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

```asm
<matnorm_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L22
L27:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L37
L7:
  mov    -0x14(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <fabs@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0x18(%rbp)
L37:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L7
  movsd  -0x8(%rbp),%xmm0
  comisd -0x10(%rbp),%xmm0
  jbe    L5
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L5:
  addl   $0x1,-0x14(%rbp)
L22:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L27
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double matnorm_1(const double *x, const int n) {
  int i, j;
  double norm = 0;
  double tmp;
  for (j = 0; j < n; j++) {
    tmp = 0;
    for (i = 0; i < n; i++)
      tmp += fabs(x[i + j * n]);
    if (tmp > norm)
      norm = tmp;
  }
  return norm;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<matnorm_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  jle    L9
  mov    %esi,%r13d
  movslq %esi,%r15
  shl    $0x3,%r15
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
  pxor   %xmm5,%xmm5
  movsd  %xmm5,0x8(%rsp)
  mov    $0x0,%r12d
  mov    %esi,%r14d
  neg    %r14
  shl    $0x3,%r14
L27:
  lea    (%r14,%rbp,1),%rbx
  pxor   %xmm4,%xmm4
  movsd  %xmm4,(%rsp)
L26:
  movsd  (%rbx),%xmm0
  call   <fabs@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L26
  movapd %xmm0,%xmm2
  maxsd  0x8(%rsp),%xmm2
  movsd  %xmm2,0x8(%rsp)
  add    $0x1,%r12d
  add    %r15,%rbp
  cmp    %r12d,%r13d
  jne    L27
L4:
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  pxor   %xmm6,%xmm6
  movsd  %xmm6,0x8(%rsp)
  jmp    L4
```

assistant: ```c
double matnorm_1(const double *x, const int n) {
  int i, j;
  double norm = 0;
  double tmp;
  for (j = 0; j < n; j++) {
    tmp = 0;
    for (i = 0; i < n; i++)
      tmp += fabs(x[i + j * n]);
    if (tmp > norm)
      norm = tmp;
  }
  return norm;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<matnorm_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  jle    L8
  lea    -0x1(%rsi),%eax
  movslq %esi,%r15
  mov    $0xfffffffffffffff8,%r14
  mov    %esi,%r13d
  movq   $0x0,0x8(%rsp)
  shl    $0x3,%rax
  shl    $0x3,%r15
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,1),%rbp
  sub    %rax,%r14
  cs nopw 0x0(%rax,%rax,1)
L16:
  lea    (%r14,%rbp,1),%rbx
  pxor   %xmm1,%xmm1
  nopl   0x0(%rax,%rax,1)
L18:
  movsd  (%rbx),%xmm0
  movsd  %xmm1,(%rsp)
  add    $0x8,%rbx
  call   <fabs@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm1
  cmp    %rbp,%rbx
  jne    L18
  maxsd  0x8(%rsp),%xmm1
  add    $0x1,%r12d
  add    %r15,%rbp
  movsd  %xmm1,0x8(%rsp)
  cmp    %r12d,%r13d
  jne    L16
L25:
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L25
```

assistant: ```c
double matnorm_1(const double *x, const int n) {
  int i, j;
  double norm = 0;
  double tmp;
  for (j = 0; j < n; j++) {
    tmp = 0;
    for (i = 0; i < n; i++)
      tmp += fabs(x[i + j * n]);
    if (tmp > norm)
      norm = tmp;
  }
  return norm;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<matnorm_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  jle    L26
  lea    -0x1(%rsi),%eax
  movslq %esi,%r15
  mov    $0xfffffffffffffff8,%r14
  mov    %esi,%r13d
  movq   $0x0,0x8(%rsp)
  shl    $0x3,%rax
  shl    $0x3,%r15
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,1),%rbp
  sub    %rax,%r14
  cs nopw 0x0(%rax,%rax,1)
L34:
  lea    (%r14,%rbp,1),%rbx
  pxor   %xmm1,%xmm1
  nopl   0x0(%rax,%rax,1)
L18:
  movsd  (%rbx),%xmm0
  movsd  %xmm1,(%rsp)
  add    $0x8,%rbx
  call   <fabs@plt>
  pxor   %xmm0,%xmm0
  movsd  (%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm1
  cmp    %rbp,%rbx
  jne    L18
  maxsd  0x8(%rsp),%xmm1
  add    $0x1,%r12d
  add    %r15,%rbp
  movsd  %xmm1,0x8(%rsp)
  cmp    %r12d,%r13d
  jne    L34
L11:
  movsd  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x8(%rsp)
  jmp    L11
```

assistant: ```c
double matnorm_1(const double *x, const int n) {
  int i, j;
  double norm = 0;
  double tmp;
  for (j = 0; j < n; j++) {
    tmp = 0;
    for (i = 0; i < n; i++)
      tmp += fabs(x[i + j * n]);
    if (tmp > norm)
      norm = tmp;
  }
  return norm;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<char_in_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L4:
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x14(%rbp)
  jne    L24
  mov    -0x4(%rbp),%eax
  jmp    L10
L24:
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    %rax,-0x28(%rbp)
  ja     L4
  mov    $0xffffffff,%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int char_in_array(char c, const char a[], size_t size) {
  unsigned int i = 0;
  while (i < size) {
    if (c == a[i]) {
      return i;
    }
    ++i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<char_in_array>:
  endbr64
  test   %rdx,%rdx
  je     L25
  mov    $0x0,%eax
  mov    $0x0,%ecx
L27:
  cmp    %dil,(%rsi,%rcx,1)
  je     L23
  add    $0x1,%eax
  mov    %eax,%ecx
  cmp    %rdx,%rcx
  jb     L27
  mov    $0xffffffff,%eax
  ret
L25:
  mov    $0xffffffff,%eax
L23:
  ret
```

assistant: ```c
int char_in_array(char c, const char a[], size_t size) {
  unsigned int i = 0;
  while (i < size) {
    if (c == a[i]) {
      return i;
    }
    ++i;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<char_in_array>:
  endbr64
  test   %rdx,%rdx
  je     L31
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L12
  nop
L29:
  lea    0x1(%rax),%ecx
  mov    %rcx,%rax
  cmp    %rdx,%rcx
  jae    L31
L12:
  cmp    %dil,(%rsi,%rcx,1)
  jne    L29
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int char_in_array(char c, const char a[], size_t size) {
  unsigned int i = 0;
  while (i < size) {
    if (c == a[i]) {
      return i;
    }
    ++i;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<char_in_array>:
  endbr64
  test   %rdx,%rdx
  je     L22
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L28
  nop
L13:
  lea    0x1(%rax),%ecx
  mov    %rcx,%rax
  cmp    %rdx,%rcx
  jae    L22
L28:
  cmp    %dil,(%rsi,%rcx,1)
  jne    L13
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int char_in_array(char c, const char a[], size_t size) {
  unsigned int i = 0;
  while (i < size) {
    if (c == a[i]) {
      return i;
    }
    ++i;
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x4,-0xc(%rbp)
  movl   $0x4,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  const int a = 4;
  const struct A {
    int a;
    int b;
  } X;
  struct B {
    const int a;
    int b;
  } Y;
  Y.b = 4;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main() {
  const int a = 4;
  const struct A {
    int a;
    int b;
  } X;
  struct B {
    const int a;
    int b;
  } Y;
  Y.b = 4;
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
int main() {
  const int a = 4;
  const struct A {
    int a;
    int b;
  } X;
  struct B {
    const int a;
    int b;
  } Y;
  Y.b = 4;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

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

assistant: ```c
int main() {
  const int a = 4;
  const struct A {
    int a;
    int b;
  } X;
  struct B {
    const int a;
    int b;
  } Y;
  Y.b = 4;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fromuint64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    $0x20,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fromuint64(uint64_t v, int *fd, int *id) {
  *fd = (int)(v >> 32);
  *id = (int)(v & 0xffffffff);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fromuint64>:
  endbr64
  mov    %edi,%eax
  mov    $0x20,%ecx
  sar    %cl,%eax
  mov    %eax,(%rsi)
  mov    %edi,(%rdx)
  ret
```

assistant: ```c
void fromuint64(uint64_t v, int *fd, int *id) {
  *fd = (int)(v >> 32);
  *id = (int)(v & 0xffffffff);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fromuint64>:
  endbr64
  mov    %edi,%eax
  mov    $0x20,%ecx
  sar    %cl,%eax
  mov    %eax,(%rsi)
  mov    %edi,(%rdx)
  ret
```

assistant: ```c
void fromuint64(uint64_t v, int *fd, int *id) {
  *fd = (int)(v >> 32);
  *id = (int)(v & 0xffffffff);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fromuint64>:
  endbr64
  mov    %edi,%eax
  mov    $0x20,%ecx
  sar    %cl,%eax
  mov    %eax,(%rsi)
  mov    %edi,(%rdx)
  ret
```

assistant: ```c
void fromuint64(uint64_t v, int *fd, int *id) {
  *fd = (int)(v >> 32);
  *id = (int)(v & 0xffffffff);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D26
datatype: string
name: read_data
arguments: label: D22
datatype: string
name: read_data
arguments: label: D4
datatype: string

tool: D26: Running ps with system


tool: D22: ps -eaf &

tool: D4: Done.


assistant: ```c
int main() {
  printf("Running ps with system\n");
  system("ps -eaf &");
  printf("Done.\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <system@plt>
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
type_guessed: string
name: read_data
arguments: data_tag: D1
type_guessed: string
name: read_data
arguments: data_tag: D22
type_guessed: string

tool: D6: Running ps with system


tool: D1: ps -eaf &

tool: D22: Done.


assistant: ```c
int main() {
  printf("Running ps with system\n");
  system("ps -eaf &");
  printf("Done.\n");
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <system@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D12
type: string
name: parse_data
arguments: label: D20
type: string
name: parse_data
arguments: label: D29
type: string

tool: D12: Running ps with system


tool: D20: ps -eaf &

tool: D29: Done.


assistant: ```c
int main() {
  printf("Running ps with system\n");
  system("ps -eaf &");
  printf("Done.\n");
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <system@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D11
type_guessed: string
name: unpack_data
arguments: tag: D20
type_guessed: string
name: unpack_data
arguments: tag: D21
type_guessed: string

tool: D11: Running ps with system


tool: D20: ps -eaf &

tool: D21: Done.


assistant: ```c
int main() {
  printf("Running ps with system\n");
  system("ps -eaf &");
  printf("Done.\n");
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <file1@plt>
  mov    $0x0,%eax
  call   <file2@plt>
  mov    $0x0,%eax
  call   <file3@plt>
  mov    $0x0,%eax
  call   <file4@plt>
  mov    $0x0,%eax
  call   <file5@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
type_guessed: string

tool: D2: All Functions Executed


assistant: ```c
int main(void) {
  file1();
  file2();
  file3();
  file4();
  file5();
  printf("All Functions Executed\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <file1@plt>
  mov    $0x0,%eax
  call   <file2@plt>
  mov    $0x0,%eax
  call   <file3@plt>
  mov    $0x0,%eax
  call   <file4@plt>
  mov    $0x0,%eax
  call   <file5@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string

tool: D9: All Functions Executed


assistant: ```c
int main(void) {
  file1();
  file2();
  file3();
  file4();
  file5();
  printf("All Functions Executed\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <file1@plt>
  xor    %eax,%eax
  call   <file2@plt>
  xor    %eax,%eax
  call   <file3@plt>
  xor    %eax,%eax
  call   <file4@plt>
  xor    %eax,%eax
  call   <file5@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
data_type: string

tool: D8: All Functions Executed


assistant: ```c
int main(void) {
  file1();
  file2();
  file3();
  file4();
  file5();
  printf("All Functions Executed\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <file1@plt>
  xor    %eax,%eax
  call   <file2@plt>
  xor    %eax,%eax
  call   <file3@plt>
  xor    %eax,%eax
  call   <file4@plt>
  xor    %eax,%eax
  call   <file5@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_infered: string

tool: D9: All Functions Executed


assistant: ```c
int main(void) {
  file1();
  file2();
  file3();
  file4();
  file5();
  printf("All Functions Executed\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ho2cuLoopf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %rdx,-0x98(%rbp)
  mov    %ecx,-0x9c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x3,-0x60(%rbp)
  movq   $0x3,-0x58(%rbp)
  movq   $0x0,-0x68(%rbp)
  jmp    L67
L16:
  mov    -0x68(%rbp),%rax
  imul   -0x60(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  mov    -0x68(%rbp),%rax
  imul   -0x58(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  movq   $0x0,-0x70(%rbp)
  jmp    L15
L41:
  mov    -0x50(%rbp),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    -0x70(%rbp),%rax
  movsd  %xmm0,-0x40(%rbp,%rax,8)
  addq   $0x1,-0x70(%rbp)
L15:
  mov    -0x70(%rbp),%rax
  cmp    -0x60(%rbp),%rax
  jb     L41
  mov    -0x9c(%rbp),%edx
  lea    -0x20(%rbp),%rcx
  lea    -0x40(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ho2cu@plt>
  mov    %eax,-0x74(%rbp)
  movq   $0x0,-0x70(%rbp)
  jmp    L11
L56:
  mov    -0x70(%rbp),%rax
  movsd  -0x20(%rbp,%rax,8),%xmm0
  mov    -0x48(%rbp),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x90(%rbp),%rax
  add    %rdx,%rax
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0x70(%rbp)
L11:
  mov    -0x70(%rbp),%rax
  cmp    -0x58(%rbp),%rax
  jb     L56
  addq   $0x1,-0x68(%rbp)
L67:
  mov    -0x68(%rbp),%rax
  cmp    -0x98(%rbp),%rax
  jb     L16
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int ho2cuLoopf(float *ho, float *cu, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, hostep = 3, custep = 3;
  double hoFlt[3], cuFlt[3];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, hoFlt, cuFlt) firstprivate(hostep, custep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * hostep;
    iOut = i * custep;
    for (j = 0; j < hostep; j++) {
      hoFlt[j] = (double)ho[iIn + j];
    }
    trash = ho2cu(hoFlt, cuFlt, p);
    for (j = 0; j < custep; j++) {
      cu[iOut + j] = (float)cuFlt[j];
    }
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ho2cuLoopf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L43
  mov    %ecx,%r14d
  mov    %rsi,%rbp
  mov    %rdi,%rbx
  lea    (%rdx,%rdx,2),%r15
  mov    $0x0,%r13d
  lea    0x20(%rsp),%r12
  jmp    L3
L25:
  add    $0x3,%r13
  add    $0xc,%rbp
  add    $0xc,%rbx
  cmp    %r15,%r13
  je     L43
L3:
  mov    $0x0,%eax
L10:
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx,%rax,4),%xmm0
  movsd  %xmm0,(%rsp,%rax,8)
  add    $0x1,%rax
  cmp    $0x3,%rax
  jne    L10
  mov    %rsp,%rdi
  mov    %r14d,%edx
  mov    %r12,%rsi
  call   <ho2cu@plt>
  mov    $0x0,%eax
L0:
  pxor   %xmm0,%xmm0
  cvtsd2ss (%r12,%rax,8),%xmm0
  movss  %xmm0,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x3,%rax
  jne    L0
  jmp    L25
L43:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x1,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ho2cuLoopf(float *ho, float *cu, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, hostep = 3, custep = 3;
  double hoFlt[3], cuFlt[3];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, hoFlt, cuFlt) firstprivate(hostep, custep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * hostep;
    iOut = i * custep;
    for (j = 0; j < hostep; j++) {
      hoFlt[j] = (double)ho[iIn + j];
    }
    trash = ho2cu(hoFlt, cuFlt, p);
    for (j = 0; j < custep; j++) {
      cu[iOut + j] = (float)cuFlt[j];
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ho2cuLoopf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L11
  lea    (%rdx,%rdx,2),%rax
  mov    %ecx,%ebx
  mov    %rsi,%r13
  mov    %rdi,%r12
  mov    %rax,0x8(%rsp)
  xor    %ebp,%ebp
  lea    0x10(%rsp),%r15
  lea    0x30(%rsp),%r14
  nopl   0x0(%rax)
L9:
  xor    %eax,%eax
L10:
  pxor   %xmm0,%xmm0
  cvtss2sd (%r12,%rax,4),%xmm0
  movsd  %xmm0,(%r15,%rax,8)
  add    $0x1,%rax
  cmp    $0x3,%rax
  jne    L10
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <ho2cu@plt>
  xor    %eax,%eax
L3:
  pxor   %xmm0,%xmm0
  cvtsd2ss (%r14,%rax,8),%xmm0
  movss  %xmm0,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    $0x3,%rax
  jne    L3
  add    $0x3,%rbp
  add    $0xc,%r13
  add    $0xc,%r12
  cmp    %rbp,0x8(%rsp)
  jne    L9
L11:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x58,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ho2cuLoopf(float *ho, float *cu, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, hostep = 3, custep = 3;
  double hoFlt[3], cuFlt[3];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, hoFlt, cuFlt) firstprivate(hostep, custep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * hostep;
    iOut = i * custep;
    for (j = 0; j < hostep; j++) {
      hoFlt[j] = (double)ho[iIn + j];
    }
    trash = ho2cu(hoFlt, cuFlt, p);
    for (j = 0; j < custep; j++) {
      cu[iOut + j] = (float)cuFlt[j];
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ho2cuLoopf>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L25
  mov    %ecx,%r13d
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  xor    %r12d,%r12d
  lea    0x30(%rsp),%r15
  lea    0x10(%rsp),%r14
  nopw   0x0(%rax,%rax,1)
L2:
  movq   0x0(%rbp),%xmm0
  mov    %r13d,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  add    $0x1,%r12
  add    $0xc,%rbp
  add    $0xc,%rbx
  cvtps2pd %xmm0,%xmm0
  movaps %xmm0,0x10(%rsp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x4(%rbp),%xmm0
  movsd  %xmm0,0x20(%rsp)
  call   <ho2cu@plt>
  cvtpd2ps 0x30(%rsp),%xmm0
  movlps %xmm0,-0xc(%rbx)
  pxor   %xmm0,%xmm0
  cvtsd2ss 0x40(%rsp),%xmm0
  movss  %xmm0,-0x4(%rbx)
  cmp    %r12,0x8(%rsp)
  jne    L2
L25:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x58,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ho2cuLoopf(float *ho, float *cu, unsigned long long n, int p) {
  int trash;
  unsigned long long j, i, iIn, iOut, hostep = 3, custep = 3;
  double hoFlt[3], cuFlt[3];
#pragma omp parallel for default(shared) private(j, i, iIn, iOut, trash, hoFlt, cuFlt) firstprivate(hostep, custep) schedule(static)
  for (i = 0; i < n; i++) {
    iIn = i * hostep;
    iOut = i * custep;
    for (j = 0; j < hostep; j++) {
      hoFlt[j] = (double)ho[iIn + j];
    }
    trash = ho2cu(hoFlt, cuFlt, p);
    for (j = 0; j < custep; j++) {
      cu[iOut + j] = (float)cuFlt[j];
    }
  }
  return 1;
}
```