ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<readfile>:
  endbr64
  push   %r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  lea    D0(%rip),%rsi
  push   %rbx
  sub    $0x18,%rsp
  movl   $0x0,(%r9)
  mov    %r9,0x8(%rsp)
  call   <fopen@plt>
  test   %rax,%rax
  je     L21
  mov    (%r12),%edi
  test   %edi,%edi
  jle    L32
  movslq 0x0(%rbp),%rdx
  mov    %rax,%rbx
  xor    %r15d,%r15d
  jmp    L34
  nop
L3:
  mov    (%r12),%edi
  add    $0x1,%r15d
  cmp    %r15d,%edi
  jle    L32
L34:
  imul   (%r14),%edi
  mov    %rbx,%rcx
  mov    $0x1,%esi
  movslq %edi,%rdi
  add    %rdx,%rdi
  add    %r13,%rdi
  call   <fread@plt>
  movslq 0x0(%rbp),%rdx
  cmp    %rdx,%rax
  je     L3
  mov    0x8(%rsp),%rax
  mov    %r15d,(%rax)
L32:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  mov    0x8(%rsp),%rax
  movl   $0xffffffff,(%rax)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
datatype: string

tool: D0: rb

assistant: ```c
void readfile(const char *filename, const int *m, const int *n, char *A, const int *ldA, int *info) {
  *info = 0;
  FILE *fin = fopen(filename, "rb");
  if (!fin) {
    *info = -1;
    return;
  }
  int j;
  for (j = 0; j < *n; j++)
    if (fread(A + *m + *n * *ldA, sizeof(char), (size_t)*m, fin) != (size_t)*m) {
      *info = j;
      return;
    }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    $0x2,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
int *fx(int *b, int *e) {
  long unsigned int p = e - b;
  return b + p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fx>:
  endbr64
  mov    %rsi,%rax
  ret
```

assistant: ```c
int *fx(int *b, int *e) {
  long unsigned int p = e - b;
  return b + p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fx>:
  endbr64
  mov    %rsi,%rax
  ret
```

assistant: ```c
int *fx(int *b, int *e) {
  long unsigned int p = e - b;
  return b + p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fx>:
  endbr64
  mov    %rsi,%rax
  ret
```

assistant: ```c
int *fx(int *b, int *e) {
  long unsigned int p = e - b;
  return b + p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<proc_names>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L13
L0:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  je     L38
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L39
L38:
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <add_name@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  jmp    L27
L13:
  nop
L27:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
data_type: string

tool: D1:  

assistant: ```c
void proc_names(const char *chan, char *names) {
  char *p;
  if (!(p = strtok(names, " ")))
    return;
  do {
    if (*p == '@' || *p == '+')
      p++;
    add_name(chan, p);
  } while ((p = strtok(NULL, " ")));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<proc_names>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  lea    D4(%rip),%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L43
  mov    %rax,%rsi
  lea    D4(%rip),%rbp
  jmp    L31
L41:
  add    $0x1,%rsi
L32:
  mov    %rbx,%rdi
  call   <add_name@plt>
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L43
L31:
  movzbl (%rsi),%eax
  cmp    $0x40,%al
  je     L41
  cmp    $0x2b,%al
  jne    L32
  jmp    L41
L43:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type_infered: string

tool: D4:  

assistant: ```c
void proc_names(const char *chan, char *names) {
  char *p;
  if (!(p = strtok(names, " ")))
    return;
  do {
    if (*p == '@' || *p == '+')
      p++;
    add_name(chan, p);
  } while ((p = strtok(NULL, " ")));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<proc_names>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L30
  mov    %rax,%rsi
  jmp    L3
  nopl   0x0(%rax)
L32:
  mov    %rbx,%rdi
  call   <add_name@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L30
L3:
  movzbl (%rsi),%eax
  cmp    $0x40,%al
  je     L27
  cmp    $0x2b,%al
  jne    L32
L27:
  add    $0x1,%rsi
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type: string

tool: D5:  

assistant: ```c
void proc_names(const char *chan, char *names) {
  char *p;
  if (!(p = strtok(names, " ")))
    return;
  do {
    if (*p == '@' || *p == '+')
      p++;
    add_name(chan, p);
  } while ((p = strtok(NULL, " ")));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<proc_names>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L47
  mov    %rax,%rsi
  jmp    L21
  nopl   0x0(%rax)
L19:
  mov    %rbx,%rdi
  call   <add_name@plt>
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L47
L21:
  movzbl (%rsi),%eax
  cmp    $0x40,%al
  je     L48
  cmp    $0x2b,%al
  jne    L19
L48:
  add    $0x1,%rsi
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L47:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string

tool: D2:  

assistant: ```c
void proc_names(const char *chan, char *names) {
  char *p;
  if (!(p = strtok(names, " ")))
    return;
  do {
    if (*p == '@' || *p == '+')
      p++;
    add_name(chan, p);
  } while ((p = strtok(NULL, " ")));
}
```
user: What is the c source code of the assembly code below?

```asm
<isincr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0xb,-0x4(%rbp)
  jmp    L10
L26:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,-0x4(%rbp)
  jge    L12
  mov    $0x0,%eax
  jmp    L15
L12:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
L10:
  cmpl   $0x0,-0x14(%rbp)
  jne    L26
  mov    $0x1,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int isincr(int num) {
  int prev;
  prev = 11;
  while (num) {
    if (prev < (num % 10))
      return 0;
    prev = num % 10;
    num /= 10;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<isincr>:
  endbr64
  mov    %edi,%ecx
  test   %edi,%edi
  je     L2
  movslq %edi,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %ecx,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  lea    (%rdi,%rdi,4),%eax
  add    %eax,%eax
  mov    %ecx,%edi
  sub    %eax,%edi
L20:
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  je     L32
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  cmp    %edi,%eax
  jg     L15
  mov    %eax,%edi
  jmp    L20
L32:
  mov    $0x1,%eax
  ret
L2:
  mov    $0x1,%eax
  ret
L15:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int isincr(int num) {
  int prev;
  prev = 11;
  while (num) {
    if (prev < (num % 10))
      return 0;
    prev = num % 10;
    num /= 10;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<isincr>:
  endbr64
  mov    %edi,%ecx
  test   %edi,%edi
  je     L11
  movslq %edi,%rax
  mov    %ecx,%edx
  imul   $0x66666667,%rax,%rdi
  sar    $0x1f,%edx
  sar    $0x22,%rdi
  sub    %edx,%edi
  lea    (%rdi,%rdi,4),%edx
  mov    %ecx,%edi
  add    %edx,%edx
  sub    %edx,%edi
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L37:
  movslq %ecx,%rax
  mov    %ecx,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    %esi,%esi
  sub    %esi,%edx
  cmp    %edi,%edx
  jg     L2
  mov    %edx,%edi
L26:
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%ecx
  jne    L37
L11:
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int isincr(int num) {
  int prev;
  prev = 11;
  while (num) {
    if (prev < (num % 10))
      return 0;
    prev = num % 10;
    num /= 10;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isincr>:
  endbr64
  mov    %edi,%ecx
  test   %edi,%edi
  je     L6
  movslq %edi,%rax
  mov    %ecx,%edx
  imul   $0x66666667,%rax,%rdi
  sar    $0x1f,%edx
  sar    $0x22,%rdi
  sub    %edx,%edi
  lea    (%rdi,%rdi,4),%edx
  mov    %ecx,%edi
  add    %edx,%edx
  sub    %edx,%edi
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L20:
  movslq %ecx,%rax
  mov    %ecx,%esi
  imul   $0x66666667,%rax,%rdx
  sar    $0x1f,%esi
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  mov    %ecx,%edx
  add    %esi,%esi
  sub    %esi,%edx
  cmp    %edi,%edx
  jg     L4
  mov    %edx,%edi
L32:
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%ecx
  jne    L20
L6:
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int isincr(int num) {
  int prev;
  prev = 11;
  while (num) {
    if (prev < (num % 10))
      return 0;
    prev = num % 10;
    num /= 10;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rbx
  mov    -0x18(%rbp),%r12
  cmp    -0x20(%rbp),%rbx
  jne    L76
  cmp    -0x18(%rbp),%r12
  jne    L76
  jmp    L9
L20:
  mov    %rbx,%rdx
  lea    0x4(%rdx),%rbx
  mov    %r12,%rax
  lea    0x4(%rax),%r12
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  subl   $0x8,-0x24(%rbp)
L9:
  cmpl   $0x7,-0x24(%rbp)
  jg     L20
  cmpl   $0x0,-0x24(%rbp)
  je     L63
  jmp    L76
L31:
  mov    -0x20(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x18(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L76:
  subl   $0x1,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L31
  jmp    L43
L63:
  nop
L43:
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
memcpy(to, from, count) char *from, *to;
{
  register *f, *t;
  f = (int *)from;
  t = (int *)to;
  if ((char *)f == from && (char *)t == to) {
    while (count >= 8)
      *t++ = *f++, count -= 8;
    if (!count)
      return;
  }
  while (--count >= 0)
    *to++ = *from++;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<memcpy>:
  endbr64
  cmp    $0x7,%edx
  jle    L36
  lea    -0x8(%rdx),%r9d
  shr    $0x3,%r9d
  lea    0x1(%r9),%r8d
  mov    %r8d,%r8d
  shl    $0x2,%r8
  mov    $0x0,%eax
L3:
  mov    (%rsi,%rax,1),%ecx
  mov    %ecx,(%rdi,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L3
  neg    %r9d
  lea    -0x8(%rdx,%r9,8),%edx
L36:
  test   %edx,%edx
  jle    L10
  mov    %edx,%edx
  mov    $0x0,%eax
L32:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L32
L10:
  ret
```

assistant: ```c
memcpy(to, from, count) char *from, *to;
{
  register *f, *t;
  f = (int *)from;
  t = (int *)to;
  if ((char *)f == from && (char *)t == to) {
    while (count >= 8)
      *t++ = *f++, count -= 8;
    if (!count)
      return;
  }
  while (--count >= 0)
    *to++ = *from++;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rcx
  cmp    $0x7,%edx
  jle    L38
  sub    $0x8,%edx
  xor    %eax,%eax
  mov    %edx,%r9d
  shr    $0x3,%r9d
  lea    0x1(%r9),%r8d
  shl    $0x2,%r8
L2:
  mov    (%rsi,%rax,1),%edi
  mov    %edi,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L2
  neg    %r9d
  lea    (%rdx,%r9,8),%edx
L38:
  test   %edx,%edx
  jle    L25
  movslq %edx,%rdx
  xor    %eax,%eax
  nop
L26:
  movzbl (%rsi,%rax,1),%edi
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L26
L25:
  ret
```

assistant: ```c
memcpy(to, from, count) char *from, *to;
{
  register *f, *t;
  f = (int *)from;
  t = (int *)to;
  if ((char *)f == from && (char *)t == to) {
    while (count >= 8)
      *t++ = *f++, count -= 8;
    if (!count)
      return;
  }
  while (--count >= 0)
    *to++ = *from++;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<signal_handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <debugf@plt>
  mov    $0x0,%eax
  call   <kill_all@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <kill@plt>
  mov    $0x40,%edi
  call   <_exit@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type_infered: string

tool: D7: reaper: signal %d

assistant: ```c
void signal_handler(int signum) {
  debugf("reaper: signal %d", signum);
  kill_all();
  kill(getpid(), signum);
  _exit(64);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<signal_handler>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    %edi,%esi
  lea    D3(%rip),%rdi
  call   <debugf@plt>
  mov    $0x0,%eax
  call   <kill_all@plt>
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  mov    %ebx,%esi
  call   <kill@plt>
  mov    $0x40,%edi
  call   <_exit@plt>
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type: string

tool: D3: reaper: signal %d

assistant: ```c
void signal_handler(int signum) {
  debugf("reaper: signal %d", signum);
  kill_all();
  kill(getpid(), signum);
  _exit(64);
}
```
user: Translate the below assembly code into C programming language:

```asm
<signal_handler>:
  endbr64
  push   %rbp
  mov    %edi,%esi
  mov    %edi,%ebp
  lea    D4(%rip),%rdi
  call   <debugf@plt>
  xor    %eax,%eax
  call   <kill_all@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <kill@plt>
  mov    $0x40,%edi
  pop    %rbp
  jmp    1090 <_exit@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_guessed: string

tool: D4: reaper: signal %d

assistant: ```c
void signal_handler(int signum) {
  debugf("reaper: signal %d", signum);
  kill_all();
  kill(getpid(), signum);
  _exit(64);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<signal_handler>:
  endbr64
  push   %rbp
  mov    %edi,%esi
  mov    %edi,%ebp
  lea    D7(%rip),%rdi
  call   <debugf@plt>
  xor    %eax,%eax
  call   <kill_all@plt>
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %ebp,%esi
  mov    %eax,%edi
  call   <kill@plt>
  mov    $0x40,%edi
  pop    %rbp
  jmp    1090 <_exit@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D7
datatype: string

tool: D7: reaper: signal %d

assistant: ```c
void signal_handler(int signum) {
  debugf("reaper: signal %d", signum);
  kill_all();
  kill(getpid(), signum);
  _exit(64);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x3,-0x4(%rbp)
  jmp    L4
L3:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    $0x2a,%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,(%rcx)
L4:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void f(int *a) {
  int i;
  for (i = 3; --i;)
    a[i] = 42 / i;
}
```
user: What is the c source code of the assembly code below:

```asm
<f>:
  endbr64
  movl   $0x15,0x8(%rdi)
  movl   $0x2a,0x4(%rdi)
  ret
```

assistant: ```c
void f(int *a) {
  int i;
  for (i = 3; --i;)
    a[i] = 42 / i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f>:
  endbr64
  movabs $0x150000002a,%rax
  mov    %rax,0x4(%rdi)
  ret
```

assistant: ```c
void f(int *a) {
  int i;
  for (i = 3; --i;)
    a[i] = 42 / i;
}
```
user: What is the c source code of the assembly code below?

```asm
<f>:
  endbr64
  mov    D2(%rip),%rax
  mov    %rax,0x4(%rdi)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: byte[8]

tool: {"D2": [0x2a,0x00,0x00,0x00,0x15,0x00,0x00,0x00]}

assistant: ```c
void f(int *a) {
  int i;
  for (i = 3; --i;)
    a[i] = 42 / i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eps_qos_bit_rate_ext_value>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L37
  cmpl   $0x4a,-0x4(%rbp)
  jg     L37
  mov    -0x4(%rbp),%eax
  imul   $0x64,%eax,%eax
  add    $0x2198,%eax
  jmp    L45
L37:
  cmpl   $0x4a,-0x4(%rbp)
  jle    L75
  cmpl   $0xba,-0x4(%rbp)
  jg     L75
  mov    -0x4(%rbp),%eax
  sub    $0x3a,%eax
  shl    $0xa,%eax
  jmp    L45
L75:
  cmpl   $0xba,-0x4(%rbp)
  jle    L79
  cmpl   $0xfa,-0x4(%rbp)
  jg     L79
  mov    -0x4(%rbp),%eax
  sub    $0x7a,%eax
  shl    $0xb,%eax
  jmp    L45
L79:
  mov    $0xffffffff,%eax
L45:
  pop    %rbp
  ret
```

assistant: ```c
int eps_qos_bit_rate_ext_value(uint8_t br) {
  if ((br > 0b00000000) && (br < 0b01001011)) {
    return (8600 + br * 100);
  } else if ((br > 0b01001010) && (br < 0b10111011)) {
    return (16384 + (br - 0b01001010) * 1024);
  } else if ((br > 0b10111010) && (br < 0b11111011)) {
    return (131072 + (br - 0b10111010) * 2048);
  } else {
    return (-1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eps_qos_bit_rate_ext_value>:
  endbr64
  lea    -0x1(%rdi),%eax
  cmp    $0x49,%eax
  jbe    L22
  lea    -0x4b(%rdi),%eax
  cmp    $0x6f,%eax
  jbe    L25
  lea    -0xbb(%rdi),%eax
  cmp    $0x3f,%eax
  ja     L10
  lea    -0x7a(%rdi),%eax
  shl    $0xb,%eax
  ret
L22:
  imul   $0x64,%edi,%edi
  lea    0x2198(%rdi),%eax
  ret
L25:
  lea    -0x3a(%rdi),%eax
  shl    $0xa,%eax
  ret
L10:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int eps_qos_bit_rate_ext_value(uint8_t br) {
  if ((br > 0b00000000) && (br < 0b01001011)) {
    return (8600 + br * 100);
  } else if ((br > 0b01001010) && (br < 0b10111011)) {
    return (16384 + (br - 0b01001010) * 1024);
  } else if ((br > 0b10111010) && (br < 0b11111011)) {
    return (131072 + (br - 0b10111010) * 2048);
  } else {
    return (-1);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<eps_qos_bit_rate_ext_value>:
  endbr64
  lea    -0x1(%rdi),%eax
  cmp    $0x49,%eax
  jbe    L16
  lea    -0x4b(%rdi),%eax
  cmp    $0x6f,%eax
  jbe    L24
  lea    -0xbb(%rdi),%eax
  cmp    $0x3f,%eax
  ja     L21
  lea    -0x7a(%rdi),%eax
  shl    $0xb,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L24:
  lea    -0x3a(%rdi),%eax
  shl    $0xa,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  imul   $0x64,%edi,%edi
  lea    0x2198(%rdi),%eax
  ret
L21:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int eps_qos_bit_rate_ext_value(uint8_t br) {
  if ((br > 0b00000000) && (br < 0b01001011)) {
    return (8600 + br * 100);
  } else if ((br > 0b01001010) && (br < 0b10111011)) {
    return (16384 + (br - 0b01001010) * 1024);
  } else if ((br > 0b10111010) && (br < 0b11111011)) {
    return (131072 + (br - 0b10111010) * 2048);
  } else {
    return (-1);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<eps_qos_bit_rate_ext_value>:
  endbr64
  lea    -0x1(%rdi),%eax
  cmp    $0x49,%eax
  jbe    L20
  lea    -0x4b(%rdi),%eax
  cmp    $0x6f,%eax
  jbe    L2
  lea    -0xbb(%rdi),%eax
  cmp    $0x3f,%eax
  ja     L27
  lea    -0x7a(%rdi),%eax
  shl    $0xb,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  lea    -0x3a(%rdi),%eax
  shl    $0xa,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  imul   $0x64,%edi,%edi
  lea    0x2198(%rdi),%eax
  ret
L27:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int eps_qos_bit_rate_ext_value(uint8_t br) {
  if ((br > 0b00000000) && (br < 0b01001011)) {
    return (8600 + br * 100);
  } else if ((br > 0b01001010) && (br < 0b10111011)) {
    return (16384 + (br - 0b01001010) * 1024);
  } else if ((br > 0b10111010) && (br < 0b11111011)) {
    return (131072 + (br - 0b10111010) * 2048);
  } else {
    return (-1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<turn_backward>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <left_motor@plt>
  mov    -0x4(%rbp),%eax
  neg    %eax
  mov    %eax,%edi
  call   <right_motor@plt>
  nop
  leave
  ret
```

assistant: ```c
void turn_backward(int speed) {
  left_motor(speed);
  right_motor(-speed);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<turn_backward>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <left_motor@plt>
  neg    %ebx
  mov    %ebx,%edi
  call   <right_motor@plt>
  pop    %rbx
  ret
```

assistant: ```c
void turn_backward(int speed) {
  left_motor(speed);
  right_motor(-speed);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<turn_backward>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  neg    %ebx
  call   <left_motor@plt>
  mov    %ebx,%edi
  pop    %rbx
  jmp    1060 <right_motor@plt>
```

assistant: ```c
void turn_backward(int speed) {
  left_motor(speed);
  right_motor(-speed);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<turn_backward>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  neg    %ebx
  call   <left_motor@plt>
  mov    %ebx,%edi
  pop    %rbx
  jmp    1060 <right_motor@plt>
```

assistant: ```c
void turn_backward(int speed) {
  left_motor(speed);
  right_motor(-speed);
}
```
user: Translate the below assembly code into C programming language:

```asm
<dxl_hal_open>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <USART1_Configuration@plt>
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int dxl_hal_open(int devIndex, int baudrate) {
  USART1_Configuration(baudrate);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dxl_hal_open>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%edi
  call   <USART1_Configuration@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int dxl_hal_open(int devIndex, int baudrate) {
  USART1_Configuration(baudrate);
  return 1;
}
```