ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<modele_frottements>:
  endbr64
  movsd  D3(%rip),%xmm0
  movapd %xmm0,%xmm1
  mulsd  (%rsi),%xmm1
  movsd  %xmm1,(%rdx)
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
data_type: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

assistant: ```c
void modele_frottements(const double q[2], const double qdot[2], double F[2]) {
  F[0] = 50 * qdot[0];
  F[1] = 50 * qdot[1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<modele_frottements>:
  endbr64
  movsd  (%rsi),%xmm1
  movsd  D9(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
type_guessed: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

assistant: ```c
void modele_frottements(const double q[2], const double qdot[2], double F[2]) {
  F[0] = 50 * qdot[0];
  F[1] = 50 * qdot[1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<modele_frottements>:
  endbr64
  movsd  (%rsi),%xmm1
  movsd  D4(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  ret
```

tool_calls: name: read_data
arguments: label: D4
type: byte[8]

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x49,0x40]}

assistant: ```c
void modele_frottements(const double q[2], const double qdot[2], double F[2]) {
  F[0] = 50 * qdot[0];
  F[1] = 50 * qdot[1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L29
L13:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  movzbl %al,%eax
  mov    -0x24(%rbp),%edx
  add    %edx,%edx
  movslq %edx,%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0xf,%eax
  mov    %eax,%ecx
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  movzbl -0x20(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x24(%rbp)
L29:
  mov    -0x24(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L13
  mov    -0x24(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
void dump_hex(const char *data, char *buf, int datasize) {
  const char HEXSTR[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  int i;
  for (i = 0; i < datasize; ++i) {
    buf[i * 2] = HEXSTR[(unsigned char)data[i] >> 4];
    buf[i * 2 + 1] = HEXSTR[(unsigned char)data[i] & 0x0f];
  }
  buf[i * 2] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_hex>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%r9
  movabs $0x6665646362613938,%r10
  mov    %r9,(%rsp)
  mov    %r10,0x8(%rsp)
  test   %edx,%edx
  jle    L1
  mov    %rdi,%r8
  mov    %edx,%edi
L26:
  movzbl (%r8,%rax,1),%ecx
  shr    $0x4,%cl
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,(%rsi,%rax,2)
  movzbl (%r8,%rax,1),%ecx
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,0x1(%rsi,%rax,2)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L26
L23:
  add    %edx,%edx
  movslq %edx,%rdx
  movb   $0x0,(%rsi,%rdx,1)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x28,%rsp
  ret
L1:
  mov    $0x0,%edx
  jmp    L23
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void dump_hex(const char *data, char *buf, int datasize) {
  const char HEXSTR[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  int i;
  for (i = 0; i < datasize; ++i) {
    buf[i * 2] = HEXSTR[(unsigned char)data[i] >> 4];
    buf[i * 2 + 1] = HEXSTR[(unsigned char)data[i] & 0x0f];
  }
  buf[i * 2] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dump_hex>:
  endbr64
  sub    $0x28,%rsp
  movdqa D3(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  test   %edx,%edx
  jle    L19
  mov    %rdi,%r8
  movslq %edx,%rdi
  xchg   %ax,%ax
L9:
  movzbl (%r8,%rax,1),%ecx
  shr    $0x4,%cl
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,(%rsi,%rax,2)
  movzbl (%r8,%rax,1),%ecx
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,0x1(%rsi,%rax,2)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L9
  add    %edx,%edx
  movslq %edx,%rdx
  add    %rdx,%rsi
L19:
  movb   $0x0,(%rsi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x28,%rsp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: byte[16]

tool: {"D3": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void dump_hex(const char *data, char *buf, int datasize) {
  const char HEXSTR[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  int i;
  for (i = 0; i < datasize; ++i) {
    buf[i * 2] = HEXSTR[(unsigned char)data[i] >> 4];
    buf[i * 2 + 1] = HEXSTR[(unsigned char)data[i] & 0x0f];
  }
  buf[i * 2] = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_hex>:
  endbr64
  sub    $0x28,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  test   %edx,%edx
  jle    L24
  mov    %rdi,%r8
  movslq %edx,%rdi
  xchg   %ax,%ax
L23:
  movzbl (%r8,%rax,1),%ecx
  shr    $0x4,%cl
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,(%rsi,%rax,2)
  movzbl (%r8,%rax,1),%ecx
  and    $0xf,%ecx
  movzbl (%rsp,%rcx,1),%ecx
  mov    %cl,0x1(%rsi,%rax,2)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L23
  add    %edx,%edx
  movslq %edx,%rdx
  add    %rdx,%rsi
L24:
  movb   $0x0,(%rsi)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x28,%rsp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D1
type_guessed: byte[16]

tool: {"D1": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void dump_hex(const char *data, char *buf, int datasize) {
  const char HEXSTR[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f'};
  int i;
  for (i = 0; i < datasize; ++i) {
    buf[i * 2] = HEXSTR[(unsigned char)data[i] >> 4];
    buf[i * 2 + 1] = HEXSTR[(unsigned char)data[i] & 0x0f];
  }
  buf[i * 2] = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cec_msg_text_view_on>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x2,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    0x4(%rax),%rdx
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cec_msg_text_view_on(struct cec_msg *msg) {
  msg->len = 2;
  msg->msg[1] = CEC_MSG_TEXT_VIEW_ON;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cec_msg_text_view_on>:
  endbr64
  movl   $0x2,(%rdi)
  mov    0x8(%rdi),%rax
  mov    D2(%rip),%rdx
  mov    (%rdx),%edx
  mov    %edx,0x4(%rax)
  ret
```

assistant: ```c
void cec_msg_text_view_on(struct cec_msg *msg) {
  msg->len = 2;
  msg->msg[1] = CEC_MSG_TEXT_VIEW_ON;
}
```
user: What is the c source code of the assembly code below?

```asm
<cec_msg_text_view_on>:
  endbr64
  mov    D6(%rip),%rdx
  mov    0x8(%rdi),%rax
  movl   $0x2,(%rdi)
  mov    (%rdx),%edx
  mov    %edx,0x4(%rax)
  ret
```

assistant: ```c
void cec_msg_text_view_on(struct cec_msg *msg) {
  msg->len = 2;
  msg->msg[1] = CEC_MSG_TEXT_VIEW_ON;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cec_msg_text_view_on>:
  endbr64
  mov    D4(%rip),%rdx
  mov    0x8(%rdi),%rax
  movl   $0x2,(%rdi)
  mov    (%rdx),%edx
  mov    %edx,0x4(%rax)
  ret
```

assistant: ```c
void cec_msg_text_view_on(struct cec_msg *msg) {
  msg->len = 2;
  msg->msg[1] = CEC_MSG_TEXT_VIEW_ON;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<date_get_min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <localtime@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int date_get_min(void) {
  time_t t = time(NULL);
  struct tm *lt = localtime(&t);
  return lt->tm_min;
}
```
user: What is the c source code of the assembly code below?

```asm
<date_get_min>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <localtime@plt>
  mov    (%rax),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int date_get_min(void) {
  time_t t = time(NULL);
  struct tm *lt = localtime(&t);
  return lt->tm_min;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<date_get_min>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  mov    (%rax),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int date_get_min(void) {
  time_t t = time(NULL);
  struct tm *lt = localtime(&t);
  return lt->tm_min;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<date_get_min>:
  endbr64
  sub    $0x18,%rsp
  xor    %edi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  lea    0x4(%rsp),%rdi
  mov    %eax,0x4(%rsp)
  call   <localtime@plt>
  mov    (%rax),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int date_get_min(void) {
  time_t t = time(NULL);
  struct tm *lt = localtime(&t);
  return lt->tm_min;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ustrvprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x8,-0x14(%rbp)
L7:
  shll   -0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <urealloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <va_copy@plt>
  mov    -0x10(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x14(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L7
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *ustrvprintf(const char *fmt, va_list l) {
  char *buf = NULL;
  int len = 8, ret;
  do {
    va_list lcp;
    len *= 2;
    buf = urealloc(buf, len);
    va_copy(lcp, l);
    ret = vsnprintf(buf, len, fmt, lcp);
    va_end(lcp);
  } while (ret >= len);
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ustrvprintf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r15
  mov    %esi,%r14d
  mov    $0x8,%ebp
  mov    $0x0,%ebx
L5:
  add    %ebp,%ebp
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <urealloc@plt>
  mov    %rax,%rbx
  mov    %r14d,%esi
  mov    $0x0,%r13d
  mov    %r13d,%edi
  call   <va_copy@plt>
  mov    %r13d,%ecx
  mov    %r15,%rdx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vsnprintf@plt>
  mov    %eax,%r12d
  mov    %r13d,%edi
  call   <va_end@plt>
  cmp    %r12d,%ebp
  jle    L5
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *ustrvprintf(const char *fmt, va_list l) {
  char *buf = NULL;
  int len = 8, ret;
  do {
    va_list lcp;
    len *= 2;
    buf = urealloc(buf, len);
    va_copy(lcp, l);
    ret = vsnprintf(buf, len, fmt, lcp);
    va_end(lcp);
  } while (ret >= len);
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ustrvprintf>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x8,%ebx
  sub    $0x8,%rsp
  nop
L8:
  add    %ebx,%ebx
  mov    %rbp,%rdi
  xor    %r15d,%r15d
  mov    %ebx,%esi
  call   <urealloc@plt>
  mov    %r13d,%esi
  mov    %r15d,%edi
  mov    %rax,%rbp
  call   <va_copy@plt>
  mov    %rbp,%rdi
  mov    %r15d,%ecx
  mov    %r14,%rdx
  mov    %ebx,%esi
  call   <vsnprintf@plt>
  mov    %r15d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  cmp    %r12d,%ebx
  jle    L8
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *ustrvprintf(const char *fmt, va_list l) {
  char *buf = NULL;
  int len = 8, ret;
  do {
    va_list lcp;
    len *= 2;
    buf = urealloc(buf, len);
    va_copy(lcp, l);
    ret = vsnprintf(buf, len, fmt, lcp);
    va_end(lcp);
  } while (ret >= len);
  return buf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ustrvprintf>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x8,%ebx
  sub    $0x8,%rsp
  nop
L3:
  add    %ebx,%ebx
  mov    %rbp,%rdi
  xor    %r15d,%r15d
  mov    %ebx,%esi
  call   <urealloc@plt>
  mov    %r13d,%esi
  mov    %r15d,%edi
  mov    %rax,%rbp
  call   <va_copy@plt>
  mov    %rbp,%rdi
  mov    %r15d,%ecx
  mov    %r14,%rdx
  mov    %ebx,%esi
  call   <vsnprintf@plt>
  mov    %r15d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  cmp    %r12d,%ebx
  jle    L3
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *ustrvprintf(const char *fmt, va_list l) {
  char *buf = NULL;
  int len = 8, ret;
  do {
    va_list lcp;
    len *= 2;
    buf = urealloc(buf, len);
    va_copy(lcp, l);
    ret = vsnprintf(buf, len, fmt, lcp);
    va_end(lcp);
  } while (ret >= len);
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<divisionwithmodule>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfd0,%rsp
  mov    %edi,-0xfc4(%rbp)
  mov    %esi,-0xfc8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0xfc4(%rbp),%eax
  mov    %eax,-0xfbc(%rbp)
  movl   $0x0,-0xfb4(%rbp)
  movq   $0x0,-0xfb0(%rbp)
  movq   $0x0,-0xfa8(%rbp)
  lea    -0xfa0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x1f2,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0xfb8(%rbp)
  jmp    L4
L48:
  mov    -0xfbc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0xfbc(%rbp)
  mov    -0xfbc(%rbp),%eax
  cltd
  idivl  -0xfc8(%rbp)
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    %dl,-0xfb0(%rbp,%rax,1)
  mov    -0xfb8(%rbp),%eax
  add    $0x1,%eax
  cltq
  movb   $0x0,-0xfb0(%rbp,%rax,1)
  mov    -0xfbc(%rbp),%eax
  cltd
  idivl  -0xfc8(%rbp)
  mov    %edx,-0xfbc(%rbp)
  addl   $0x1,-0xfb8(%rbp)
L4:
  cmpl   $0x0,-0xfbc(%rbp)
  je     L1
  cmpl   $0x0,-0xfb4(%rbp)
  jne    L1
  cmpl   $0xf9f,-0xfb8(%rbp)
  jle    L48
L1:
  lea    -0xfb0(%rbp),%rax
  mov    %rax,%rdi
  call   <maxRepetition@plt>
  mov    %eax,-0xfb4(%rbp)
  mov    -0xfb4(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
int divisionwithmodule(int dividend, int divisor) {
  int aux = dividend;
  int rep = 0;
  char decimals[4000] = {'\x0'};
  int i = 0;
  while (aux && !rep && i < 4000) {
    aux *= 10;
    decimals[i] = (aux / divisor + '0');
    decimals[i + 1] = '\x0';
    aux = aux % divisor;
    i++;
  }
  rep = maxRepetition(decimals);
  return rep;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<divisionwithmodule>:
  endbr64
  sub    $0xfb8,%rsp
  mov    %edi,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x1f2,%ecx
  rep stos %rax,%es:(%rdi)
  test   %r8d,%r8d
  je     L29
  mov    %rsp,%rax
  mov    %rax,%rcx
  mov    $0x1,%r9d
  sub    %eax,%r9d
L7:
  lea    (%r8,%r8,4),%eax
  add    %eax,%eax
  cltd
  idiv   %esi
  add    $0x30,%eax
  mov    %al,(%rcx)
  movb   $0x0,0x1(%rcx)
  mov    %edx,%r8d
  lea    (%r9,%rcx,1),%eax
  cmp    $0xf9f,%eax
  setle  %dil
  test   %edx,%edx
  setne  %al
  add    $0x1,%rcx
  test   %al,%dil
  jne    L7
L29:
  mov    %rsp,%rdi
  call   <maxRepetition@plt>
  mov    0xfa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0xfb8,%rsp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int divisionwithmodule(int dividend, int divisor) {
  int aux = dividend;
  int rep = 0;
  char decimals[4000] = {'\x0'};
  int i = 0;
  while (aux && !rep && i < 4000) {
    aux *= 10;
    decimals[i] = (aux / divisor + '0');
    decimals[i + 1] = '\x0';
    aux = aux % divisor;
    i++;
  }
  rep = maxRepetition(decimals);
  return rep;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<divisionwithmodule>:
  endbr64
  sub    $0xfb8,%rsp
  mov    %edi,%edx
  pxor   %xmm0,%xmm0
  mov    $0x1f2,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movaps %xmm0,(%rsp)
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%rdi
  mov    %rdi,%rcx
  test   %edx,%edx
  je     L3
  nop
L9:
  lea    (%rdx,%rdx,4),%eax
  movb   $0x0,0x1(%rcx)
  add    $0x1,%rcx
  add    %eax,%eax
  cltd
  idiv   %esi
  add    $0x30,%eax
  mov    %al,-0x1(%rcx)
  test   %edx,%edx
  jne    L9
L3:
  call   <maxRepetition@plt>
  mov    0xfa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0xfb8,%rsp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int divisionwithmodule(int dividend, int divisor) {
  int aux = dividend;
  int rep = 0;
  char decimals[4000] = {'\x0'};
  int i = 0;
  while (aux && !rep && i < 4000) {
    aux *= 10;
    decimals[i] = (aux / divisor + '0');
    decimals[i + 1] = '\x0';
    aux = aux % divisor;
    i++;
  }
  rep = maxRepetition(decimals);
  return rep;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divisionwithmodule>:
  endbr64
  sub    $0xfb8,%rsp
  mov    %edi,%edx
  pxor   %xmm0,%xmm0
  mov    $0x1f2,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  movaps %xmm0,(%rsp)
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%rdi
  mov    %rdi,%rcx
  test   %edx,%edx
  je     L7
  nop
L15:
  lea    (%rdx,%rdx,4),%eax
  movb   $0x0,0x1(%rcx)
  add    $0x1,%rcx
  add    %eax,%eax
  cltd
  idiv   %esi
  add    $0x30,%eax
  mov    %al,-0x1(%rcx)
  test   %edx,%edx
  jne    L15
L7:
  call   <maxRepetition@plt>
  mov    0xfa8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0xfb8,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int divisionwithmodule(int dividend, int divisor) {
  int aux = dividend;
  int rep = 0;
  char decimals[4000] = {'\x0'};
  int i = 0;
  while (aux && !rep && i < 4000) {
    aux *= 10;
    decimals[i] = (aux / divisor + '0');
    decimals[i + 1] = '\x0';
    aux = aux % divisor;
    i++;
  }
  rep = maxRepetition(decimals);
  return rep;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<namematches>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L42
L56:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5b,%al
  jne    L43
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L43
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <charmatches@plt>
  test   %rax,%rax
  je     L51
  mov    $0x1,%eax
  jmp    L80
L43:
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L51
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jne    L51
  mov    $0x1,%eax
  jmp    L80
L51:
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x7c,%al
  jne    L42
  addq   $0x1,-0x20(%rbp)
L42:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <subexplen@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L56
  mov    $0x0,%eax
L80:
  leave
  ret
```

assistant: ```c
int namematches(const char *name, const char *pattern) {
  int n;
  while (n = subexplen(pattern)) {
    if (*pattern == '[' && !name[1]) {
      if (charmatches(*name, pattern))
        return 1;
    } else if (!strncmp(name, pattern, n) && n == strlen(name))
      return 1;
    pattern += n;
    if (*pattern == '|')
      pattern++;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<namematches>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  jmp    L48
L6:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  je     L10
L16:
  movslq %ebx,%rbx
  add    %rbx,%rbp
  cmpb   $0x7c,0x0(%rbp)
  sete   %al
  movzbl %al,%eax
  add    %rax,%rbp
L48:
  mov    %rbp,%rdi
  call   <subexplen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L44
  cmpb   $0x5b,0x0(%rbp)
  jne    L6
  cmpb   $0x0,0x1(%r12)
  jne    L6
  movsbl (%r12),%edi
  mov    %rbp,%rsi
  call   <charmatches@plt>
  test   %rax,%rax
  je     L16
  mov    $0x1,%ebx
  jmp    L44
L10:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jne    L16
  mov    $0x1,%ebx
L44:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int namematches(const char *name, const char *pattern) {
  int n;
  while (n = subexplen(pattern)) {
    if (*pattern == '[' && !name[1]) {
      if (charmatches(*name, pattern))
        return 1;
    } else if (!strncmp(name, pattern, n) && n == strlen(name))
      return 1;
    pattern += n;
    if (*pattern == '|')
      pattern++;
  }
  return 0;
}
```