ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<pow_poly>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  je     L19
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edx,%r12d
  mov    $0x1,%r13d
  jmp    L21
  nopl   0x0(%rax)
L29:
  mov    %ebp,%esi
  mov    %ebp,%edi
  mov    %r12d,%edx
  call   <mul_poly@plt>
  shr    %ebx
  mov    %eax,%ebp
  je     L46
L21:
  test   $0x1,%bl
  je     L29
  mov    %r13d,%edi
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <mul_poly@plt>
  mov    %eax,%r13d
  jmp    L29
  nopl   0x0(%rax)
L19:
  mov    $0x1,%r13d
L46:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned int pow_poly(unsigned int a, unsigned int n, unsigned int poly) {
  unsigned int r;
  r = 1;
  while (n) {
    if (n & 1)
      r = mul_poly(r, a, poly);
    a = mul_poly(a, a, poly);
    n >>= 1;
  }
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<pow_poly>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  je     L1
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    %edx,%r12d
  mov    $0x1,%r13d
  jmp    L0
  nopl   0x0(%rax)
L17:
  mov    %ebp,%esi
  mov    %ebp,%edi
  mov    %r12d,%edx
  call   <mul_poly@plt>
  shr    %ebx
  mov    %eax,%ebp
  je     L2
L0:
  test   $0x1,%bl
  je     L17
  mov    %r13d,%edi
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <mul_poly@plt>
  mov    %eax,%r13d
  jmp    L17
  nopl   0x0(%rax)
L1:
  mov    $0x1,%r13d
L2:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
unsigned int pow_poly(unsigned int a, unsigned int n, unsigned int poly) {
  unsigned int r;
  r = 1;
  while (n) {
    if (n & 1)
      r = mul_poly(r, a, poly);
    a = mul_poly(a, a, poly);
    n >>= 1;
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Substring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L12
L1:
  movl   $0x0,-0xc(%rbp)
  jmp    L29
L43:
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0xc(%rbp)
L29:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L43
  movl   $0x0,-0xc(%rbp)
  jmp    L6
L9:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0xc(%rbp)
L6:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L9
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L12:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: string

tool: D3: Unable to allocate memory.


assistant: ```c
char *Substring(char *string, int position, int length) {
  char *pointer;
  int c;
  pointer = malloc(length + 1);
  if (pointer == NULL) {
    printf("Unable to allocate memory.\n");
    return NULL;
  }
  for (c = 0; c < position - 1; c++)
    string++;
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Substring>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    %edx,%r12d
  lea    0x1(%rdx),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L27
  lea    -0x1(%r13),%edx
  mov    $0x0,%eax
  cmp    $0x1,%r13d
  jle    L67
L41:
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L41
  lea    -0x2(%r13),%eax
  lea    0x1(%rbp,%rax,1),%rbp
L67:
  test   %r12d,%r12d
  jle    L16
  mov    %r12d,%ecx
  mov    $0x0,%eax
L3:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L3
L36:
  movslq %r12d,%r12
  movb   $0x0,(%rbx,%r12,1)
L45:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L27:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L45
L16:
  mov    $0x0,%r12d
  jmp    L36
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
type: string

tool: D7: Unable to allocate memory.


assistant: ```c
char *Substring(char *string, int position, int length) {
  char *pointer;
  int c;
  pointer = malloc(length + 1);
  if (pointer == NULL) {
    printf("Unable to allocate memory.\n");
    return NULL;
  }
  for (c = 0; c < position - 1; c++)
    string++;
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Substring>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L38
  cmp    $0x1,%r13d
  jle    L29
  lea    -0x2(%r13),%eax
  lea    0x1(%rbp,%rax,1),%rbp
L29:
  test   %ebx,%ebx
  jle    L1
  movslq %ebx,%rdx
  xor    %eax,%eax
  xchg   %ax,%ax
L54:
  movzbl 0x0(%rbp,%rax,1),%ecx
  mov    %cl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L54
  add    %r12,%rdx
L36:
  movb   $0x0,(%rdx)
L27:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %r12,%rdx
  jmp    L36
L38:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L27
```

tool_calls: name: read_data
arguments: label: D4
data_type: string

tool: D4: Unable to allocate memory.


assistant: ```c
char *Substring(char *string, int position, int length) {
  char *pointer;
  int c;
  pointer = malloc(length + 1);
  if (pointer == NULL) {
    printf("Unable to allocate memory.\n");
    return NULL;
  }
  for (c = 0; c < position - 1; c++)
    string++;
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Substring>:
  endbr64
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L130
  cmp    $0x1,%r13d
  jle    L83
  lea    -0x2(%r13),%eax
  lea    0x1(%rbp,%rax,1),%rbp
L83:
  test   %ebx,%ebx
  jle    L16
  lea    0x1(%rbp),%rcx
  mov    %r12,%rdx
  lea    -0x1(%rbx),%eax
  mov    %ebx,%esi
  sub    %rcx,%rdx
  cmp    $0xe,%rdx
  jbe    L78
  cmp    $0x6,%eax
  jbe    L78
  cmp    $0xe,%eax
  jbe    L184
  mov    %ebx,%edx
  xor    %eax,%eax
  shr    $0x4,%edx
  shl    $0x4,%rdx
  nopw   0x0(%rax,%rax,1)
L179:
  movdqu 0x0(%rbp,%rax,1),%xmm0
  movups %xmm0,(%r12,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L179
  mov    %ebx,%eax
  and    $0xfffffff0,%eax
  mov    %eax,%ecx
  lea    0x0(%rbp,%rcx,1),%rdx
  cmp    %eax,%ebx
  je     L63
  mov    %ebx,%esi
  sub    %eax,%esi
  lea    -0x1(%rsi),%edi
  cmp    $0x6,%edi
  jbe    L125
L61:
  mov    0x0(%rbp,%rcx,1),%rdi
  mov    %rdi,(%r12,%rcx,1)
  mov    %esi,%ecx
  and    $0xfffffff8,%ecx
  mov    %ecx,%edi
  add    %ecx,%eax
  add    %rdi,%rdx
  cmp    %ecx,%esi
  je     L63
L125:
  movzbl (%rdx),%esi
  movslq %eax,%rcx
  mov    %sil,(%r12,%rcx,1)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L63
  movzbl 0x1(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r12,%rcx,1)
  lea    0x2(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L63
  movzbl 0x2(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r12,%rcx,1)
  lea    0x3(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L63
  movzbl 0x3(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r12,%rcx,1)
  lea    0x4(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L63
  movzbl 0x4(%rdx),%esi
  movslq %ecx,%rcx
  mov    %sil,(%r12,%rcx,1)
  lea    0x5(%rax),%ecx
  cmp    %ecx,%ebx
  jle    L63
  movzbl 0x5(%rdx),%esi
  movslq %ecx,%rcx
  add    $0x6,%eax
  mov    %sil,(%r12,%rcx,1)
  cmp    %eax,%ebx
  jle    L63
  movzbl 0x6(%rdx),%edx
  cltq
  mov    %dl,(%r12,%rax,1)
L63:
  movslq %ebx,%rbx
  add    %r12,%rbx
L197:
  movb   $0x0,(%rbx)
L85:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L78:
  mov    %ebx,%ecx
  xor    %eax,%eax
  nopl   0x0(%rax)
L46:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L46
  jmp    L63
  nopl   0x0(%rax)
L16:
  mov    %r12,%rbx
  jmp    L197
L184:
  mov    %rbp,%rdx
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L61
L130:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L85
```

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

tool: D3: Unable to allocate memory.


assistant: ```c
char *Substring(char *string, int position, int length) {
  char *pointer;
  int c;
  pointer = malloc(length + 1);
  if (pointer == NULL) {
    printf("Unable to allocate memory.\n");
    return NULL;
  }
  for (c = 0; c < position - 1; c++)
    string++;
  for (c = 0; c < length; c++) {
    *(pointer + c) = *string;
    string++;
  }
  *(pointer + c) = '\0';
  return pointer;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<quick_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L5
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <partition@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
L5:
  nop
  leave
  ret
```

assistant: ```c
void quick_sort(float *array, int low, int high) {
  if (low < high) {
    int p = partition(array, low, high);
    quick_sort(array, low, p - 1);
    quick_sort(array, p + 1, high);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<quick_sort>:
  endbr64
  cmp    %edx,%esi
  jl     L7
  ret
L7:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  call   <partition@plt>
  mov    %eax,%ebp
  lea    -0x1(%rax),%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <quick_sort>
  lea    0x1(%rbp),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <quick_sort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void quick_sort(float *array, int low, int high) {
  if (low < high) {
    int p = partition(array, low, high);
    quick_sort(array, low, p - 1);
    quick_sort(array, p + 1, high);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<segvhld>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x60(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <backtrace@plt>
  mov    %rax,-0x70(%rbp)
  mov    -0x70(%rbp),%rdx
  lea    -0x60(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <backtrace_symbols@plt>
  mov    %rax,-0x68(%rbp)
  mov    -0x70(%rbp),%rcx
  lea    -0x60(%rbp),%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <backtrace_symbols_fd@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void segvhld(int no) {
  void *array[10];
  size_t size;
  char **strings;
  int i;
  size = backtrace(array, 10);
  strings = backtrace_symbols(array, size);
  backtrace_symbols_fd(array, size, 1);
  exit(1);
}
```
user: What is the c source code of the assembly code below:

```asm
<segvhld>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <backtrace@plt>
  mov    %rax,%rbx
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <backtrace_symbols@plt>
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <backtrace_symbols_fd@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void segvhld(int no) {
  void *array[10];
  size_t size;
  char **strings;
  int i;
  size = backtrace(array, 10);
  strings = backtrace_symbols(array, size);
  backtrace_symbols_fd(array, size, 1);
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<segvhld>:
  endbr64
  push   %r12
  mov    $0xa,%esi
  push   %rbp
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <backtrace@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <backtrace_symbols@plt>
  mov    %r12,%rdi
  mov    $0x1,%edx
  mov    %rbp,%rsi
  call   <backtrace_symbols_fd@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x68,%rsp
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void segvhld(int no) {
  void *array[10];
  size_t size;
  char **strings;
  int i;
  size = backtrace(array, 10);
  strings = backtrace_symbols(array, size);
  backtrace_symbols_fd(array, size, 1);
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<segvhld>:
  endbr64
  push   %r12
  mov    $0xa,%esi
  push   %rbp
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <backtrace@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <backtrace_symbols@plt>
  mov    %r12,%rdi
  mov    $0x1,%edx
  mov    %rbp,%rsi
  call   <backtrace_symbols_fd@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x68,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void segvhld(int no) {
  void *array[10];
  size_t size;
  char **strings;
  int i;
  size = backtrace(array, 10);
  strings = backtrace_symbols(array, size);
  backtrace_symbols_fd(array, size, 1);
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<migstep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r15
  mov    %edx,%r12d
  mov    %ecx,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %ecx,%rax
  lea    0x0(,%rax,8),%rbx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rsi
  mov    %rsi,-0x48(%rbp)
  imul   %rsi,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffffff0,%rdx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L64:
  cmp    %rcx,%rsp
  je     L56
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L64
L56:
  mov    %rdx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L47
  orq    $0x0,-0x8(%rsp,%rax,1)
L47:
  mov    %rsp,%rdi
  mov    %r13d,%eax
  imul   %r12d,%eax
  shl    $0x3,%eax
  mov    %eax,-0x54(%rbp)
  mov    %eax,%esi
  mov    %rdi,-0x50(%rbp)
  call   <bzero@plt>
  test   %r12d,%r12d
  jle    L15
  mov    %r15,%r9
  lea    -0x1(%r12),%eax
  lea    0x8(%r15,%rax,8),%rcx
  mov    -0x50(%rbp),%r15
  lea    -0x1(%r13),%eax
  lea    0x8(%r14,%rax,8),%r10
  mov    $0x0,%r11d
  jmp    L50
L45:
  mov    %rsi,%r8
  movsd  (%rsi),%xmm1
  mov    %rdi,%rdx
  mov    %r9,%rax
L18:
  movsd  (%rax),%xmm0
  mulsd  (%rdx),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x8,%rax
  add    %rbx,%rdx
  cmp    %rcx,%rax
  jne    L18
  movsd  %xmm1,(%r8)
  add    $0x8,%rsi
  add    $0x8,%rdi
  cmp    %r10,%rdi
  jne    L45
L77:
  add    $0x1,%r11d
  mov    -0x48(%rbp),%rax
  add    %rax,%r9
  add    %rax,%rcx
  add    %rbx,%r15
  cmp    %r11d,%r12d
  je     L15
L50:
  mov    %r14,%rdi
  mov    %r15,%rsi
  test   %r13d,%r13d
  jg     L45
  jmp    L77
L15:
  mov    -0x54(%rbp),%edx
  mov    -0x50(%rbp),%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void migstep(double *vec, const double *mig, const int nislands, const int nhap) {
  int i, j, k;
  double vec2[nislands][nhap];
  bzero(vec2, sizeof(vec2));
  for (i = 0; i < nislands; ++i)
    for (j = 0; j < nhap; ++j) {
      for (k = 0; k < nislands; ++k)
        vec2[i][j] += ((double(*)[nislands])mig)[i][k] * ((double(*)[nhap])vec)[k][j];
    }
  memcpy(vec, (double *)vec2, sizeof(double) * nislands * nhap);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<migstep>:
  endbr64
  push   %rbp
  mov    %ecx,%r8d
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %r13,%r15
  shl    $0x3,%r13
  mov    %rsi,%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %ecx,%rax
  mov    %rsp,%rcx
  lea    0x0(,%rax,8),%rbx
  imul   %r13,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L75
L98:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L98
L75:
  mov    %rdx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L34
L66:
  mov    %r8d,%eax
  mov    %rsp,%rdi
  mov    %r8d,-0x50(%rbp)
  imul   %r15d,%eax
  mov    %rdi,-0x48(%rbp)
  shl    $0x3,%eax
  mov    %eax,%esi
  mov    %eax,-0x4c(%rbp)
  call   <bzero@plt>
  test   %r15d,%r15d
  jle    L8
  mov    -0x50(%rbp),%r8d
  lea    -0x1(%r15),%eax
  mov    -0x48(%rbp),%r11
  xor    %r10d,%r10d
  lea    0x8(%r12,%rax,8),%rcx
  lea    -0x1(%r8),%eax
  lea    0x8(%r14,%rax,8),%r9
  nopl   0x0(%rax)
L53:
  mov    %r14,%rdi
  mov    %r11,%rsi
  test   %r8d,%r8d
  jle    L96
  nopl   0x0(%rax,%rax,1)
L77:
  movsd  (%rsi),%xmm1
  mov    %rdi,%rdx
  mov    %r12,%rax
  nopw   0x0(%rax,%rax,1)
L41:
  movsd  (%rax),%xmm0
  mulsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    %rbx,%rdx
  addsd  %xmm0,%xmm1
  cmp    %rcx,%rax
  jne    L41
  add    $0x8,%rdi
  movsd  %xmm1,(%rsi)
  add    $0x8,%rsi
  cmp    %r9,%rdi
  jne    L77
L96:
  add    $0x1,%r10d
  add    %r13,%r12
  add    %r13,%rcx
  add    %rbx,%r11
  cmp    %r10d,%r15d
  jne    L53
L8:
  mov    -0x4c(%rbp),%edx
  mov    -0x48(%rbp),%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L34:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L66
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void migstep(double *vec, const double *mig, const int nislands, const int nhap) {
  int i, j, k;
  double vec2[nislands][nhap];
  bzero(vec2, sizeof(vec2));
  for (i = 0; i < nislands; ++i)
    for (j = 0; j < nhap; ++j) {
      for (k = 0; k < nislands; ++k)
        vec2[i][j] += ((double(*)[nislands])mig)[i][k] * ((double(*)[nhap])vec)[k][j];
    }
  memcpy(vec, (double *)vec2, sizeof(double) * nislands * nhap);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<migstep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  movslq %edx,%r12
  mov    %ecx,%edx
  push   %rbx
  mov    %r12,%r14
  shl    $0x3,%r12
  sub    $0x28,%rsp
  mov    %rsi,-0x50(%rbp)
  mov    %rsp,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %ecx,%rax
  lea    0x0(,%rax,8),%rbx
  imul   %r12,%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L64
L59:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L59
L64:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L57
L9:
  mov    %edx,%eax
  mov    %rsp,%r15
  mov    %edx,-0x48(%rbp)
  imul   %r14d,%eax
  mov    %r15,%rdi
  shl    $0x3,%eax
  mov    %eax,%esi
  mov    %eax,-0x44(%rbp)
  call   <bzero@plt>
  test   %r14d,%r14d
  jle    L27
  mov    -0x48(%rbp),%edx
  test   %edx,%edx
  jle    L27
  mov    -0x50(%rbp),%r8
  lea    -0x1(%r14),%eax
  mov    %r15,%r11
  xor    %r10d,%r10d
  lea    0x8(%r8,%rax,8),%rcx
  lea    -0x1(%rdx),%eax
  lea    0x8(%r13,%rax,8),%r9
  xchg   %ax,%ax
L71:
  mov    %r13,%rdi
  mov    %r11,%rsi
  cs nopw 0x0(%rax,%rax,1)
L2:
  movsd  (%rsi),%xmm1
  mov    %rdi,%rdx
  mov    %r8,%rax
  nopw   0x0(%rax,%rax,1)
L14:
  movsd  (%rax),%xmm0
  mulsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    %rbx,%rdx
  addsd  %xmm0,%xmm1
  cmp    %rcx,%rax
  jne    L14
  add    $0x8,%rdi
  movsd  %xmm1,(%rsi)
  add    $0x8,%rsi
  cmp    %rdi,%r9
  jne    L2
  add    $0x1,%r10d
  add    %r12,%r8
  add    %r12,%rcx
  add    %rbx,%r11
  cmp    %r10d,%r14d
  jne    L71
L27:
  mov    -0x44(%rbp),%edx
  mov    %r15,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L57:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L9
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void migstep(double *vec, const double *mig, const int nislands, const int nhap) {
  int i, j, k;
  double vec2[nislands][nhap];
  bzero(vec2, sizeof(vec2));
  for (i = 0; i < nislands; ++i)
    for (j = 0; j < nhap; ++j) {
      for (k = 0; k < nislands; ++k)
        vec2[i][j] += ((double(*)[nislands])mig)[i][k] * ((double(*)[nhap])vec)[k][j];
    }
  memcpy(vec, (double *)vec2, sizeof(double) * nislands * nhap);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ircproto_privmsg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x2e0,%rsp
  mov    %rdi,-0x2d8(%rbp)
  mov    %rsi,-0x2e0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L16
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L16:
  mov    %fs:0x28,%rax
  mov    %rax,-0xb8(%rbp)
  xor    %eax,%eax
  mov    -0x2e0(%rbp),%rdx
  mov    -0x2c4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0x2c4(%rbp),%ecx
  mov    -0x2e0(%rbp),%rdx
  lea    -0x2c0(%rbp),%rax
  mov    $0x200,%esi
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    -0x2c4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  lea    -0x2c0(%rbp),%rdx
  mov    -0x2d8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <sock_sendline@plt>
  nop
  mov    -0xb8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
type: string

tool: D7: PRIVMSG %s :%s

assistant: ```c
void ircproto_privmsg(char *recipient, char *format, ...) {
  va_list args;
  char message[512];
  va_start(args, format);
  vsnprintf(message, sizeof(message), format, args);
  va_end(args);
  sock_sendline("PRIVMSG %s :%s", recipient, message);
}
```