ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<qarn>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  jle    L15
  sub    $0x1,%edx
  movsd  D2(%rip),%xmm5
  movsd  D11(%rip),%xmm4
  movsd  D7(%rip),%xmm2
  movsd  D37(%rip),%xmm3
  lea    0x8(%rsi,%rdx,8),%rcx
  nopl   0x0(%rax,%rax,1)
L12:
  movsd  (%rax),%xmm0
  add    $0x8,%rsi
  mulsd  %xmm5,%xmm0
  addsd  %xmm4,%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm2,%xmm1
  cvttsd2si %xmm1,%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %edx,%xmm1
  mulsd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,-0x8(%rsi)
  cmp    %rsi,%rcx
  jne    L12
L15:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
datatype: f64
name: disassemble_data
arguments: tag: D11
datatype: f64
name: disassemble_data
arguments: tag: D7
datatype: byte[8]
name: disassemble_data
arguments: tag: D37
datatype: f64

tool: D2: 2053

tool: D11: 13849

tool: {"D7": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3e]}

tool: D37: 65536

assistant: ```c
void qarn(double *r, double p[], int n) {
  int i, m;
  double s, u, v;
  s = 65536.0;
  u = 2053.0;
  v = 13849.0;
  for (i = 0; i <= n - 1; i++) {
    *r = u * (*r) + v;
    m = (int)(*r / s);
    *r = *r - m * s;
    p[i] = *r / s;
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<qarn>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  jle    L13
  sub    $0x1,%edx
  movsd  D8(%rip),%xmm5
  movsd  D23(%rip),%xmm4
  movsd  D28(%rip),%xmm2
  movsd  D24(%rip),%xmm3
  lea    0x8(%rsi,%rdx,8),%rcx
  nopl   0x0(%rax,%rax,1)
L0:
  movsd  (%rax),%xmm0
  add    $0x8,%rsi
  mulsd  %xmm5,%xmm0
  addsd  %xmm4,%xmm0
  movapd %xmm0,%xmm1
  mulsd  %xmm2,%xmm1
  cvttsd2si %xmm1,%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %edx,%xmm1
  mulsd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,-0x8(%rsi)
  cmp    %rcx,%rsi
  jne    L0
L13:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type: f64
name: unpack_data
arguments: data_tag: D23
type: f64
name: unpack_data
arguments: data_tag: D28
type: byte[8]
name: unpack_data
arguments: data_tag: D24
type: f64

tool: D8: 2053

tool: D23: 13849

tool: {"D28": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3e]}

tool: D24: 65536

assistant: ```c
void qarn(double *r, double p[], int n) {
  int i, m;
  double s, u, v;
  s = 65536.0;
  u = 2053.0;
  v = 13849.0;
  for (i = 0; i <= n - 1; i++) {
    *r = u * (*r) + v;
    m = (int)(*r / s);
    *r = *r - m * s;
    p[i] = *r / s;
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<FA_match>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L18
L32:
  mov    -0x24(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  sub    $0x61,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jne    L27
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  sub    -0x28(%rbp),%eax
  jmp    L29
L27:
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L32
  mov    $0xffffffff,%eax
L29:
  leave
  ret
```

assistant: ```c
int FA_match(char *text, int *array, int numchars, int receive) {
  int n = strlen(text);
  int q = 0;
  int i;
  for (i = 0; i < n; i++) {
    int index = numchars * q + text[i] - 'a';
    q = array[index];
    if (q == receive)
      return i + 1 - receive;
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<FA_match>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %edx,%r13d
  mov    %ecx,%ebp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L11
  mov    %eax,%eax
  mov    $0x0,%ecx
  mov    $0x0,%edx
L0:
  imul   %r13d,%edx
  movsbl (%rbx,%rcx,1),%esi
  lea    -0x61(%rdx,%rsi,1),%edx
  movslq %edx,%rdx
  mov    (%r12,%rdx,4),%edx
  cmp    %ebp,%edx
  je     L45
  add    $0x1,%rcx
  cmp    %rax,%rcx
  jne    L0
  mov    $0xffffffff,%eax
  jmp    L27
L45:
  lea    0x1(%rcx),%eax
  sub    %ebp,%eax
L27:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  mov    $0xffffffff,%eax
  jmp    L27
```

assistant: ```c
int FA_match(char *text, int *array, int numchars, int receive) {
  int n = strlen(text);
  int q = 0;
  int i;
  for (i = 0; i < n; i++) {
    int index = numchars * q + text[i] - 'a';
    q = array[index];
    if (q == receive)
      return i + 1 - receive;
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<FA_match>:
  endbr64
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L8
  lea    0x1(%rax),%esi
  mov    $0x1,%ecx
  xor    %edx,%edx
  jmp    L11
  nop
L1:
  add    $0x1,%rcx
  cmp    %rcx,%rsi
  je     L8
L11:
  imul   %r12d,%edx
  movsbl -0x1(%rbx,%rcx,1),%eax
  lea    -0x61(%rdx,%rax,1),%eax
  cltq
  mov    0x0(%rbp,%rax,4),%edx
  mov    %ecx,%eax
  cmp    %r13d,%edx
  jne    L1
  add    $0x8,%rsp
  sub    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int FA_match(char *text, int *array, int numchars, int receive) {
  int n = strlen(text);
  int q = 0;
  int i;
  for (i = 0; i < n; i++) {
    int index = numchars * q + text[i] - 'a';
    q = array[index];
    if (q == receive)
      return i + 1 - receive;
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<FA_match>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %eax,%eax
  jle    L4
  lea    0x1(%rax),%esi
  mov    $0x1,%edx
  xor    %ecx,%ecx
  cmp    $0x1,%r13d
  je     L80
  jmp    L30
  nopl   0x0(%rax)
L74:
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  je     L4
L80:
  movsbl -0x1(%rbx,%rdx,1),%eax
  lea    -0x61(%rcx,%rax,1),%eax
  cltq
  mov    (%r12,%rax,4),%ecx
  mov    %edx,%eax
  cmp    %ecx,%ebp
  jne    L74
  sub    %ebp,%eax
L38:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  mov    $0x1,%ecx
  xor    %edx,%edx
  jmp    L6
  nopl   0x0(%rax)
L54:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L4
L6:
  imul   %r13d,%edx
  movsbl -0x1(%rbx,%rcx,1),%eax
  lea    -0x61(%rdx,%rax,1),%eax
  cltq
  mov    (%r12,%rax,4),%edx
  mov    %ecx,%eax
  cmp    %ebp,%edx
  jne    L54
  sub    %ebp,%eax
  jmp    L38
```

assistant: ```c
int FA_match(char *text, int *array, int numchars, int receive) {
  int n = strlen(text);
  int q = 0;
  int i;
  for (i = 0; i < n; i++) {
    int index = numchars * q + text[i] - 'a';
    q = array[index];
    if (q == receive)
      return i + 1 - receive;
  }
  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
<uart_puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L16
L6:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <uart_putc@plt>
L16:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  nop
  nop
  leave
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  while (*str) {
    uart_putc(*str++);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L0
L7:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <uart_putc@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L7
L0:
  pop    %rbx
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  while (*str) {
    uart_putc(*str++);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L6
L0:
  call   <uart_putc@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L0
L6:
  pop    %rbx
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  while (*str) {
    uart_putc(*str++);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<uart_puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L4
L15:
  call   <uart_putc@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L15
L4:
  pop    %rbx
  ret
```

assistant: ```c
void uart_puts(const char *str) {
  while (*str) {
    uart_putc(*str++);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rec_merge_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x1,-0x30(%rbp)
  jne    L15
  mov    -0x28(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <copy_array@plt>
  jmp    L8
L15:
  mov    -0x30(%rbp),%rax
  shr    %rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rec_merge_sort>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <rec_merge_sort>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <merge@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
L8:
  leave
  ret
```

assistant: ```c
double *rec_merge_sort(double *array, size_t size) {
  if (size == 1) {
    return copy_array(array, 1);
  } else {
    size_t size2 = size / 2;
    double *s0 = rec_merge_sort(array, size2);
    double *s1 = rec_merge_sort(array + size2, size - size2);
    double *res = merge(s0, size2, s1, size - size2);
    free(s0);
    free(s1);
    return res;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rec_merge_sort>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  cmp    $0x1,%rsi
  je     L7
  mov    %rsi,%rbx
  mov    %rsi,%r13
  shr    %r13
  mov    %r13,%rsi
  call   <rec_merge_sort>
  mov    %rax,%r12
  sub    %r13,%rbx
  lea    0x0(%rbp,%r13,8),%rdi
  mov    %rbx,%rsi
  call   <rec_merge_sort>
  mov    %rax,%rbp
  mov    %rbx,%rcx
  mov    %rax,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <merge@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L4:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  mov    $0x1,%esi
  call   <copy_array@plt>
  mov    %rax,%rbx
  jmp    L4
```

assistant: ```c
double *rec_merge_sort(double *array, size_t size) {
  if (size == 1) {
    return copy_array(array, 1);
  } else {
    size_t size2 = size / 2;
    double *s0 = rec_merge_sort(array, size2);
    double *s1 = rec_merge_sort(array + size2, size - size2);
    double *res = merge(s0, size2, s1, size - size2);
    free(s0);
    free(s1);
    return res;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rec_merge_sort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  cmp    $0x1,%rsi
  je     L4
  mov    %rsi,%r14
  mov    %rsi,%r12
  shr    %r14
  sub    %r14,%r12
  mov    %r14,%rsi
  call   <rec_merge_sort>
  lea    0x0(%rbp,%r14,8),%rdi
  mov    %r12,%rsi
  mov    %rax,%r13
  call   <rec_merge_sort>
  mov    %r12,%rcx
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %rax,%rdx
  mov    %rax,%rbp
  call   <merge@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x8,%rsp
  mov    $0x1,%esi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <copy_array@plt>
```

assistant: ```c
double *rec_merge_sort(double *array, size_t size) {
  if (size == 1) {
    return copy_array(array, 1);
  } else {
    size_t size2 = size / 2;
    double *s0 = rec_merge_sort(array, size2);
    double *s1 = rec_merge_sort(array + size2, size - size2);
    double *res = merge(s0, size2, s1, size - size2);
    free(s0);
    free(s1);
    return res;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<shift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %al,-0x14(%rbp)
  movsbl -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cmp    $0x5a,%eax
  jle    L13
  mov    -0x18(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x14(%rbp),%eax
  add    %edx,%eax
  sub    $0x1a,%eax
  jmp    L15
L13:
  mov    -0x18(%rbp),%eax
  mov    %eax,%edx
  movzbl -0x14(%rbp),%eax
  add    %edx,%eax
L15:
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
char shift(char c, int key) {
  char d = ((c + key) > 'Z') ? ((c + key) - 26) : c + key;
  return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<shift>:
  endbr64
  movsbl %dil,%eax
  add    %esi,%eax
  lea    -0x1a(%rsi,%rdi,1),%edx
  add    %esi,%edi
  cmp    $0x5b,%eax
  mov    %edx,%eax
  cmovl  %edi,%eax
  ret
```

assistant: ```c
char shift(char c, int key) {
  char d = ((c + key) > 'Z') ? ((c + key) - 26) : c + key;
  return d;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<shift>:
  endbr64
  movsbl %dil,%eax
  lea    -0x1a(%rsi,%rdi,1),%edx
  add    %esi,%edi
  add    %esi,%eax
  cmp    $0x5b,%eax
  mov    %edx,%eax
  cmovl  %edi,%eax
  ret
```

assistant: ```c
char shift(char c, int key) {
  char d = ((c + key) > 'Z') ? ((c + key) - 26) : c + key;
  return d;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<shift>:
  endbr64
  movsbl %dil,%eax
  lea    -0x1a(%rsi,%rdi,1),%edx
  add    %esi,%edi
  add    %esi,%eax
  cmp    $0x5b,%eax
  mov    %edx,%eax
  cmovl  %edi,%eax
  ret
```

assistant: ```c
char shift(char c, int key) {
  char d = ((c + key) > 'Z') ? ((c + key) - 26) : c + key;
  return d;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<evalR0_0fs4d>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r14
  mov    %esi,%ebx
  mov    %rdx,%rbp
  mov    %rcx,%r12
  mov    %r8,0x10(%rsp)
  mov    %r9,0x18(%rsp)
  lea    0x0(,%rsi,8),%r15d
  mov    %r15d,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    %ebx,%r13d
  lea    0x0(,%r13,8),%rax
  mov    %rax,0x8(%rsp)
  add    %rax,%r14
  test   %ebx,%ebx
  je     L19
  mov    %r13,%rax
  shl    $0x5,%rax
  mov    %rax,0x20(%rsp)
  shl    $0x4,%r13
  add    %r14,%r13
  mov    %ebx,%eax
  mov    %rax,0x28(%rsp)
  mov    $0x0,%ebx
L16:
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x10(%rsp),%rax
  subsd  (%rax,%rbx,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x10(%rsp),%rax
  addsd  (%rax,%rbx,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  mov    0x8(%rsp),%rax
  lea    (%rax,%r14,1),%rdi
  mov    %r15d,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x18(%rsp),%rcx
  subsd  (%rcx,%rbx,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  mov    %r15d,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  movsd  (%r12,%rbx,8),%xmm0
  mov    0x18(%rsp),%rcx
  addsd  (%rcx,%rbx,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  mov    0x8(%rsp),%rax
  lea    (%rax,%r13,1),%rdi
  mov    %r15d,%edx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  mov    0x20(%rsp),%rax
  add    %rax,%r14
  movsd  (%r12,%rbx,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
  add    %rax,%r13
  cmp    0x28(%rsp),%rbx
  jne    L16
L19:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void evalR0_0fs4d(double *pts, unsigned dim, double *p, const double *c, const double *r1, const double *r2) {
  unsigned i;
  memcpy(pts, p, sizeof(double) * dim);
  pts += dim;
  for (i = 0; i < dim; i++) {
    p[i] = c[i] - r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] - r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<evalR0_0fs4d>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rsi,8),%ebp
  push   %rbx
  mov    %ebp,%edx
  mov    %esi,%ebx
  mov    %r13,%rsi
  sub    $0x38,%rsp
  mov    %r8,0x10(%rsp)
  mov    %r9,0x18(%rsp)
  call   <memcpy@plt>
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rcx
  mov    %rax,0x28(%rsp)
  mov    %rcx,0x8(%rsp)
  add    %rcx,%r12
  test   %ebx,%ebx
  je     L3
  mov    %rax,%r9
  shl    $0x4,%rax
  xor    %r15d,%r15d
  mov    %rax,%rbx
  shl    $0x5,%r9
  add    %r12,%rbx
  nopl   0x0(%rax,%rax,1)
L9:
  mov    0x10(%rsp),%rax
  movsd  (%r14,%r15,8),%xmm0
  mov    %r12,%rdi
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %r9,0x20(%rsp)
  subsd  (%rax,%r15,8),%xmm0
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    %ebp,%edx
  mov    %r13,%rsi
  movsd  (%r14,%r15,8),%xmm0
  addsd  (%rax,%r15,8),%xmm0
  mov    0x8(%rsp),%rax
  lea    (%rax,%r12,1),%rdi
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x18(%rsp),%rcx
  mov    %rbx,%rdi
  mov    %ebp,%edx
  movsd  (%r14,%r15,8),%xmm0
  mov    %r13,%rsi
  subsd  (%rcx,%r15,8),%xmm0
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x18(%rsp),%rcx
  mov    %ebp,%edx
  mov    %r13,%rsi
  movsd  (%r14,%r15,8),%xmm0
  mov    0x8(%rsp),%rax
  addsd  (%rcx,%r15,8),%xmm0
  lea    (%rax,%rbx,1),%rdi
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x20(%rsp),%r9
  movsd  (%r14,%r15,8),%xmm0
  add    %r9,%r12
  movsd  %xmm0,0x0(%r13,%r15,8)
  add    %r9,%rbx
  add    $0x1,%r15
  cmp    0x28(%rsp),%r15
  jne    L9
L3:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void evalR0_0fs4d(double *pts, unsigned dim, double *p, const double *c, const double *r1, const double *r2) {
  unsigned i;
  memcpy(pts, p, sizeof(double) * dim);
  pts += dim;
  for (i = 0; i < dim; i++) {
    p[i] = c[i] - r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] - r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<evalR0_0fs4d>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rsi,8),%ebp
  push   %rbx
  mov    %ebp,%edx
  mov    %esi,%ebx
  mov    %r13,%rsi
  sub    $0x38,%rsp
  mov    %r8,0x10(%rsp)
  mov    %r9,0x18(%rsp)
  call   <memcpy@plt>
  mov    %ebx,%eax
  lea    0x0(,%rax,8),%rcx
  mov    %rax,0x28(%rsp)
  mov    %rcx,0x8(%rsp)
  add    %rcx,%r12
  test   %ebx,%ebx
  je     L10
  mov    %rax,%r9
  shl    $0x4,%rax
  xor    %r15d,%r15d
  mov    %rax,%rbx
  shl    $0x5,%r9
  add    %r12,%rbx
  nopl   0x0(%rax,%rax,1)
L5:
  mov    0x10(%rsp),%rax
  movsd  (%r14,%r15,8),%xmm0
  mov    %r12,%rdi
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %r9,0x20(%rsp)
  subsd  (%rax,%r15,8),%xmm0
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x10(%rsp),%rax
  mov    %ebp,%edx
  mov    %r13,%rsi
  movsd  (%r14,%r15,8),%xmm0
  addsd  (%rax,%r15,8),%xmm0
  mov    0x8(%rsp),%rax
  lea    (%rax,%r12,1),%rdi
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x18(%rsp),%rcx
  mov    %rbx,%rdi
  mov    %ebp,%edx
  movsd  (%r14,%r15,8),%xmm0
  mov    %r13,%rsi
  subsd  (%rcx,%r15,8),%xmm0
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x18(%rsp),%rcx
  mov    %ebp,%edx
  mov    %r13,%rsi
  movsd  (%r14,%r15,8),%xmm0
  mov    0x8(%rsp),%rax
  addsd  (%rcx,%r15,8),%xmm0
  lea    (%rax,%rbx,1),%rdi
  movsd  %xmm0,0x0(%r13,%r15,8)
  call   <memcpy@plt>
  mov    0x20(%rsp),%r9
  movsd  (%r14,%r15,8),%xmm0
  add    %r9,%r12
  movsd  %xmm0,0x0(%r13,%r15,8)
  add    %r9,%rbx
  add    $0x1,%r15
  cmp    %r15,0x28(%rsp)
  jne    L5
L10:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void evalR0_0fs4d(double *pts, unsigned dim, double *p, const double *c, const double *r1, const double *r2) {
  unsigned i;
  memcpy(pts, p, sizeof(double) * dim);
  pts += dim;
  for (i = 0; i < dim; i++) {
    p[i] = c[i] - r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r1[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] - r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i] + r2[i];
    memcpy(pts, p, sizeof(double) * dim);
    pts += dim;
    p[i] = c[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<chirp_ticket_isticketsubject>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x34(%rbp),%rsi
  lea    -0x30(%rbp),%rcx
  lea    -0x38(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rsi,%r8
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  jne    L4
  mov    -0x34(%rbp),%eax
  mov    -0x38(%rbp),%edx
  sub    %edx,%eax
  cmp    $0x20,%eax
  jne    L4
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  cmp    %rdx,%rax
  jne    L4
  mov    -0x38(%rbp),%eax
  test   %eax,%eax
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rax,%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  jmp    L2
L4:
  mov    $0x0,%eax
L2:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string

tool: D9: ticket:%n%32[0123456789abcdefABCDEF]%n

assistant: ```c
int chirp_ticket_isticketsubject(const char *ticket_subject, const char **digest) {
  int n1, n2;
  char buffer[((16 << 1)) + 1];
  if ((sscanf(ticket_subject,
              "ticket:"
              "%n%32[0123456789abcdefABCDEF]%n",
              &n1, buffer, &n2) == 1) &&
      ((n2 - n1) == ((16 << 1))) && (strlen(ticket_subject) == (size_t)n2)) {
    assert(n1 > 0);
    *digest = ticket_subject + n1;
    return 1;
  } else {
    return 0;
  }
}
```