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

```asm
<main>:
  endbr64
  push   %rbp
  lea    D61(%rip),%rdi
  lea    D56(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movss  (%rsp),%xmm0
  addss  0x4(%rsp),%xmm0
  lea    D50(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  subss  0x4(%rsp),%xmm0
  lea    D12(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  movss  (%rsp),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D6(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D52(%rip),%rdi
  mov    $0x1,%eax
  movss  0x4(%rsp),%xmm0
  divss  (%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D61
datatype: string
name: read_data
arguments: label: D56
datatype: string
name: read_data
arguments: label: D27
datatype: string
name: read_data
arguments: label: D50
datatype: string
name: read_data
arguments: label: D12
datatype: string
name: read_data
arguments: label: D6
datatype: string
name: read_data
arguments: label: D52
datatype: string

tool: D61: What is your favorite number? 

tool: D56: %f

tool: D27: How old are you? 

tool: D50: The sum of your favorite number and your age is %.1f 


tool: D12: The difference between your favorite number and your age is %.1f 


tool: D6: The product of your favorite number and your age is %.1f 


tool: D52: The quotient between your favorite number and your age is %f 


assistant: ```c
int main() {
  float favoritenumber, userage;
  float sum, product, quotient, difference;
  printf("What is your favorite number? \n");
  scanf("%f", &favoritenumber);
  printf("How old are you? \n");
  scanf("%f", &userage);
  sum = (favoritenumber + userage);
  printf("The sum of your favorite number and your age is %.1f \n", sum);
  difference = (favoritenumber - userage);
  printf("The difference between your favorite number and your age is %.1f \n", difference);
  product = (favoritenumber * userage);
  printf("The product of your favorite number and your age is %.1f \n", product);
  quotient = (userage / favoritenumber);
  printf("The quotient between your favorite number and your age is %f \n", quotient);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strlwr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L9
L15:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x14(%rbp)
L9:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void strlwr(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; s[i] = tolower(s[i]), i++)
    ;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strlwr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L14
L7:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L7
L14:
  pop    %rbx
  ret
```

assistant: ```c
void strlwr(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; s[i] = tolower(s[i]), i++)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strlwr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L5
L8:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L8
L5:
  pop    %rbx
  ret
```

assistant: ```c
void strlwr(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; s[i] = tolower(s[i]), i++)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlwr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L15
L0:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L0
L15:
  pop    %rbx
  ret
```

assistant: ```c
void strlwr(char *s) {
  int i;
  for (i = 0; s[i] != '\0'; s[i] = tolower(s[i]), i++)
    ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<part>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L25:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L12
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L11
L46:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x10(%rbp)
L11:
  cmpl   $0x0,-0x10(%rbp)
  jg     L46
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L12:
  addl   $0x1,-0xc(%rbp)
L33:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L25
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void part(int *array, int n) {
  int temp0 = array[0];
  int j = 0;
  for (int i = 0; i < n; ++i) {
    if (array[i] < temp0) {
      int tempi = array[i];
      for (j = i; j >= 1; --j) {
        array[j] = array[j - 1];
      }
      array[j] = tempi;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<part>:
  endbr64
  mov    (%rdi),%r10d
  test   %esi,%esi
  jle    L64
  lea    0x4(%rdi),%r8
  mov    $0x0,%ecx
  mov    $0x0,%r11d
  jmp    L9
L53:
  mov    %ecx,%eax
L0:
  cltq
  mov    %r9d,(%rdi,%rax,4)
L38:
  add    $0x4,%r8
L9:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  je     L64
  mov    (%r8),%r9d
  cmp    %r10d,%r9d
  jge    L38
  test   %ecx,%ecx
  jle    L53
  mov    %r8,%rax
L42:
  mov    -0x4(%rax),%edx
  mov    %edx,(%rax)
  sub    $0x4,%rax
  cmp    %rdi,%rax
  jne    L42
  mov    %r11d,%eax
  jmp    L0
L64:
  ret
```

assistant: ```c
void part(int *array, int n) {
  int temp0 = array[0];
  int j = 0;
  for (int i = 0; i < n; ++i) {
    if (array[i] < temp0) {
      int tempi = array[i];
      for (j = i; j >= 1; --j) {
        array[j] = array[j - 1];
      }
      array[j] = tempi;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<part>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%r14d
  test   %esi,%esi
  jle    L47
  movslq %esi,%rsi
  mov    %rdi,%r12
  lea    0x4(%rdi),%rcx
  mov    $0x4,%ebx
  lea    0x0(,%rsi,4),%r13
  jmp    L30
  nopl   0x0(%rax)
L32:
  add    $0x4,%rbx
L30:
  cmp    %rbx,%r13
  je     L47
  mov    (%r12,%rbx,1),%ebp
  cmp    %r14d,%ebp
  jge    L32
  mov    %rcx,%rdi
  mov    %rbx,%rdx
  mov    %r12,%rsi
  call   <memmove@plt>
  mov    %ebp,(%r12)
  mov    %rax,%rcx
  jmp    L32
  nopl   0x0(%rax)
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void part(int *array, int n) {
  int temp0 = array[0];
  int j = 0;
  for (int i = 0; i < n; ++i) {
    if (array[i] < temp0) {
      int tempi = array[i];
      for (j = i; j >= 1; --j) {
        array[j] = array[j - 1];
      }
      array[j] = tempi;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<part>:
  endbr64
  push   %r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  shl    $0x2,%r13
  push   %rbp
  push   %rbx
  mov    (%rdi),%r14d
  cmp    $0x1,%esi
  jle    L10
  mov    %rdi,%r12
  mov    $0x4,%ebx
  lea    0x4(%rdi),%rcx
  jmp    L7
  nopl   0x0(%rax)
L44:
  add    $0x4,%rbx
  cmp    %rbx,%r13
  je     L10
L7:
  mov    (%r12,%rbx,1),%ebp
  cmp    %r14d,%ebp
  jge    L44
  mov    %rbx,%rdx
  mov    %rcx,%rdi
  mov    %r12,%rsi
  add    $0x4,%rbx
  call   <memmove@plt>
  mov    %ebp,(%r12)
  mov    %rax,%rcx
  cmp    %rbx,%r13
  jne    L7
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void part(int *array, int n) {
  int temp0 = array[0];
  int j = 0;
  for (int i = 0; i < n; ++i) {
    if (array[i] < temp0) {
      int tempi = array[i];
      for (j = i; j >= 1; --j) {
        array[j] = array[j - 1];
      }
      array[j] = tempi;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  jmp    L29
L20:
  mov    -0x18(%rbp),%eax
  cltq
  movl   $0x0,-0x14(%rbp,%rax,4)
  addl   $0x1,-0x18(%rbp)
L29:
  cmpl   $0x2,-0x18(%rbp)
  jle    L20
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int main() {
  int a[3];
  int i;
  for (i = 0; i < 3; i++)
    a[i] = 0;
  return a[0];
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int a[3];
  int i;
  for (i = 0; i < 3; i++)
    a[i] = 0;
  return a[0];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int a[3];
  int i;
  for (i = 0; i < 3; i++)
    a[i] = 0;
  return a[0];
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int a[3];
  int i;
  for (i = 0; i < 3; i++)
    a[i] = 0;
  return a[0];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<isneg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L12
L33:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  pxor   %xmm0,%xmm0
  comisd %xmm1,%xmm0
  jbe    L39
  mov    $0x1,%eax
  jmp    L5
L39:
  addq   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L33
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int32_t isneg(double *y, int32_t sz) {
  for (int32_t i = 0; i < sz; i++)
    if (y[i] < 0)
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<isneg>:
  endbr64
  test   %rsi,%rsi
  je     L14
  mov    $0x0,%edx
  pxor   %xmm0,%xmm0
L5:
  comisd (%rdi,%rdx,8),%xmm0
  ja     L1
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L5
  mov    $0x0,%eax
  ret
L14:
  mov    %rsi,%rax
  ret
L1:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int32_t isneg(double *y, int32_t sz) {
  for (int32_t i = 0; i < sz; i++)
    if (y[i] < 0)
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<isneg>:
  endbr64
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L34
  pxor   %xmm0,%xmm0
  jmp    L26
  nopl   0x0(%rax)
L39:
  add    $0x1,%rax
  cmp    %rax,%rsi
  je     L12
L26:
  comisd (%rdi,%rax,8),%xmm0
  jbe    L39
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L12:
  xor    %eax,%eax
  ret
L34:
  ret
```

assistant: ```c
int32_t isneg(double *y, int32_t sz) {
  for (int32_t i = 0; i < sz; i++)
    if (y[i] < 0)
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isneg>:
  endbr64
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L32
  pxor   %xmm0,%xmm0
  jmp    L2
  nopl   0x0(%rax)
L16:
  add    $0x1,%rax
  cmp    %rax,%rsi
  je     L35
L2:
  comisd (%rdi,%rax,8),%xmm0
  jbe    L16
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L35:
  xor    %eax,%eax
  ret
L32:
  ret
```

assistant: ```c
int32_t isneg(double *y, int32_t sz) {
  for (int32_t i = 0; i < sz; i++)
    if (y[i] < 0)
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <sin@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm1
  mulsd  -0x18(%rbp),%xmm1
  movsd  %xmm1,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double f(double x) {
  double value = x * sin(x) + cos(x);
  return value;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<f>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  movq   %xmm0,%rbx
  lea    0x8(%rsp),%rdi
  mov    %rsp,%rsi
  call   <sincos@plt>
  movq   %rbx,%xmm0
  mulsd  0x8(%rsp),%xmm0
  addsd  (%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
double f(double x) {
  double value = x * sin(x) + cos(x);
  return value;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<f>:
  endbr64
  sub    $0x28,%rsp
  lea    0x18(%rsp),%rdi
  lea    0x10(%rsp),%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <sincos@plt>
  movsd  0x8(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  add    $0x28,%rsp
  ret
```

assistant: ```c
double f(double x) {
  double value = x * sin(x) + cos(x);
  return value;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f>:
  endbr64
  sub    $0x28,%rsp
  lea    0x18(%rsp),%rdi
  lea    0x10(%rsp),%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <sincos@plt>
  movsd  0x8(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  addsd  0x10(%rsp),%xmm0
  add    $0x28,%rsp
  ret
```

assistant: ```c
double f(double x) {
  double value = x * sin(x) + cos(x);
  return value;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<copmr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9,-0x38(%rbp)
  mov    -0x28(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x28(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%esi
  mov    -0x38(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <verbatim_copy@plt>
  mov    -0x28(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%esi
  mov    -0x20(%rbp),%rax
  mov    %edx,%ecx
  mov    0x10(%rbp),%rdx
  mov    %rax,%rdi
  call   <verbatim_copy@plt>
  mov    -0x28(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L13
L2:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    0x10(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L2
  nop
  leave
  ret
```

assistant: ```c
void copmr(double *matrixInR, double *matrixInI, int sigInLength, int InRow, int InCol, double *matrixOutR, double *matrixOutI) {
  int i;
  verbatim_copy(matrixInR, InRow * InCol, matrixOutR, InRow * InCol);
  verbatim_copy(matrixInI, InRow * InCol, matrixOutI, InRow * InCol);
  for (i = InRow * InCol; i < sigInLength; i++) {
    matrixOutR[i] = matrixInR[i] + matrixInI[i];
    matrixOutI[i] = matrixInR[i] - matrixInI[i];
  }
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copmr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  mov    %r9,%r13
  mov    0x40(%rsp),%r14
  imul   %r8d,%ecx
  mov    %ecx,%r15d
  mov    %r9,%rdx
  mov    %ecx,%esi
  call   <verbatim_copy@plt>
  mov    %r15d,%ecx
  mov    %r14,%rdx
  mov    %r15d,%esi
  mov    %rbp,%rdi
  call   <verbatim_copy@plt>
  cmp    %r12d,%r15d
  jge    L9
  movslq %r15d,%rcx
L19:
  movsd  (%rbx,%rcx,8),%xmm0
  addsd  0x0(%rbp,%rcx,8),%xmm0
  movsd  %xmm0,0x0(%r13,%rcx,8)
  movsd  (%rbx,%rcx,8),%xmm0
  subsd  0x0(%rbp,%rcx,8),%xmm0
  movsd  %xmm0,(%r14,%rcx,8)
  add    $0x1,%rcx
  cmp    %ecx,%r12d
  jg     L19
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void copmr(double *matrixInR, double *matrixInI, int sigInLength, int InRow, int InCol, double *matrixOutR, double *matrixOutI) {
  int i;
  verbatim_copy(matrixInR, InRow * InCol, matrixOutR, InRow * InCol);
  verbatim_copy(matrixInI, InRow * InCol, matrixOutI, InRow * InCol);
  for (i = InRow * InCol; i < sigInLength; i++) {
    matrixOutR[i] = matrixInR[i] + matrixInI[i];
    matrixOutI[i] = matrixInR[i] - matrixInI[i];
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<copmr>:
  endbr64
  push   %r15
  push   %r14
  mov    %r9,%r14
  push   %r13
  mov    %edx,%r13d
  mov    %r9,%rdx
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  imul   %r8d,%ebx
  sub    $0x8,%rsp
  mov    0x40(%rsp),%r15
  mov    %ebx,%ecx
  mov    %ebx,%esi
  call   <verbatim_copy@plt>
  mov    %ebx,%ecx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    %r15,%rdx
  call   <verbatim_copy@plt>
  cmp    %r13d,%ebx
  jge    L1
  movslq %ebx,%rcx
  nopl   0x0(%rax)
L16:
  movsd  0x0(%rbp,%rcx,8),%xmm0
  addsd  (%r12,%rcx,8),%xmm0
  movsd  %xmm0,(%r14,%rcx,8)
  movsd  0x0(%rbp,%rcx,8),%xmm0
  subsd  (%r12,%rcx,8),%xmm0
  movsd  %xmm0,(%r15,%rcx,8)
  add    $0x1,%rcx
  cmp    %ecx,%r13d
  jg     L16
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void copmr(double *matrixInR, double *matrixInI, int sigInLength, int InRow, int InCol, double *matrixOutR, double *matrixOutI) {
  int i;
  verbatim_copy(matrixInR, InRow * InCol, matrixOutR, InRow * InCol);
  verbatim_copy(matrixInI, InRow * InCol, matrixOutI, InRow * InCol);
  for (i = InRow * InCol; i < sigInLength; i++) {
    matrixOutR[i] = matrixInR[i] + matrixInI[i];
    matrixOutI[i] = matrixInR[i] - matrixInI[i];
  }
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<copmr>:
  endbr64
  push   %r15
  imul   %r8d,%ecx
  push   %r14
  mov    %edx,%r14d
  mov    %r9,%rdx
  push   %r13
  mov    %r9,%r13
  push   %r12
  mov    %rsi,%r12
  mov    %ecx,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%r15
  call   <verbatim_copy@plt>
  mov    %ebx,%ecx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    %r15,%rdx
  call   <verbatim_copy@plt>
  cmp    %r14d,%ebx
  jge    L46
  mov    %r14d,%eax
  sub    %ebx,%eax
  mov    %eax,0x4(%rsp)
  movslq %ebx,%rax
  lea    0x10(,%rax,8),%r8
  lea    0x0(,%rax,8),%rdx
  lea    -0x8(%r8),%r10
  lea    0x0(%r13,%r8,1),%rsi
  lea    0x0(%rbp,%r10,1),%r9
  mov    %rsi,0x8(%rsp)
  lea    (%r15,%rdx,1),%rsi
  lea    0x0(%r13,%r10,1),%r11
  lea    0x0(%r13,%rdx,1),%rdi
  lea    0x0(%rbp,%rdx,1),%rcx
  add    %r12,%rdx
  cmp    %r9,%rsi
  setne  %r9b
  cmp    %r11,%rsi
  setne  %r11b
  and    %r9d,%r11d
  cmpl   $0x1,0x4(%rsp)
  setne  %r9b
  add    %r12,%r10
  and    %r11d,%r9d
  cmp    %r10,%rsi
  setne  %r10b
  and    %r10d,%r9d
  lea    0x0(%rbp,%r8,1),%r10
  cmp    %r10,%rdi
  setae  %r10b
  cmp    %rcx,0x8(%rsp)
  setbe  %r11b
  or     %r11d,%r10d
  test   %r10b,%r9b
  je     L39
  add    %r12,%r8
  cmp    %r8,%rdi
  setae  %r8b
  cmp    %rdx,0x8(%rsp)
  setbe  %r9b
  or     %r9b,%r8b
  je     L39
  mov    0x4(%rsp),%r8d
  xor    %eax,%eax
  shr    %r8d
  shl    $0x4,%r8
  nopl   (%rax)
L63:
  movupd (%rcx,%rax,1),%xmm0
  movupd (%rdx,%rax,1),%xmm1
  addpd  %xmm1,%xmm0
  movups %xmm0,(%rdi,%rax,1)
  movupd (%rcx,%rax,1),%xmm0
  movupd (%rdx,%rax,1),%xmm2
  subpd  %xmm2,%xmm0
  movups %xmm0,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %r8,%rax
  jne    L63
  mov    0x4(%rsp),%edi
  mov    %edi,%eax
  and    $0xfffffffe,%eax
  add    %eax,%ebx
  cmp    %eax,%edi
  je     L46
  movslq %ebx,%rbx
  lea    0x0(,%rbx,8),%rax
  add    %rax,%rbp
  add    %rax,%r12
  movsd  0x0(%rbp),%xmm0
  addsd  (%r12),%xmm0
  movsd  %xmm0,0x0(%r13,%rbx,8)
  movsd  0x0(%rbp),%xmm0
  subsd  (%r12),%xmm0
  movsd  %xmm0,(%r15,%rbx,8)
L46:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  movsd  0x0(%rbp,%rax,8),%xmm0
  addsd  (%r12,%rax,8),%xmm0
  movsd  %xmm0,0x0(%r13,%rax,8)
  movsd  0x0(%rbp,%rax,8),%xmm0
  subsd  (%r12,%rax,8),%xmm0
  movsd  %xmm0,(%r15,%rax,8)
  add    $0x1,%rax
  cmp    %eax,%r14d
  jg     L39
  jmp    L46
```

assistant: ```c
void copmr(double *matrixInR, double *matrixInI, int sigInLength, int InRow, int InCol, double *matrixOutR, double *matrixOutI) {
  int i;
  verbatim_copy(matrixInR, InRow * InCol, matrixOutR, InRow * InCol);
  verbatim_copy(matrixInI, InRow * InCol, matrixOutI, InRow * InCol);
  for (i = InRow * InCol; i < sigInLength; i++) {
    matrixOutR[i] = matrixInR[i] + matrixInI[i];
    matrixOutI[i] = matrixInR[i] - matrixInI[i];
  }
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

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

tool: D6: Usage: huffpuff [--character-map=FILE]
                [--table-output=FILE] [--data-output=FILE]
                [--table-label=LABEL] [--node-label-prefix=PREFIX]
                [--string-label-prefix=PREFIX]
                [--generate-string-table] [--append-byte=VALUE]
                [--ignore-case] [--verbose]
                [--help] [--usage] [--version]
                FILE


assistant: ```c
void usage() {
  printf("Usage: huffpuff [--character-map=FILE]\n"
         "                [--table-output=FILE] [--data-output=FILE]\n"
         "                [--table-label=LABEL] [--node-label-prefix=PREFIX]\n"
         "                [--string-label-prefix=PREFIX]\n"
         "                [--generate-string-table] [--append-byte=VALUE]\n"
         "                [--ignore-case] [--verbose]\n"
         "                [--help] [--usage] [--version]\n"
         "                FILE\n");
  exit(0);
}
```