L12
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %eax,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *untrim(char *str) {
  unsigned int i;
  char *nstr = (char *)malloc((strlen(str) * 2) + 2);
  nstr[0] = '\0';
  for (i = 0; i < (strlen(str)); i++) {
    nstr[i * 2] = ' ';
    nstr[(i * 2) + 1] = str[i];
  }
  nstr[strlen(str) * 2] = ' ';
  nstr[(strlen(str) * 2) + 1] = '\0';
  return nstr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<untrim>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    0x2(%rax,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  movb   $0x0,(%rax)
  mov    $0x0,%ebx
  jmp    L17
L5:
  lea    (%rbx,%rbx,1),%eax
  mov    %eax,%edx
  movb   $0x20,0x0(%rbp,%rdx,1)
  mov    %ebx,%edx
  movzbl (%r12,%rdx,1),%edx
  lea    0x1(%rax),%eax
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%ebx
L17:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  ja     L5
  mov    %r12,%rdi
  call   <strlen@plt>
  add    %eax,%eax
  cltq
  movb   $0x20,0x0(%rbp,%rax,1)
  mov    %r12,%rdi
  call   <strlen@plt>
  add    %eax,%eax
  cltq
  movb   $0x0,0x1(%rbp,%rax,1)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *untrim(char *str) {
  unsigned int i;
  char *nstr = (char *)malloc((strlen(str) * 2) + 2);
  nstr[0] = '\0';
  for (i = 0; i < (strlen(str)); i++) {
    nstr[i * 2] = ' ';
    nstr[(i * 2) + 1] = str[i];
  }
  nstr[strlen(str) * 2] = ' ';
  nstr[(strlen(str) * 2) + 1] = '\0';
  return nstr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<untrim>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    0x2(%rax,%rax,1),%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r12
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L11:
  lea    (%rbx,%rbx,1),%edx
  movb   $0x20,(%r12,%rdx,1)
  mov    %rdx,%rax
  mov    %ebx,%edx
  add    $0x1,%ebx
  movzbl 0x0(%rbp,%rdx,1),%edx
  add    $0x1,%eax
  mov    %dl,(%r12,%rax,1)
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  ja     L11
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  add    %eax,%eax
  cltq
  movb   $0x20,(%r12,%rax,1)
  call   <strlen@plt>
  add    %eax,%eax
  cltq
  movb   $0x0,0x1(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *untrim(char *str) {
  unsigned int i;
  char *nstr = (char *)malloc((strlen(str) * 2) + 2);
  nstr[0] = '\0';
  for (i = 0; i < (strlen(str)); i++) {
    nstr[i * 2] = ' ';
    nstr[(i * 2) + 1] = str[i];
  }
  nstr[strlen(str) * 2] = ' ';
  nstr[(strlen(str) * 2) + 1] = '\0';
  return nstr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<untrim>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    0x2(%rax,%rax,1),%edi
  call   <malloc@plt>
  movb   $0x0,(%rax)
  mov    %rax,%r12
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L3:
  lea    (%rbx,%rbx,1),%edx
  movb   $0x20,(%r12,%rdx,1)
  mov    %rdx,%rax
  mov    %ebx,%edx
  add    $0x1,%ebx
  movzbl 0x0(%rbp,%rdx,1),%edx
  add    $0x1,%eax
  mov    %dl,(%r12,%rax,1)
L17:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  ja     L3
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  add    %eax,%eax
  cltq
  movb   $0x20,(%r12,%rax,1)
  call   <strlen@plt>
  add    %eax,%eax
  cltq
  movb   $0x0,0x1(%r12,%rax,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *untrim(char *str) {
  unsigned int i;
  char *nstr = (char *)malloc((strlen(str) * 2) + 2);
  nstr[0] = '\0';
  for (i = 0; i < (strlen(str)); i++) {
    nstr[i * 2] = ' ';
    nstr[(i * 2) + 1] = str[i];
  }
  nstr[strlen(str) * 2] = ' ';
  nstr[(strlen(str) * 2) + 1] = '\0';
  return nstr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<copyStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *copyStr(char *src) {
  int slen = strlen(src);
  char *dst = (char *)malloc(sizeof(char) * (slen + 1));
  memcpy(dst, src, slen + 1);
  return dst;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<copyStr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *copyStr(char *src) {
  int slen = strlen(src);
  char *dst = (char *)malloc(sizeof(char) * (slen + 1));
  memcpy(dst, src, slen + 1);
  return dst;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<copyStr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *copyStr(char *src) {
  int slen = strlen(src);
  char *dst = (char *)malloc(sizeof(char) * (slen + 1));
  memcpy(dst, src, slen + 1);
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bufout_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L39
L8:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <bufout_putc@plt>
  test   %eax,%eax
  jne    L13
  mov    $0x0,%eax
  jmp    L2
L13:
  addq   $0x1,-0x10(%rbp)
L39:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    $0x1,%eax
L2:
  leave
  ret
```

assistant: ```c
int bufout_puts(struct bufout *b, const char *s) {
  while (*s) {
    if (!bufout_putc(b, *s))
      return 0;
    s++;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bufout_puts>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  movzbl (%rsi),%esi
  test   %sil,%sil
  je     L13
L24:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <bufout_putc@plt>
  test   %eax,%eax
  je     L3
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L24
  mov    $0x1,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  mov    $0x1,%eax
  jmp    L3
```

assistant: ```c
int bufout_puts(struct bufout *b, const char *s) {
  while (*s) {
    if (!bufout_putc(b, *s))
      return 0;
    s++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bufout_puts>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rsi),%esi
  test   %sil,%sil
  je     L8
  mov    %rdi,%rbp
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L34:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  je     L8
L3:
  mov    %rbp,%rdi
  call   <bufout_putc@plt>
  test   %eax,%eax
  jne    L34
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int bufout_puts(struct bufout *b, const char *s) {
  while (*s) {
    if (!bufout_putc(b, *s))
      return 0;
    s++;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bufout_puts>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rsi),%esi
  test   %sil,%sil
  je     L20
  mov    %rdi,%rbp
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L9:
  movsbl 0x1(%rbx),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  je     L20
L37:
  mov    %rbp,%rdi
  call   <bufout_putc@plt>
  test   %eax,%eax
  jne    L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L20:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int bufout_puts(struct bufout *b, const char *s) {
  while (*s) {
    if (!bufout_putc(b, *s))
      return 0;
    s++;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<updateCarry>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  subq   $0x8,-0x10(%rbp)
  cmpb   $0x0,-0x4(%rbp)
  je     L13
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  or     $0x20000000,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L19
L13:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  and    $0xdfffffff,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
L19:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void updateCarry(unsigned char cyf, unsigned int *reg) {
  reg -= 2;
  if (cyf) {
    *reg |= (1 << 29);
  } else {
    *reg &= ~(1 << 29);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<updateCarry>:
  endbr64
  test   %dil,%dil
  je     L19
  mov    -0x8(%rsi),%eax
  or     $0x20000000,%eax
L1:
  mov    %eax,-0x8(%rsi)
  ret
L19:
  mov    -0x8(%rsi),%eax
  and    $0xdfffffff,%eax
  jmp    L1
```

assistant: ```c
void updateCarry(unsigned char cyf, unsigned int *reg) {
  reg -= 2;
  if (cyf) {
    *reg |= (1 << 29);
  } else {
    *reg &= ~(1 << 29);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<updateCarry>:
  endbr64
  mov    -0x8(%rsi),%eax
  mov    %eax,%edx
  and    $0xdfffffff,%eax
  or     $0x20000000,%edx
  test   %dil,%dil
  cmovne %edx,%eax
  mov    %eax,-0x8(%rsi)
  ret
```

assistant: ```c
void updateCarry(unsigned char cyf, unsigned int *reg) {
  reg -= 2;
  if (cyf) {
    *reg |= (1 << 29);
  } else {
    *reg &= ~(1 << 29);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<updateCarry>:
  endbr64
  mov    -0x8(%rsi),%eax
  mov    %eax,%edx
  and    $0xdfffffff,%eax
  or     $0x20000000,%edx
  test   %dil,%dil
  cmovne %edx,%eax
  mov    %eax,-0x8(%rsi)
  ret
```

assistant: ```c
void updateCarry(unsigned char cyf, unsigned int *reg) {
  reg -= 2;
  if (cyf) {
    *reg |= (1 << 29);
  } else {
    *reg &= ~(1 << 29);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<http_parse_querystring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L95
  mov    $0x0,%eax
  jmp    L120
L95:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L117
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L117
L42:
  cmpb   $0x3d,-0x21(%rbp)
  jne    L86
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L122
L86:
  cmpb   $0x2b,-0x21(%rbp)
  jne    L0
  mov    -0x20(%rbp),%rax
  movb   $0x20,(%rax)
  jmp    L122
L0:
  cmpb   $0x25,-0x21(%rbp)
  jne    L27
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <convert_query_hexchar@plt>
  mov    %rax,-0x20(%rbp)
  jmp    L122
L27:
  cmpb   $0x26,-0x21(%rbp)
  jne    L122
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x40(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
L122:
  addq   $0x1,-0x20(%rbp)
L117:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
  cmpb   $0x0,-0x21(%rbp)
  jne    L42
  mov    -0x20(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  je     L112
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x40(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
L112:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L120:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type_guessed: string

tool: D3: Out of memory


assistant: ```c
int http_parse_querystring(const char *txt_, void (*fn)(const char *name, const char *value)) {
  char *o, *t, *txt, *value = NULL, c;
  if (!txt_)
    return 0;
  o = t = txt = strdup(txt_);
  if (t == NULL) {
    printf("Out of memory\n");
    exit(1);
  }
  while ((c = *t) != '\0') {
    if (c == '=') {
      *t = '\0';
      value = t + 1;
    } else if (c == '+') {
      *t = ' ';
    } else if (c == '%') {
      t = convert_query_hexchar(t);
    } else if (c == '&') {
      *t = '\0';
      (*fn)(txt, value);
      txt = t + 1;
      value = NULL;
    }
    t++;
  }
  if (t != txt)
    (*fn)(txt, value);
  free(o);
  return 0;
}
```