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

```asm
<trace_log_int>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <htonl@plt>
  lea    0x4(%rsp),%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %eax,0x4(%rsp)
  mov    D2(%rip),%rax
  mov    (%rax),%ecx
  call   <fwrite@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void trace_log_int(uint32_t v) {
  uint32_t sw;
  sw = htonl(v);
  fwrite(&sw, 4, 1, stdout);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<trace_log_int>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <htonl@plt>
  lea    0x4(%rsp),%rdi
  mov    $0x1,%edx
  mov    $0x4,%esi
  mov    %eax,0x4(%rsp)
  mov    D7(%rip),%rax
  mov    (%rax),%ecx
  call   <fwrite@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void trace_log_int(uint32_t v) {
  uint32_t sw;
  sw = htonl(v);
  fwrite(&sw, 4, 1, stdout);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0xc(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x5,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  lea    (%rax,%rax,1),%esi
  mov    -0xc(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string

tool: D5:  %d %d %d 


assistant: ```c
main() {
  int a = 10, b = 3, c = 5, d;
  printf(" %d %d %d \n", a << 1, a = b, a >> 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%ecx
  mov    $0x3,%edx
  mov    $0x6,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
datatype: string

tool: D1:  %d %d %d 


assistant: ```c
main() {
  int a = 10, b = 3, c = 5, d;
  printf(" %d %d %d \n", a << 1, a = b, a >> 1);
}
```
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
  sub    $0x8,%rsp
  mov    $0x5,%ecx
  mov    $0x3,%edx
  mov    $0x6,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type: string

tool: D1:  %d %d %d 


assistant: ```c
main() {
  int a = 10, b = 3, c = 5, d;
  printf(" %d %d %d \n", a << 1, a = b, a >> 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%ecx
  mov    $0x3,%edx
  mov    $0x6,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D1:  %d %d %d 


assistant: ```c
main() {
  int a = 10, b = 3, c = 5, d;
  printf(" %d %d %d \n", a << 1, a = b, a >> 1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D6(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <encrypt@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
datatype: string

tool: D6: CONVERT THIS MORSE CODE

assistant: ```c
int main() {
  char *msg = "CONVERT THIS MORSE CODE";
  encrypt(msg);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <encrypt@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type: string

tool: D4: CONVERT THIS MORSE CODE

assistant: ```c
int main() {
  char *msg = "CONVERT THIS MORSE CODE";
  encrypt(msg);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <encrypt@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_infered: string

tool: D5: CONVERT THIS MORSE CODE

assistant: ```c
int main() {
  char *msg = "CONVERT THIS MORSE CODE";
  encrypt(msg);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <encrypt@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
datatype: string

tool: D4: CONVERT THIS MORSE CODE

assistant: ```c
int main() {
  char *msg = "CONVERT THIS MORSE CODE";
  encrypt(msg);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<close_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L25
  mov    D11(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  je     L25
  mov    D19(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x8(%rbp)
  je     L25
  cmpl   $0x0,-0xc(%rbp)
  je     L27
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pclose@plt>
  jmp    L25
L27:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
L25:
  nop
  leave
  ret
```

assistant: ```c
void close_file(FILE *fp, int zip) {
  if (fp && fp != stdin && fp != stdout) {
    if (zip)
      pclose(fp);
    else
      fclose(fp);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<close_file>:
  endbr64
  test   %rdi,%rdi
  je     L46
  mov    D0(%rip),%rax
  cmp    %rdi,(%rax)
  je     L46
  mov    D12(%rip),%rax
  cmp    %rdi,(%rax)
  je     L46
  sub    $0x8,%rsp
  test   %esi,%esi
  je     L34
  call   <pclose@plt>
L18:
  add    $0x8,%rsp
  ret
L34:
  call   <fclose@plt>
  jmp    L18
L46:
  ret
```

assistant: ```c
void close_file(FILE *fp, int zip) {
  if (fp && fp != stdin && fp != stdout) {
    if (zip)
      pclose(fp);
    else
      fclose(fp);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<close_file>:
  endbr64
  test   %rdi,%rdi
  je     L31
  mov    D1(%rip),%rax
  cmp    %rdi,(%rax)
  je     L31
  mov    D13(%rip),%rax
  cmp    %rdi,(%rax)
  je     L31
  test   %esi,%esi
  je     L17
  jmp    1070 <pclose@plt>
  nopw   0x0(%rax,%rax,1)
L31:
  ret
  nopl   0x0(%rax)
L17:
  jmp    1060 <fclose@plt>
```

assistant: ```c
void close_file(FILE *fp, int zip) {
  if (fp && fp != stdin && fp != stdout) {
    if (zip)
      pclose(fp);
    else
      fclose(fp);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<close_file>:
  endbr64
  test   %rdi,%rdi
  je     L27
  mov    D18(%rip),%rax
  cmp    %rdi,(%rax)
  je     L27
  mov    D11(%rip),%rax
  cmp    %rdi,(%rax)
  je     L27
  test   %esi,%esi
  je     L32
  jmp    1070 <pclose@plt>
  nopw   0x0(%rax,%rax,1)
L27:
  ret
  nopl   0x0(%rax)
L32:
  jmp    1060 <fclose@plt>
```

assistant: ```c
void close_file(FILE *fp, int zip) {
  if (fp && fp != stdin && fp != stdout) {
    if (zip)
      pclose(fp);
    else
      fclose(fp);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,%rcx
  mov    %edx,%eax
  mov    %rcx,%rbx
  jmp    L27
L48:
  mov    %rsi,%rdi
  lea    0x1(%rdi),%rsi
  mov    %rcx,%rdx
  lea    0x1(%rdx),%rcx
  movzbl (%rdi),%edi
  mov    %dil,(%rdx)
  movzbl (%rdx),%edx
  test   %dl,%dl
  jne    L27
  jmp    L0
L11:
  mov    %rcx,%rdx
  lea    0x1(%rdx),%rcx
  movb   $0x0,(%rdx)
L0:
  sub    $0x1,%eax
  test   %eax,%eax
  jns    L11
L27:
  sub    $0x1,%eax
  test   %eax,%eax
  jns    L48
  mov    %rbx,%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strncpy(s1, s2, n)
register char *s1, *s2;
register int n;
{
  register char *os1 = s1;
  while (--n >= 0)
    if ((*s1++ = *s2++) == '\0')
      while (--n >= 0)
        *s1++ = '\0';
  return (os1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  sub    $0x1,%edx
  js     L61
  mov    %rdi,%rcx
  jmp    L79
L70:
  mov    %edx,%edi
  sub    $0x1,%edi
  js     L20
  lea    -0x1(%rdx),%r8d
  lea    0x2(%rcx,%r8,1),%rcx
  mov    %r9,%rdx
L33:
  add    $0x1,%rdx
  movb   $0x0,-0x1(%rdx)
  cmp    %rcx,%rdx
  jne    L33
  lea    0x1(%r9,%r8,1),%rcx
  mov    $0xffffffff,%edx
L17:
  sub    $0x1,%edx
  js     L61
L79:
  add    $0x1,%rsi
  lea    0x1(%rcx),%r9
  movzbl -0x1(%rsi),%r8d
  mov    %r8b,(%rcx)
  test   %r8b,%r8b
  je     L70
  mov    %r9,%rcx
  jmp    L17
L20:
  mov    %edi,%edx
  mov    %r9,%rcx
  jmp    L17
L61:
  ret
```

assistant: ```c
char *strncpy(s1, s2, n)
register char *s1, *s2;
register int n;
{
  register char *os1 = s1;
  while (--n >= 0)
    if ((*s1++ = *s2++) == '\0')
      while (--n >= 0)
        *s1++ = '\0';
  return (os1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strncpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  sub    $0x1,%edx
  jns    L29
  jmp    L46
L28:
  cmp    $0xffffffff,%edx
  je     L46
L29:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %edx,%ecx
  sub    $0x1,%edx
  mov    %al,-0x1(%rdi)
  test   %al,%al
  jne    L28
  cmp    $0xffffffff,%edx
  je     L46
  movslq %ecx,%rdx
  xor    %esi,%esi
  call   <memset@plt>
L46:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strncpy(s1, s2, n)
register char *s1, *s2;
register int n;
{
  register char *os1 = s1;
  while (--n >= 0)
    if ((*s1++ = *s2++) == '\0')
      while (--n >= 0)
        *s1++ = '\0';
  return (os1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strncpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  sub    $0x1,%edx
  jns    L19
  jmp    L24
L7:
  cmp    $0xffffffff,%edx
  je     L24
L19:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %edx,%ecx
  sub    $0x1,%edx
  mov    %al,-0x1(%rdi)
  test   %al,%al
  jne    L7
  cmp    $0xffffffff,%edx
  je     L24
  movslq %ecx,%rdx
  xor    %esi,%esi
  call   <memset@plt>
L24:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strncpy(s1, s2, n)
register char *s1, *s2;
register int n;
{
  register char *os1 = s1;
  while (--n >= 0)
    if ((*s1++ = *s2++) == '\0')
      while (--n >= 0)
        *s1++ = '\0';
  return (os1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  sub    $0x6b,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  jmp    L13
L2:
  movzbl -0x9(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  shl    $0x3,%eax
  and    $0x78,%eax
  or     %edx,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%edx
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%ecx
  movzbl -0xa(%rbp),%eax
  sub    %ecx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  movzbl -0xa(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0xa(%rbp)
L13:
  cmpb   $0x0,-0xa(%rbp)
  je     L2
  movzbl -0x9(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      state[local1] = state[0UL] - state[local1];
      local1++;
    }
    output[0UL] = state[0UL] + (787497705UL >> (unsigned char)255);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      state[local1] = state[0UL] - state[local1];
      local1++;
    }
    output[0UL] = state[0UL] + (787497705UL >> (unsigned char)255);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RandomFunc>:
  endbr64
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      state[local1] = state[0UL] - state[local1];
      local1++;
    }
    output[0UL] = state[0UL] + (787497705UL >> (unsigned char)255);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<RandomFunc>:
  endbr64
  movb   $0x0,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = (input[0UL] - 51238316UL) - (unsigned char)191;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 3UL;
      state[local1] = state[0UL] - state[local1];
      local1++;
    }
    output[0UL] = state[0UL] + (787497705UL >> (unsigned char)255);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<brkcom>:
  endbr64
  test   %rdi,%rdi
  je     L116
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  mov    %rcx,%r13
  movq   $0x0,(%rcx)
  movq   $0x0,(%rdx)
  movq   $0x0,(%rsi)
  cmpb   $0x28,(%rdi)
  je     L15
L117:
  cmpb   $0x5b,(%rbx)
  je     L18
L58:
  cmpb   $0x7b,(%rbx)
  je     L84
L120:
  mov    $0x0,%eax
  cmpq   $0x0,0x0(%rbp)
  je     L41
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  mov    $0x1,%eax
L14:
  lea    -0x1(%rax),%r15d
  mov    %eax,%edi
  mov    %rax,%r14
  add    $0x1,%rax
  cmpb   $0x29,-0x1(%rbx,%rax,1)
  jne    L14
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(%rbp)
  lea    0x1(%rbx),%rsi
  mov    %r15d,%edx
  call   <strncpy@plt>
  mov    0x0(%rbp),%rax
  movb   $0x0,-0x1(%rax,%r14,1)
  lea    0x1(%rbx,%r14,1),%rbx
  jmp    L117
L18:
  mov    $0x1,%eax
L44:
  lea    -0x1(%rax),%r15d
  mov    %eax,%edi
  mov    %rax,%r14
  add    $0x1,%rax
  cmpb   $0x5d,-0x1(%rbx,%rax,1)
  jne    L44
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r12)
  lea    0x1(%rbx),%rsi
  mov    %r15d,%edx
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r14,1)
  lea    0x1(%rbx,%r14,1),%rbx
  jmp    L58
L84:
  mov    $0x1,%eax
L61:
  lea    -0x1(%rax),%r15d
  mov    %eax,%edi
  mov    %rax,%r14
  add    $0x1,%rax
  cmpb   $0x7d,-0x1(%rbx,%rax,1)
  jne    L61
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(%r13)
  lea    0x1(%rbx),%rsi
  mov    %r15d,%edx
  call   <strncpy@plt>
  mov    0x0(%r13),%rax
  movb   $0x0,-0x1(%rax,%r14,1)
  jmp    L120
L41:
  cmpq   $0x0,(%r12)
  jne    L22
  cmpq   $0x0,0x0(%r13)
  sete   %al
  movzbl %al,%eax
  add    %eax,%eax
  jmp    L22
L116:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int brkcom(char *com, char **sin, char **mb, char **me) {
  if (com == NULL)
    return 1;
  char *comp = com;
  int cc;
  *sin = *mb = *me = NULL;
  if (*comp == '(') {
    cc = 0;
    while (*(comp + cc) != ')')
      cc++;
    *sin = malloc(cc);
    strncpy(*sin, comp + 1, cc - 1);
    *(*sin + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '[') {
    cc = 0;
    while (*(comp + cc) != ']')
      cc++;
    *mb = malloc(cc);
    strncpy(*mb, comp + 1, cc - 1);
    *(*mb + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '{') {
    cc = 0;
    while (*(comp + cc) != '}')
      cc++;
    *me = malloc(cc);
    strncpy(*me, comp + 1, cc - 1);
    *(*me + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*sin == NULL && *mb == NULL && *me == NULL)
    return 2;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<brkcom>:
  endbr64
  test   %rdi,%rdi
  je     L53
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rcx)
  movq   $0x0,(%rdx)
  movq   $0x0,(%rsi)
  movzbl (%rdi),%eax
  cmp    $0x28,%al
  je     L108
L102:
  cmp    $0x5b,%al
  je     L20
L22:
  cmp    $0x7b,%al
  je     L86
L76:
  xor    %eax,%eax
  cmpq   $0x0,(%r12)
  je     L81
L38:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L81:
  cmpq   $0x0,0x0(%r13)
  jne    L38
  xor    %eax,%eax
  cmpq   $0x0,(%r14)
  sete   %al
  add    $0x18,%rsp
  pop    %rbx
  add    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L86:
  mov    $0x1,%eax
  nopl   (%rax)
L30:
  lea    -0x1(%rax),%r15d
  mov    %eax,%edi
  mov    %rax,%rbp
  add    $0x1,%rax
  cmpb   $0x7d,-0x1(%rbx,%rax,1)
  jne    L30
  call   <malloc@plt>
  lea    0x1(%rbx),%rsi
  mov    %r15d,%edx
  mov    %rax,(%r14)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    (%r14),%rax
  movb   $0x0,-0x1(%rax,%rbp,1)
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0x1,%ebp
  nopl   (%rax)
L31:
  lea    -0x1(%rbp),%edx
  mov    %ebp,%edi
  mov    %rbp,%r15
  add    $0x1,%rbp
  cmpb   $0x5d,-0x1(%rbx,%rbp,1)
  jne    L31
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  lea    0x1(%rbx),%rsi
  add    %rbp,%rbx
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    0x0(%r13),%rax
  movb   $0x0,-0x1(%rax,%r15,1)
  movzbl (%rbx),%eax
  jmp    L22
  nopl   (%rax)
L108:
  mov    $0x1,%ebp
  nopl   (%rax)
L4:
  lea    -0x1(%rbp),%edx
  mov    %ebp,%edi
  mov    %rbp,%r15
  add    $0x1,%rbp
  cmpb   $0x29,-0x1(%rbx,%rbp,1)
  jne    L4
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  lea    0x1(%rbx),%rsi
  add    %rbp,%rbx
  mov    %rax,(%r12)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r15,1)
  movzbl (%rbx),%eax
  jmp    L102
  nopl   (%rax)
L53:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int brkcom(char *com, char **sin, char **mb, char **me) {
  if (com == NULL)
    return 1;
  char *comp = com;
  int cc;
  *sin = *mb = *me = NULL;
  if (*comp == '(') {
    cc = 0;
    while (*(comp + cc) != ')')
      cc++;
    *sin = malloc(cc);
    strncpy(*sin, comp + 1, cc - 1);
    *(*sin + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '[') {
    cc = 0;
    while (*(comp + cc) != ']')
      cc++;
    *mb = malloc(cc);
    strncpy(*mb, comp + 1, cc - 1);
    *(*mb + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '{') {
    cc = 0;
    while (*(comp + cc) != '}')
      cc++;
    *me = malloc(cc);
    strncpy(*me, comp + 1, cc - 1);
    *(*me + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*sin == NULL && *mb == NULL && *me == NULL)
    return 2;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<brkcom>:
  endbr64
  test   %rdi,%rdi
  je     L59
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movq   $0x0,(%rcx)
  movq   $0x0,(%rdx)
  movq   $0x0,(%rsi)
  movzbl (%rdi),%eax
  cmp    $0x28,%al
  je     L77
L86:
  cmp    $0x5b,%al
  je     L74
L33:
  cmp    $0x7b,%al
  je     L92
L99:
  xor    %eax,%eax
  cmpq   $0x0,(%r12)
  je     L85
L109:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L85:
  cmpq   $0x0,0x0(%r13)
  jne    L109
  xor    %eax,%eax
  cmpq   $0x0,(%r14)
  sete   %al
  add    $0x18,%rsp
  pop    %rbx
  add    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L92:
  mov    $0x1,%eax
  nopl   (%rax)
L106:
  lea    -0x1(%rax),%r15d
  mov    %eax,%edi
  mov    %rax,%rbp
  add    $0x1,%rax
  cmpb   $0x7d,-0x1(%rbx,%rax,1)
  jne    L106
  call   <malloc@plt>
  lea    0x1(%rbx),%rsi
  mov    %r15d,%edx
  mov    %rax,(%r14)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    (%r14),%rax
  movb   $0x0,-0x1(%rax,%rbp,1)
  jmp    L99
  nopl   0x0(%rax,%rax,1)
L74:
  mov    $0x1,%ebp
  nopl   (%rax)
L80:
  lea    -0x1(%rbp),%edx
  mov    %ebp,%edi
  mov    %rbp,%r15
  add    $0x1,%rbp
  cmpb   $0x5d,-0x1(%rbx,%rbp,1)
  jne    L80
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  lea    0x1(%rbx),%rsi
  add    %rbp,%rbx
  mov    %rax,0x0(%r13)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    0x0(%r13),%rax
  movb   $0x0,-0x1(%rax,%r15,1)
  movzbl (%rbx),%eax
  jmp    L33
  nopl   (%rax)
L77:
  mov    $0x1,%ebp
  nopl   (%rax)
L5:
  lea    -0x1(%rbp),%edx
  mov    %ebp,%edi
  mov    %rbp,%r15
  add    $0x1,%rbp
  cmpb   $0x29,-0x1(%rbx,%rbp,1)
  jne    L5
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  lea    0x1(%rbx),%rsi
  add    %rbp,%rbx
  mov    %rax,(%r12)
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    (%r12),%rax
  movb   $0x0,-0x1(%rax,%r15,1)
  movzbl (%rbx),%eax
  jmp    L86
  nopl   (%rax)
L59:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int brkcom(char *com, char **sin, char **mb, char **me) {
  if (com == NULL)
    return 1;
  char *comp = com;
  int cc;
  *sin = *mb = *me = NULL;
  if (*comp == '(') {
    cc = 0;
    while (*(comp + cc) != ')')
      cc++;
    *sin = malloc(cc);
    strncpy(*sin, comp + 1, cc - 1);
    *(*sin + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '[') {
    cc = 0;
    while (*(comp + cc) != ']')
      cc++;
    *mb = malloc(cc);
    strncpy(*mb, comp + 1, cc - 1);
    *(*mb + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*comp == '{') {
    cc = 0;
    while (*(comp + cc) != '}')
      cc++;
    *me = malloc(cc);
    strncpy(*me, comp + 1, cc - 1);
    *(*me + cc - 1) = '\0';
    comp += cc + 1;
  }
  if (*sin == NULL && *mb == NULL && *me == NULL)
    return 2;
  else
    return 0;
}
```