ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test3>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x10,%edx
  mov    $0x8,%r8d
  movdqa D3(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  lea    0x20(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movabs $0x706050403020100,%rax
  movaps %xmm0,0x10(%rsp)
  movdqa D15(%rip),%xmm0
  mov    %rax,(%rsp)
  movabs $0x31d8ee87827a7c15,%rax
  movaps %xmm0,0x20(%rsp)
  movdqa D16(%rip),%xmm0
  mov    %rax,0x8(%rsp)
  lea    0x30(%rsp),%rax
  movaps %xmm0,0x30(%rsp)
  push   %rax
  push   $0x8
  lea    0x18(%rsp),%r9
  call   <run_test@plt>
  pop    %rdx
  pop    %rcx
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x58,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
data_type: byte[16]
name: retrieve_data
arguments: data_tag: D15
data_type: byte[16]
name: retrieve_data
arguments: data_tag: D16
data_type: byte[16]

tool: {"D3": [0x00,0x11,0x22,0x33,0x44,0x55,0x66,0x77,0x88,0x99,0xaa,0xbb,0xcc,0xdd,0xee,0xff]}

tool: {"D15": [0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,0x10]}

tool: {"D16": [0xed,0x33,0xf1,0x70,0x1f,0x53,0x6b,0x7e,0xc9,0xb8,0x7e,0xb5,0x04,0x86,0x2f,0x7c]}

assistant: ```c
int test3() {
  unsigned long long mlen = 8;
  const unsigned long long clen = 8;
  const unsigned long long hlen = 16;
  const unsigned char k[16] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff};
  const unsigned char h[16] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
  const unsigned char m[8] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07};
  const unsigned char c[8] = {0x15, 0x7C, 0x7A, 0x82, 0x87, 0xEE, 0xD8, 0x31};
  const unsigned char t[16] = {0xED, 0x33, 0xF1, 0x70, 0x1F, 0x53, 0x6B, 0x7E, 0xC9, 0xB8, 0x7E, 0xB5, 0x04, 0x86, 0x2F, 0x7C};
  return run_test(k, h, hlen, m, mlen, c, clen, t);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<test3>:
  endbr64
  sub    $0x58,%rsp
  mov    $0x10,%edx
  mov    $0x8,%r8d
  movdqa D13(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  lea    0x20(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movabs $0x706050403020100,%rax
  movaps %xmm0,0x10(%rsp)
  movdqa D9(%rip),%xmm0
  mov    %rax,(%rsp)
  movabs $0x31d8ee87827a7c15,%rax
  movaps %xmm0,0x20(%rsp)
  movdqa D22(%rip),%xmm0
  mov    %rax,0x8(%rsp)
  lea    0x30(%rsp),%rax
  movaps %xmm0,0x30(%rsp)
  push   %rax
  push   $0x8
  lea    0x18(%rsp),%r9
  call   <run_test@plt>
  pop    %rdx
  pop    %rcx
  mov    0x48(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x58,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D13
data_type: byte[16]
name: parse_data
arguments: data_tag: D9
data_type: byte[16]
name: parse_data
arguments: data_tag: D22
data_type: byte[16]

tool: {"D13": [0x00,0x11,0x22,0x33,0x44,0x55,0x66,0x77,0x88,0x99,0xaa,0xbb,0xcc,0xdd,0xee,0xff]}

tool: {"D9": [0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,0x10]}

tool: {"D22": [0xed,0x33,0xf1,0x70,0x1f,0x53,0x6b,0x7e,0xc9,0xb8,0x7e,0xb5,0x04,0x86,0x2f,0x7c]}

assistant: ```c
int test3() {
  unsigned long long mlen = 8;
  const unsigned long long clen = 8;
  const unsigned long long hlen = 16;
  const unsigned char k[16] = {0x00, 0x11, 0x22, 0x33, 0x44, 0x55, 0x66, 0x77, 0x88, 0x99, 0xaa, 0xbb, 0xcc, 0xdd, 0xee, 0xff};
  const unsigned char h[16] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16};
  const unsigned char m[8] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07};
  const unsigned char c[8] = {0x15, 0x7C, 0x7A, 0x82, 0x87, 0xEE, 0xD8, 0x31};
  const unsigned char t[16] = {0xED, 0x33, 0xF1, 0x70, 0x1F, 0x53, 0x6B, 0x7E, 0xC9, 0xB8, 0x7E, 0xB5, 0x04, 0x86, 0x2F, 0x7C};
  return run_test(k, h, hlen, m, mlen, c, clen, t);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<to_bigendian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  movzbl %cl,%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  movzbl %cl,%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  lea    0xc(%rax),%rdx
  mov    -0xc(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void to_bigendian(uint8_t *buffer, uint32_t n) {
  buffer[0] = (n >> 24) & 0xff;
  buffer[1] = (n >> 16) & 0xff;
  buffer[2] = (n >> 8) & 0xff;
  buffer[3] = n & 0xff;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<to_bigendian>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%edx
  shr    $0x18,%edx
  mov    %edx,(%rdi)
  mov    %esi,%edx
  sar    $0x10,%edx
  and    $0xff,%edx
  mov    %edx,0x4(%rdi)
  movzbl %ah,%edx
  mov    %edx,0x8(%rdi)
  and    $0xff,%eax
  mov    %eax,0xc(%rdi)
  ret
```

assistant: ```c
void to_bigendian(uint8_t *buffer, uint32_t n) {
  buffer[0] = (n >> 24) & 0xff;
  buffer[1] = (n >> 16) & 0xff;
  buffer[2] = (n >> 8) & 0xff;
  buffer[3] = n & 0xff;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<to_bigendian>:
  endbr64
  mov    %esi,%edx
  mov    %esi,%eax
  shr    $0x18,%edx
  mov    %edx,(%rdi)
  mov    %esi,%edx
  sar    $0x10,%edx
  and    $0xff,%edx
  mov    %edx,0x4(%rdi)
  movzbl %ah,%edx
  and    $0xff,%eax
  mov    %edx,0x8(%rdi)
  mov    %eax,0xc(%rdi)
  ret
```

assistant: ```c
void to_bigendian(uint8_t *buffer, uint32_t n) {
  buffer[0] = (n >> 24) & 0xff;
  buffer[1] = (n >> 16) & 0xff;
  buffer[2] = (n >> 8) & 0xff;
  buffer[3] = n & 0xff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<to_bigendian>:
  endbr64
  mov    %esi,%ecx
  mov    %esi,%edx
  mov    %esi,%eax
  shr    $0x18,%ecx
  sar    $0x10,%edx
  movd   %ecx,%xmm0
  movzbl %dl,%edx
  movzbl %ah,%ecx
  movzbl %sil,%eax
  movd   %ecx,%xmm1
  movd   %eax,%xmm2
  movd   %edx,%xmm3
  punpckldq %xmm2,%xmm1
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,(%rdi)
  ret
```

assistant: ```c
void to_bigendian(uint8_t *buffer, uint32_t n) {
  buffer[0] = (n >> 24) & 0xff;
  buffer[1] = (n >> 16) & 0xff;
  buffer[2] = (n >> 8) & 0xff;
  buffer[3] = n & 0xff;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpl   $0x3,-0x24(%rbp)
  jg     L8
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <terror@plt>
L8:
  mov    -0x30(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%r12
  mov    -0x30(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rbx
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,%r13d
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %r12,%rcx
  mov    %rbx,%rdx
  mov    %r13d,%esi
  mov    %eax,%edi
  call   <psort@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_guessed: string

tool: D1: USE: sortWords <max_size> <num_proc> <input_file> <output_file>


assistant: ```c
int main(int ac, char **av) {
  if (ac < 4) {
    terror("USE: sortWords <max_size> <num_proc> <input_file> <output_file>\n");
  }
  return psort(atoi(av[1]), atoi(av[2]), av[3], av[4]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  cmp    $0x3,%edi
  jle    L15
L2:
  mov    0x20(%rbx),%r13
  mov    0x18(%rbx),%r12
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  mov    0x8(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %ebp,%esi
  call   <psort@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  lea    D6(%rip),%rdi
  call   <terror@plt>
  jmp    L2
```

tool_calls: name: parse_data
arguments: data_tag: D6
type_infered: string

tool: D6: USE: sortWords <max_size> <num_proc> <input_file> <output_file>


assistant: ```c
int main(int ac, char **av) {
  if (ac < 4) {
    terror("USE: sortWords <max_size> <num_proc> <input_file> <output_file>\n");
  }
  return psort(atoi(av[1]), atoi(av[2]), av[3], av[4]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jle    L5
L6:
  mov    0x10(%rbx),%rdi
  mov    0x20(%rbx),%r13
  mov    0x18(%rbx),%r12
  call   <atoi@plt>
  mov    0x8(%rbx),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  add    $0x8,%rsp
  mov    %r13,%rcx
  mov    %r12,%rdx
  pop    %rbx
  mov    %ebp,%esi
  mov    %eax,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <psort@plt>
L5:
  lea    D8(%rip),%rdi
  call   <terror@plt>
  jmp    L6
```

tool_calls: name: parse_data
arguments: label: D8
type_infered: string

tool: D8: USE: sortWords <max_size> <num_proc> <input_file> <output_file>


assistant: ```c
int main(int ac, char **av) {
  if (ac < 4) {
    terror("USE: sortWords <max_size> <num_proc> <input_file> <output_file>\n");
  }
  return psort(atoi(av[1]), atoi(av[2]), av[3], av[4]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x3,%edi
  jle    L13
L4:
  mov    0x10(%rbx),%rdi
  mov    0x20(%rbx),%r13
  mov    0x18(%rbx),%r12
  call   <atoi@plt>
  mov    0x8(%rbx),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  add    $0x8,%rsp
  mov    %r13,%rcx
  mov    %r12,%rdx
  pop    %rbx
  mov    %ebp,%esi
  mov    %eax,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <psort@plt>
L13:
  lea    D9(%rip),%rdi
  call   <terror@plt>
  jmp    L4
```

tool_calls: name: read_data
arguments: data_tag: D9
type: string

tool: D9: USE: sortWords <max_size> <num_proc> <input_file> <output_file>


assistant: ```c
int main(int ac, char **av) {
  if (ac < 4) {
    terror("USE: sortWords <max_size> <num_proc> <input_file> <output_file>\n");
  }
  return psort(atoi(av[1]), atoi(av[2]), av[3], av[4]);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<remove_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  shl    $0x2,%eax
  cmp    %eax,-0x28(%rbp)
  jl     L15
  mov    -0x28(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
L15:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L22
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L22
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncat@plt>
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rbx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x24(%rbp),%edx
  sub    $0x1,%edx
  sub    %edx,%eax
  cltq
  lea    (%rbx,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x24(%rbp),%edx
  lea    0x1(%rdx),%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncat@plt>
  jmp    L26
L22:
  mov    -0x40(%rbp),%rax
  movl   $0x0,(%rax)
L26:
  mov    -0x28(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int remove_char(int pos, char **edit_buffer, char **temp_buffer, int alloc_size, int *fail) {
  if ((int)strlen(*edit_buffer) * 4 <= alloc_size) {
    char *realloc_buffer;
    alloc_size /= 2;
    *edit_buffer = (realloc_buffer = realloc(*edit_buffer, alloc_size));
    *temp_buffer = (realloc_buffer = realloc(*temp_buffer, alloc_size));
  }
  if (*edit_buffer != NULL && *temp_buffer != NULL) {
    strcpy(*temp_buffer, *edit_buffer + pos);
    strncat(*temp_buffer, *edit_buffer, pos - 1);
    strcpy(*edit_buffer, *temp_buffer + (strlen(*temp_buffer) - (pos - 1)));
    strncat(*edit_buffer, *temp_buffer, (strlen(*temp_buffer) - (pos + 1)));
  } else {
    *fail = 0;
  }
  return alloc_size;
}
```
user: Translate the below assembly code into C programming language:

```asm
<remove_char>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%rbx
  mov    %rdx,%r12
  mov    %ecx,%ebp
  mov    %r8,%r14
  mov    (%rsi),%rdi
  call   <strlen@plt>
  shl    $0x2,%eax
  cmp    %ebp,%eax
  jle    L34
L49:
  mov    (%rbx),%rsi
  test   %rsi,%rsi
  je     L43
  mov    (%r12),%rdi
  test   %rdi,%rdi
  je     L43
  movslq %r13d,%rax
  add    %rax,%rsi
  call   <strcpy@plt>
  lea    -0x1(%r13),%r15d
  mov    %r15d,%edx
  mov    (%rbx),%rsi
  mov    (%r12),%rdi
  call   <strncat@plt>
  mov    (%r12),%r14
  mov    %r14,%rdi
  call   <strlen@plt>
  sub    %r15d,%eax
  cltq
  lea    (%r14,%rax,1),%rsi
  mov    (%rbx),%rdi
  call   <strcpy@plt>
  mov    (%r12),%rdi
  call   <strlen@plt>
  add    $0x1,%r13d
  sub    %r13d,%eax
  mov    %eax,%edx
  mov    (%r12),%rsi
  mov    (%rbx),%rdi
  call   <strncat@plt>
  jmp    L22
L34:
  mov    %ebp,%r15d
  shr    $0x1f,%r15d
  add    %ebp,%r15d
  sar    %r15d
  mov    %r15d,%ebp
  mov    %r15d,%esi
  mov    (%rbx),%rdi
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    %r15d,%esi
  mov    (%r12),%rdi
  call   <realloc@plt>
  mov    %rax,(%r12)
  jmp    L49
L43:
  movl   $0x0,(%r14)
L22:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int remove_char(int pos, char **edit_buffer, char **temp_buffer, int alloc_size, int *fail) {
  if ((int)strlen(*edit_buffer) * 4 <= alloc_size) {
    char *realloc_buffer;
    alloc_size /= 2;
    *edit_buffer = (realloc_buffer = realloc(*edit_buffer, alloc_size));
    *temp_buffer = (realloc_buffer = realloc(*temp_buffer, alloc_size));
  }
  if (*edit_buffer != NULL && *temp_buffer != NULL) {
    strcpy(*temp_buffer, *edit_buffer + pos);
    strncat(*temp_buffer, *edit_buffer, pos - 1);
    strcpy(*edit_buffer, *temp_buffer + (strlen(*temp_buffer) - (pos - 1)));
    strncat(*edit_buffer, *temp_buffer, (strlen(*temp_buffer) - (pos + 1)));
  } else {
    *fail = 0;
  }
  return alloc_size;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<remove_char>:
  endbr64
  push   %r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <strlen@plt>
  shl    $0x2,%eax
  cmp    %r12d,%eax
  jle    L20
L3:
  mov    (%rbx),%rsi
  test   %rsi,%rsi
  je     L47
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L47
  movslq %r13d,%rax
  lea    -0x1(%r13),%r15d
  add    $0x1,%r13d
  add    %rax,%rsi
  call   <strcpy@plt>
  mov    (%rbx),%rsi
  mov    0x0(%rbp),%rdi
  mov    %r15d,%edx
  call   <strncat@plt>
  mov    0x0(%rbp),%r14
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    (%rbx),%rdi
  sub    %r15d,%eax
  cltq
  lea    (%r14,%rax,1),%rsi
  call   <strcpy@plt>
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  mov    0x0(%rbp),%rsi
  mov    (%rbx),%rdi
  sub    %r13d,%eax
  mov    %eax,%edx
  call   <strncat@plt>
  jmp    L12
  nopl   0x0(%rax)
L47:
  movl   $0x0,(%r14)
L12:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L20:
  mov    %r12d,%eax
  mov    (%rbx),%rdi
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  mov    %eax,%esi
  mov    %eax,%r12d
  call   <realloc@plt>
  mov    %r12d,%esi
  mov    %rax,(%rbx)
  mov    0x0(%rbp),%rdi
  call   <realloc@plt>
  mov    %rax,0x0(%rbp)
  jmp    L3
```

assistant: ```c
int remove_char(int pos, char **edit_buffer, char **temp_buffer, int alloc_size, int *fail) {
  if ((int)strlen(*edit_buffer) * 4 <= alloc_size) {
    char *realloc_buffer;
    alloc_size /= 2;
    *edit_buffer = (realloc_buffer = realloc(*edit_buffer, alloc_size));
    *temp_buffer = (realloc_buffer = realloc(*temp_buffer, alloc_size));
  }
  if (*edit_buffer != NULL && *temp_buffer != NULL) {
    strcpy(*temp_buffer, *edit_buffer + pos);
    strncat(*temp_buffer, *edit_buffer, pos - 1);
    strcpy(*edit_buffer, *temp_buffer + (strlen(*temp_buffer) - (pos - 1)));
    strncat(*edit_buffer, *temp_buffer, (strlen(*temp_buffer) - (pos + 1)));
  } else {
    *fail = 0;
  }
  return alloc_size;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<remove_char>:
  endbr64
  push   %r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <strlen@plt>
  shl    $0x2,%eax
  cmp    %r12d,%eax
  jle    L32
L10:
  mov    (%rbx),%rsi
  test   %rsi,%rsi
  je     L28
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L28
  movslq %r13d,%rax
  lea    -0x1(%r13),%r15d
  add    $0x1,%r13d
  add    %rax,%rsi
  call   <strcpy@plt>
  mov    (%rbx),%rsi
  mov    0x0(%rbp),%rdi
  mov    %r15d,%edx
  call   <strncat@plt>
  mov    0x0(%rbp),%r14
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    (%rbx),%rdi
  sub    %r15d,%eax
  cltq
  lea    (%r14,%rax,1),%rsi
  call   <strcpy@plt>
  mov    0x0(%rbp),%rdi
  call   <strlen@plt>
  mov    0x0(%rbp),%rsi
  mov    (%rbx),%rdi
  sub    %r13d,%eax
  mov    %eax,%edx
  call   <strncat@plt>
  jmp    L5
  nopl   0x0(%rax)
L28:
  movl   $0x0,(%r14)
L5:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L32:
  mov    %r12d,%eax
  mov    (%rbx),%rdi
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  mov    %eax,%esi
  mov    %eax,%r12d
  call   <realloc@plt>
  mov    %r12d,%esi
  mov    %rax,(%rbx)
  mov    0x0(%rbp),%rdi
  call   <realloc@plt>
  mov    %rax,0x0(%rbp)
  jmp    L10
```

assistant: ```c
int remove_char(int pos, char **edit_buffer, char **temp_buffer, int alloc_size, int *fail) {
  if ((int)strlen(*edit_buffer) * 4 <= alloc_size) {
    char *realloc_buffer;
    alloc_size /= 2;
    *edit_buffer = (realloc_buffer = realloc(*edit_buffer, alloc_size));
    *temp_buffer = (realloc_buffer = realloc(*temp_buffer, alloc_size));
  }
  if (*edit_buffer != NULL && *temp_buffer != NULL) {
    strcpy(*temp_buffer, *edit_buffer + pos);
    strncat(*temp_buffer, *edit_buffer, pos - 1);
    strcpy(*edit_buffer, *temp_buffer + (strlen(*temp_buffer) - (pos - 1)));
    strncat(*edit_buffer, *temp_buffer, (strlen(*temp_buffer) - (pos + 1)));
  } else {
    *fail = 0;
  }
  return alloc_size;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getcookie>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x1,%eax
  jg     L27
  mov    $0x0,%eax
  jmp    L16
L27:
  mov    -0x28(%rbp),%rax
  add    $0x2,%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x3b,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
L0:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L18
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x10(%rbp)
L18:
  mov    -0x18(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
char *getcookie(const char *rawcookie) {
  char *cookie = NULL;
  char *tmpcookie;
  char *cookieend;
  if (strlen(rawcookie) < 2)
    return NULL;
  tmpcookie = strdup(rawcookie + 2);
  cookieend = strchr(tmpcookie, ';');
  if (cookieend)
    *cookieend = '\0';
  cookie = strdup(tmpcookie);
  if (tmpcookie) {
    free(tmpcookie);
    tmpcookie = NULL;
  };
  return cookie;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getcookie>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    $0x0,%ebp
  cmp    $0x1,%eax
  jle    L15
  lea    0x2(%rbx),%rdi
  call   <strdup@plt>
  mov    %rax,%rbx
  mov    $0x3b,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L11
  movb   $0x0,(%rax)
L11:
  mov    %rbx,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <free@plt>
L15:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *getcookie(const char *rawcookie) {
  char *cookie = NULL;
  char *tmpcookie;
  char *cookieend;
  if (strlen(rawcookie) < 2)
    return NULL;
  tmpcookie = strdup(rawcookie + 2);
  cookieend = strchr(tmpcookie, ';');
  if (cookieend)
    *cookieend = '\0';
  cookie = strdup(tmpcookie);
  if (tmpcookie) {
    free(tmpcookie);
    tmpcookie = NULL;
  };
  return cookie;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getcookie>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    $0x1,%eax
  jle    L18
  lea    0x2(%rbx),%rdi
  call   <strdup@plt>
  mov    $0x3b,%esi
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <strchr@plt>
  test   %rax,%rax
  je     L8
  movb   $0x0,(%rax)
L8:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
L18:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *getcookie(const char *rawcookie) {
  char *cookie = NULL;
  char *tmpcookie;
  char *cookieend;
  if (strlen(rawcookie) < 2)
    return NULL;
  tmpcookie = strdup(rawcookie + 2);
  cookieend = strchr(tmpcookie, ';');
  if (cookieend)
    *cookieend = '\0';
  cookie = strdup(tmpcookie);
  if (tmpcookie) {
    free(tmpcookie);
    tmpcookie = NULL;
  };
  return cookie;
}
```