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

```asm
<dump>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L5:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0x2,-0x4(%rbp)
  jle    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string

tool: D10: %s = 

tool: D5: [%u]

assistant: ```c
void dump(char *msg, int rank, size_t *indices) {
  int j;
  printf("%s = ", msg);
  for (j = 0; j < 3; j++)
    printf("[%u]", indices[j]);
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dump>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  mov    %rdx,%rbx
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%rbx),%rsi
  lea    D16(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rbx),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rbx),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
data_type: string
name: unpack_data
arguments: data_label: D16
data_type: string

tool: D6: %s = 

tool: D16: [%u]

assistant: ```c
void dump(char *msg, int rank, size_t *indices) {
  int j;
  printf("%s = ", msg);
  for (j = 0; j < 3; j++)
    printf("[%u]", indices[j]);
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dump>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  push   %rbx
  mov    %rdx,%rbx
  lea    D0(%rip),%rbp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: tag: D2
type_guessed: string
name: read_data
arguments: tag: D0
type_guessed: string

tool: D2: %s = 

tool: D0: [%u]

assistant: ```c
void dump(char *msg, int rank, size_t *indices) {
  int j;
  printf("%s = ", msg);
  for (j = 0; j < 3; j++)
    printf("[%u]", indices[j]);
  printf("\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dump>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  xor    %eax,%eax
  lea    D5(%rip),%rdi
  push   %rbx
  mov    %rdx,%rbx
  lea    D4(%rip),%rbp
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    (%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x10(%rbx),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
datatype: string
name: parse_data
arguments: data_tag: D4
datatype: string

tool: D5: %s = 

tool: D4: [%u]

assistant: ```c
void dump(char *msg, int rank, size_t *indices) {
  int j;
  printf("%s = ", msg);
  for (j = 0; j < 3; j++)
    printf("[%u]", indices[j]);
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  leave
  ret
```

assistant: ```c
char *strcpy(char *dest, char *src) {
  int countSrc = strlen(src);
  return strncpy(dest, src, countSrc);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strcpy>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strncpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *dest, char *src) {
  int countSrc = strlen(src);
  return strncpy(dest, src, countSrc);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  mov    %eax,%edx
  pop    %r12
  jmp    1060 <strncpy@plt>
```

assistant: ```c
char *strcpy(char *dest, char *src) {
  int countSrc = strlen(src);
  return strncpy(dest, src, countSrc);
}
```
user: What is the c source code of the assembly code below:

```asm
<strcpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  mov    %eax,%edx
  pop    %r12
  jmp    1060 <strncpy@plt>
```

assistant: ```c
char *strcpy(char *dest, char *src) {
  int countSrc = strlen(src);
  return strncpy(dest, src, countSrc);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hll_pat_len>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0xadc83b19,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <MurmurHash64A@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  and    $0x3fff,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x4000,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L2
L13:
  addl   $0x1,-0xc(%rbp)
  shll   -0x10(%rbp)
L2:
  mov    -0x8(%rbp),%eax
  and    -0x10(%rbp),%eax
  test   %eax,%eax
  je     L13
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int hll_pat_len(unsigned char *ele, size_t elesize, long *regp) {
  uint64_t hash, bit, index;
  int count;
  hash = MurmurHash64A(ele, elesize, 0xadc83b19ULL);
  index = hash & ((1 << 14) - 1);
  hash |= ((uint64_t)1 << 63);
  bit = (1 << 14);
  count = 1;
  while ((hash & bit) == 0) {
    count++;
    bit <<= 1;
  }
  *regp = (int)index;
  return count;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hll_pat_len>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  mov    $0xadc83b19,%edx
  call   <MurmurHash64A@plt>
  mov    %eax,%esi
  and    $0x3fff,%esi
  test   $0x40,%ah
  jne    L20
  mov    %eax,%ecx
  mov    $0x1,%eax
  mov    $0x4000,%edx
L10:
  add    $0x1,%eax
  add    %edx,%edx
  test   %edx,%ecx
  je     L10
L0:
  movslq %esi,%rdx
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
L20:
  mov    $0x1,%eax
  jmp    L0
```

assistant: ```c
int hll_pat_len(unsigned char *ele, size_t elesize, long *regp) {
  uint64_t hash, bit, index;
  int count;
  hash = MurmurHash64A(ele, elesize, 0xadc83b19ULL);
  index = hash & ((1 << 14) - 1);
  hash |= ((uint64_t)1 << 63);
  bit = (1 << 14);
  count = 1;
  while ((hash & bit) == 0) {
    count++;
    bit <<= 1;
  }
  *regp = (int)index;
  return count;
}
```
user: What is the c source code of the assembly code below?

```asm
<hll_pat_len>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  mov    $0xadc83b19,%edx
  call   <MurmurHash64A@plt>
  mov    %eax,%edx
  and    $0x3fff,%edx
  test   $0x40,%ah
  jne    L17
  mov    %eax,%esi
  mov    $0x4000,%ecx
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L2:
  add    %ecx,%ecx
  add    $0x1,%eax
  test   %ecx,%esi
  je     L2
  movslq %edx,%rdx
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L17:
  movslq %edx,%rdx
  mov    $0x1,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int hll_pat_len(unsigned char *ele, size_t elesize, long *regp) {
  uint64_t hash, bit, index;
  int count;
  hash = MurmurHash64A(ele, elesize, 0xadc83b19ULL);
  index = hash & ((1 << 14) - 1);
  hash |= ((uint64_t)1 << 63);
  bit = (1 << 14);
  count = 1;
  while ((hash & bit) == 0) {
    count++;
    bit <<= 1;
  }
  *regp = (int)index;
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hll_pat_len>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  mov    $0xadc83b19,%edx
  call   <MurmurHash64A@plt>
  mov    %eax,%edx
  and    $0x3fff,%edx
  test   $0x40,%ah
  jne    L3
  mov    %eax,%esi
  mov    $0x4000,%ecx
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L9:
  add    %ecx,%ecx
  add    $0x1,%eax
  test   %ecx,%esi
  je     L9
  movslq %edx,%rdx
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  movslq %edx,%rdx
  mov    $0x1,%eax
  mov    %rdx,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int hll_pat_len(unsigned char *ele, size_t elesize, long *regp) {
  uint64_t hash, bit, index;
  int count;
  hash = MurmurHash64A(ele, elesize, 0xadc83b19ULL);
  index = hash & ((1 << 14) - 1);
  hash |= ((uint64_t)1 << 63);
  bit = (1 << 14);
  count = 1;
  while ((hash & bit) == 0) {
    count++;
    bit <<= 1;
  }
  *regp = (int)index;
  return count;
}
```
user: What is the c source code of the assembly code below?

```asm
<show_mainl_usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

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

tool: D5: 

Not enough parameters:

 usage: mainl in-file k_diffuse k_specular eta Light-x y z (n.n) 


assistant: ```c
int show_mainl_usage() {
  char response[80];
  printf("\n\nNot enough parameters:"
         "\n"
         "\n usage: mainl in-file k_diffuse k_specular eta"
         " Light-x y z (n.n)"
         " "
         "\n");
  return (1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<show_mainl_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type: string

tool: D9: 

Not enough parameters:

 usage: mainl in-file k_diffuse k_specular eta Light-x y z (n.n) 


assistant: ```c
int show_mainl_usage() {
  char response[80];
  printf("\n\nNot enough parameters:"
         "\n"
         "\n usage: mainl in-file k_diffuse k_specular eta"
         " Light-x y z (n.n)"
         " "
         "\n");
  return (1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<show_mainl_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type_guessed: string

tool: D6: 

Not enough parameters:

 usage: mainl in-file k_diffuse k_specular eta Light-x y z (n.n) 


assistant: ```c
int show_mainl_usage() {
  char response[80];
  printf("\n\nNot enough parameters:"
         "\n"
         "\n usage: mainl in-file k_diffuse k_specular eta"
         " Light-x y z (n.n)"
         " "
         "\n");
  return (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
<show_mainl_usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type: string

tool: D6: 

Not enough parameters:

 usage: mainl in-file k_diffuse k_specular eta Light-x y z (n.n) 


assistant: ```c
int show_mainl_usage() {
  char response[80];
  printf("\n\nNot enough parameters:"
         "\n"
         "\n usage: mainl in-file k_diffuse k_specular eta"
         " Light-x y z (n.n)"
         " "
         "\n");
  return (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
<aln2prf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x18(%rbp)
  mov    $0x0,%eax
  call   <aa2dirichlet_code_lu@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  mov    $0x14,%esi
  mov    %eax,%edi
  call   <declare_double@plt>
  mov    %rax,-0x30(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L59
L22:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L57
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movsd  -0x10(%rbp),%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
L57:
  movsd  -0x10(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L9
  movl   $0x0,-0x1c(%rbp)
  jmp    L38
L0:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  divsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L38:
  cmpl   $0x13,-0x1c(%rbp)
  jle    L0
L9:
  addl   $0x1,-0x20(%rbp)
L59:
  mov    -0x20(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L22
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x30(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
datatype: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double **aln2prf(char *seq, double **prf) {
  int *lu;
  int c, r, d;
  double tot;
  int slen = strlen(seq);
  lu = aa2dirichlet_code_lu();
  prf = declare_double(slen, 20);
  for (c = 0; c < slen; c++) {
    tot = 0;
    memset(prf[c], 0, sizeof(double) * 20);
    d = lu[seq[c]];
    if (d >= 0) {
      prf[c][d]++;
      tot++;
    }
    if (tot > 0) {
      for (r = 0; r < 20; r++) {
        prf[c][r] /= tot;
      }
    }
  }
  free(lu);
  return prf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<aln2prf>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    $0x0,%eax
  call   <aa2dirichlet_code_lu@plt>
  mov    %rax,%r12
  mov    $0x14,%esi
  mov    %r13d,%edi
  call   <declare_double@plt>
  mov    %rax,%r14
  test   %r13d,%r13d
  jle    L36
  mov    %rax,%rbx
  lea    -0x1(%r13),%eax
  lea    0x8(%r14,%rax,8),%r13
  jmp    L57
L2:
  add    $0x8,%rbx
  add    $0x1,%rbp
  cmp    %r13,%rbx
  je     L36
L57:
  mov    $0xa0,%edx
  mov    $0x0,%esi
  mov    (%rbx),%rdi
  call   <memset@plt>
  movsbq 0x0(%rbp),%rax
  mov    (%r12,%rax,4),%eax
  test   %eax,%eax
  js     L2
  cltq
  mov    (%rbx),%rdx
  lea    (%rdx,%rax,8),%rax
  movsd  (%rax),%xmm0
  addsd  D1(%rip),%xmm0
  movsd  %xmm0,(%rax)
  mov    $0x0,%eax
L48:
  add    $0x8,%rax
  cmp    $0xa0,%rax
  jne    L48
  jmp    L2
L36:
  mov    %r12,%rdi
  call   <free@plt>
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: byte[8]

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double **aln2prf(char *seq, double **prf) {
  int *lu;
  int c, r, d;
  double tot;
  int slen = strlen(seq);
  lu = aa2dirichlet_code_lu();
  prf = declare_double(slen, 20);
  for (c = 0; c < slen; c++) {
    tot = 0;
    memset(prf[c], 0, sizeof(double) * 20);
    d = lu[seq[c]];
    if (d >= 0) {
      prf[c][d]++;
      tot++;
    }
    if (tot > 0) {
      for (r = 0; r < 20; r++) {
        prf[c][r] /= tot;
      }
    }
  }
  free(lu);
  return prf;
}
```
user: What is the c source code of the assembly code below?

```asm
<aln2prf>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <aa2dirichlet_code_lu@plt>
  mov    $0x14,%esi
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <declare_double@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L27
  mov    %rax,%rbx
  lea    -0x1(%r12),%eax
  lea    0x8(%r13,%rax,8),%r12
  nopl   0x0(%rax,%rax,1)
L20:
  mov    (%rbx),%rdi
  mov    $0xa0,%edx
  xor    %esi,%esi
  call   <memset@plt>
  movsbq (%r14),%rax
  movslq 0x0(%rbp,%rax,4),%rax
  test   %eax,%eax
  js     L10
  mov    (%rbx),%rdx
  movsd  D7(%rip),%xmm0
  lea    (%rdx,%rax,8),%rax
  addsd  (%rax),%xmm0
  movsd  %xmm0,(%rax)
L10:
  add    $0x8,%rbx
  add    $0x1,%r14
  cmp    %r12,%rbx
  jne    L20
L27:
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_infered: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double **aln2prf(char *seq, double **prf) {
  int *lu;
  int c, r, d;
  double tot;
  int slen = strlen(seq);
  lu = aa2dirichlet_code_lu();
  prf = declare_double(slen, 20);
  for (c = 0; c < slen; c++) {
    tot = 0;
    memset(prf[c], 0, sizeof(double) * 20);
    d = lu[seq[c]];
    if (d >= 0) {
      prf[c][d]++;
      tot++;
    }
    if (tot > 0) {
      for (r = 0; r < 20; r++) {
        prf[c][r] /= tot;
      }
    }
  }
  free(lu);
  return prf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<aln2prf>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <strlen@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <aa2dirichlet_code_lu@plt>
  mov    $0x14,%esi
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <declare_double@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L23
  mov    %rax,%rbx
  lea    -0x1(%r12),%eax
  lea    0x8(%r13,%rax,8),%r12
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%rdi
  mov    $0xa0,%edx
  xor    %esi,%esi
  call   <memset@plt>
  movsbq (%r14),%rax
  movslq 0x0(%rbp,%rax,4),%rax
  test   %eax,%eax
  js     L20
  mov    (%rbx),%rdx
  movsd  D8(%rip),%xmm0
  lea    (%rdx,%rax,8),%rax
  addsd  (%rax),%xmm0
  movsd  %xmm0,(%rax)
L20:
  add    $0x8,%rbx
  add    $0x1,%r14
  cmp    %r12,%rbx
  jne    L14
L23:
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type_infered: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
double **aln2prf(char *seq, double **prf) {
  int *lu;
  int c, r, d;
  double tot;
  int slen = strlen(seq);
  lu = aa2dirichlet_code_lu();
  prf = declare_double(slen, 20);
  for (c = 0; c < slen; c++) {
    tot = 0;
    memset(prf[c], 0, sizeof(double) * 20);
    d = lu[seq[c]];
    if (d >= 0) {
      prf[c][d]++;
      tot++;
    }
    if (tot > 0) {
      for (r = 0; r < 20; r++) {
        prf[c][r] /= tot;
      }
    }
  }
  free(lu);
  return prf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<udpwrite>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %r8d,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x200,-0x20(%rbp)
  jle    L16
  mov    $0xffffffff,%eax
  jmp    L11
L16:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  lea    -0xc(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <sockaddrnumfill@plt>
  lea    -0xc(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    $0x4,%r9d
  mov    %rcx,%r8
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <sendto@plt>
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
int udpwrite(int sock, uint32_t ip, uint16_t port, const void *buff, uint16_t leng) {
  struct sockaddr_in sa;
  if (leng > 512) {
    return -1;
  }
  sockaddrnumfill(&sa, ip, port);
  return sendto(sock, buff, leng, 0, (struct sockaddr *)&sa, sizeof(struct sockaddr_in));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<udpwrite>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x200,%r8d
  jg     L8
  mov    %edi,%ebp
  mov    %rcx,%r12
  mov    %r8d,%ebx
  lea    0x4(%rsp),%r13
  mov    %r13,%rdi
  call   <sockaddrnumfill@plt>
  mov    $0x4,%r9d
  mov    %r13,%r8
  mov    $0x0,%ecx
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <sendto@plt>
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    $0xffffffff,%eax
  jmp    L2
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int udpwrite(int sock, uint32_t ip, uint16_t port, const void *buff, uint16_t leng) {
  struct sockaddr_in sa;
  if (leng > 512) {
    return -1;
  }
  sockaddrnumfill(&sa, ip, port);
  return sendto(sock, buff, leng, 0, (struct sockaddr *)&sa, sizeof(struct sockaddr_in));
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<udpwrite>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x200,%r8d
  jg     L17
  lea    0x4(%rsp),%rbx
  mov    %rcx,%r13
  mov    %r8d,%r12d
  mov    %edi,%ebp
  mov    %rbx,%rdi
  call   <sockaddrnumfill@plt>
  mov    $0x4,%r9d
  mov    %rbx,%r8
  xor    %ecx,%ecx
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <sendto@plt>
L2:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L17:
  mov    $0xffffffff,%eax
  jmp    L2
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int udpwrite(int sock, uint32_t ip, uint16_t port, const void *buff, uint16_t leng) {
  struct sockaddr_in sa;
  if (leng > 512) {
    return -1;
  }
  sockaddrnumfill(&sa, ip, port);
  return sendto(sock, buff, leng, 0, (struct sockaddr *)&sa, sizeof(struct sockaddr_in));
}
```