ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  test   %edi,%edi
  je     L11
  mov    %edi,%edx
  mov    $0x0,%edi
L156:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  add    $0x1,%edi
  test   %eax,%eax
  jne    L156
  movslq %edi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L159:
  cmp    %rdx,%rsp
  je     L64
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L159
L10:
  mov    0x0(,%r9,4),%eax
  add    $0x2,%eax
  mov    %ecx,%edi
  cmp    $0x4,%eax
  ja     L11
  jmp    L135
L102:
  lea    0x8ca6bff(%rdi),%r8d
  cmp    $0x1194d7fd,%r8d
  ja     L19
L45:
  sub    $0x4,%rsi
  sub    $0x1,%edx
L1:
  mov    %eax,%r8d
  imul   (%rsi),%r8d
  add    %r8d,%edi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  mov    %r9d,%r11d
  or     %edx,%r11d
  je     L102
  test   %edx,%edx
  jns    L45
L11:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L125
  mov    %edi,%eax
  leave
  ret
L98:
  mov    %ecx,%edi
  jmp    L11
L19:
  mov    %ecx,%edi
  jmp    L11
L64:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L48
  orq    $0x0,-0x8(%rsp,%rax,1)
L48:
  lea    0x3(%rsp),%r10
  mov    %r10,%r9
  shr    $0x2,%r9
  and    $0xfffffffffffffffc,%r10
  mov    $0x0,%esi
L26:
  mov    %esi,%edx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%eax
  lea    (%rax,%rax,4),%r8d
  add    %r8d,%r8d
  sub    %r8d,%ecx
  mov    %ecx,(%r10,%rsi,4)
  mov    %eax,%ecx
  add    $0x1,%rsi
  test   %eax,%eax
  jne    L26
  cmp    $0xa,%edi
  je     L10
L135:
  test   %edx,%edx
  js     L98
  mov    0x0(,%r9,4),%r9d
  add    $0x2,%r9d
  and    $0xfffffffb,%r9d
  movslq %edx,%rax
  lea    (%r10,%rax,4),%rsi
  sub    $0x1,%edx
  mov    %ecx,%edi
  mov    $0x1,%eax
  jmp    L1
L125:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(int x) {
  int count = 0;
  int left = x;
  while (left != 0) {
    left = left / 10;
    count++;
  }
  int reverse[count];
  int i = 0;
  left = x;
  while (left != 0) {
    reverse[i++] = left % 10;
    left = left / 10;
  }
  int result = 0;
  int j = 1;
  int k = i - 1;
  int to_break = 0;
  if (count == 10 && (reverse[0] >= 3 || reverse[0] <= -3)) {
    result = 0;
    to_break = 1;
  }
  if (to_break == 0) {
    for (; k >= 0; k--) {
      result += reverse[k] * j;
      j = j * 10;
      if ((reverse[0] == 2 || reverse[0] == -2) && k == 1 && (result >= 147483647 || result <= -147483648)) {
        result = 0;
        break;
      }
    }
  }
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  test   %edi,%edi
  je     L111
  mov    %edi,%r9d
  mov    %edi,%edx
  xor    %r10d,%r10d
  nopl   0x0(%rax,%rax,1)
L121:
  movslq %edx,%rax
  mov    %edx,%ecx
  add    $0x1,%r10d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  mov    %eax,%edx
  jne    L121
  movslq %r10d,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L77
L37:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L37
L77:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L113
L43:
  lea    0x3(%rsp),%r8
  xor    %edx,%edx
  mov    %r8,%r11
  and    $0xfffffffffffffffc,%r8
  shr    $0x2,%r11
  nopl   0x0(%rax)
L18:
  movslq %r9d,%rax
  mov    %r9d,%esi
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%edi
  mov    %eax,%esi
  mov    %r9d,%eax
  add    %edi,%edi
  mov    %esi,%r9d
  sub    %edi,%eax
  mov    %eax,(%r8,%rdx,4)
  add    $0x1,%rdx
  test   %esi,%esi
  jne    L18
  mov    0x0(,%r11,4),%esi
  add    $0x2,%esi
  cmp    $0xa,%r10d
  je     L138
L140:
  and    $0xfffffffb,%esi
  sub    $0x1,%ecx
  xor    %r9d,%r9d
  mov    $0x1,%edx
  jmp    L114
  nopl   0x0(%rax)
L79:
  movslq %ecx,%rax
  cmp    $0xffffffff,%ecx
  je     L20
L52:
  mov    (%r8,%rax,4),%eax
  sub    $0x1,%ecx
L114:
  imul   %edx,%eax
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  add    %eax,%r9d
  mov    %esi,%eax
  or     %ecx,%eax
  jne    L79
  lea    0x8ca6bff(%r9),%eax
  cmp    $0x1194d7fd,%eax
  ja     L111
  xor    %eax,%eax
  jmp    L52
  nopw   0x0(%rax,%rax,1)
L138:
  cmp    $0x4,%esi
  jbe    L140
L111:
  xor    %r9d,%r9d
L20:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L130
  leave
  mov    %r9d,%eax
  ret
L113:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L43
L130:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(int x) {
  int count = 0;
  int left = x;
  while (left != 0) {
    left = left / 10;
    count++;
  }
  int reverse[count];
  int i = 0;
  left = x;
  while (left != 0) {
    reverse[i++] = left % 10;
    left = left / 10;
  }
  int result = 0;
  int j = 1;
  int k = i - 1;
  int to_break = 0;
  if (count == 10 && (reverse[0] >= 3 || reverse[0] <= -3)) {
    result = 0;
    to_break = 1;
  }
  if (to_break == 0) {
    for (; k >= 0; k--) {
      result += reverse[k] * j;
      j = j * 10;
      if ((reverse[0] == 2 || reverse[0] == -2) && k == 1 && (result >= 147483647 || result <= -147483648)) {
        result = 0;
        break;
      }
    }
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  test   %edi,%edi
  je     L56
  mov    %edi,%edx
  mov    %edi,%ecx
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L59:
  movslq %ecx,%rax
  mov    %ecx,%esi
  add    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  mov    %eax,%ecx
  jne    L59
  movslq %r8d,%rax
  mov    %rsp,%rsi
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L81
L80:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L80
L81:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L26
L37:
  lea    0x3(%rsp),%r9
  xor    %esi,%esi
  mov    %r9,%r10
  and    $0xfffffffffffffffc,%r9
  shr    $0x2,%r10
  nopl   0x0(%rax)
L112:
  movslq %edx,%rax
  mov    %edx,%edi
  mov    %esi,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x22,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%edx
  mov    %edx,(%r9,%rsi,4)
  mov    %eax,%edx
  add    $0x1,%rsi
  test   %eax,%eax
  jne    L112
  mov    0x0(,%r10,4),%edi
  add    $0x2,%edi
  cmp    $0xa,%r8d
  je     L49
L15:
  movslq %ecx,%rax
  and    $0xfffffffb,%edi
  lea    -0x1(%rcx),%edx
  xor    %r8d,%r8d
  lea    (%r9,%rax,4),%rsi
  mov    $0x1,%eax
  jmp    L58
  nopl   0x0(%rax)
L55:
  cmp    $0xffffffff,%edx
  je     L9
L66:
  sub    $0x4,%rsi
  sub    $0x1,%edx
L58:
  mov    (%rsi),%ecx
  imul   %eax,%ecx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  add    %ecx,%r8d
  mov    %edi,%ecx
  or     %edx,%ecx
  jne    L55
  lea    0x8ca6bff(%r8),%ecx
  cmp    $0x1194d7fd,%ecx
  jbe    L66
L56:
  xor    %r8d,%r8d
L9:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L145
  leave
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L49:
  cmp    $0x4,%edi
  ja     L56
  jmp    L15
L26:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L37
L145:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int reverse(int x) {
  int count = 0;
  int left = x;
  while (left != 0) {
    left = left / 10;
    count++;
  }
  int reverse[count];
  int i = 0;
  left = x;
  while (left != 0) {
    reverse[i++] = left % 10;
    left = left / 10;
  }
  int result = 0;
  int j = 1;
  int k = i - 1;
  int to_break = 0;
  if (count == 10 && (reverse[0] >= 3 || reverse[0] <= -3)) {
    result = 0;
    to_break = 1;
  }
  if (to_break == 0) {
    for (; k >= 0; k--) {
      result += reverse[k] * j;
      j = j * 10;
      if ((reverse[0] == 2 || reverse[0] == -2) && k == 1 && (result >= 147483647 || result <= -147483648)) {
        result = 0;
        break;
      }
    }
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__f90_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <__f90_set_args@plt>
  mov    $0x0,%eax
  call   <MAIN__@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int __f90_main(int argc, char **argv, char **arge) {
  __f90_set_args(argc, argv);
  MAIN__();
  exit(0);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<__f90_main>:
  endbr64
  sub    $0x8,%rsp
  call   <__f90_set_args@plt>
  mov    $0x0,%eax
  call   <MAIN__@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __f90_main(int argc, char **argv, char **arge) {
  __f90_set_args(argc, argv);
  MAIN__();
  exit(0);
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<__f90_main>:
  endbr64
  sub    $0x8,%rsp
  call   <__f90_set_args@plt>
  xor    %eax,%eax
  call   <MAIN__@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __f90_main(int argc, char **argv, char **arge) {
  __f90_set_args(argc, argv);
  MAIN__();
  exit(0);
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__f90_main>:
  endbr64
  sub    $0x8,%rsp
  call   <__f90_set_args@plt>
  xor    %eax,%eax
  call   <MAIN__@plt>
  xor    %edi,%edi
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int __f90_main(int argc, char **argv, char **arge) {
  __f90_set_args(argc, argv);
  MAIN__();
  exit(0);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<crossprod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L50
L43:
  movl   $0x0,-0x8(%rbp)
  jmp    L23
L57:
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L4
L44:
  mov    -0xc(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  mov    -0xc(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L44
  addl   $0x1,-0x8(%rbp)
L23:
  mov    -0x8(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L57
  addl   $0x1,-0xc(%rbp)
L50:
  mov    -0xc(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L43
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void crossprod(double *x, double *y, double *z, int m, int n, int p) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      z[i * p + j] = x[k * n + i] * y[k * p + j];
      for (k = 1; k < m; k++)
        z[i * p + j] += x[k * n + i] * y[k * p + j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<crossprod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,-0x8(%rsp)
  mov    %r9d,-0x20(%rsp)
  test   %r8d,%r8d
  jle    L64
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %ecx,%edi
  mov    %r9d,%esi
  movslq %r8d,%r9
  shl    $0x3,%r9
  lea    (%rbx,%r9,1),%r14
  mov    %r8d,%eax
  mov    %rax,-0x18(%rsp)
  mov    %esi,%eax
  movslq %esi,%r8
  shl    $0x3,%r8
  lea    0x0(%rbp,%r8,1),%rsi
  mov    %rsi,-0x10(%rsp)
  movl   $0x0,-0x1c(%rsp)
  mov    $0x0,%r13d
  mov    %eax,%r15d
  jmp    L37
L46:
  movslq -0x1c(%rsp),%rax
  mov    -0x8(%rsp),%rcx
  lea    (%rcx,%rax,8),%r11
  mov    -0x10(%rsp),%r12
  mov    $0x0,%r10d
L35:
  mov    %r11,%rsi
  movsd  (%rbx,%r13,8),%xmm0
  mulsd  0x0(%rbp,%r10,8),%xmm0
  movsd  %xmm0,(%r11)
  cmp    $0x1,%edi
  jle    L51
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    $0x1,%eax
L12:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx),%xmm0
  addsd  (%rsi),%xmm0
  movsd  %xmm0,(%rsi)
  add    $0x1,%eax
  add    %r9,%rdx
  add    %r8,%rcx
  cmp    %eax,%edi
  jne    L12
L51:
  add    $0x1,%r10
  add    $0x8,%r11
  add    $0x8,%r12
  cmp    %r15,%r10
  jne    L35
L41:
  add    $0x1,%r13
  mov    -0x20(%rsp),%edx
  add    %edx,-0x1c(%rsp)
  add    $0x8,%r14
  cmp    -0x18(%rsp),%r13
  je     L64
L37:
  cmpl   $0x0,-0x20(%rsp)
  jg     L46
  jmp    L41
L64:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void crossprod(double *x, double *y, double *z, int m, int n, int p) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      z[i * p + j] = x[k * n + i] * y[k * p + j];
      for (k = 1; k < m; k++)
        z[i * p + j] += x[k * n + i] * y[k * p + j];
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<crossprod>:
  endbr64
  push   %r15
  movslq %r8d,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,-0x8(%rsp)
  mov    %r9d,-0x1c(%rsp)
  test   %eax,%eax
  jle    L33
  mov    %rsi,%rbp
  mov    %r9d,%esi
  mov    %rax,-0x18(%rsp)
  mov    %rdi,%rbx
  movslq %esi,%r8
  lea    0x0(,%rax,8),%r9
  mov    %ecx,%edi
  xor    %r15d,%r15d
  shl    $0x3,%r8
  lea    (%rbx,%r9,1),%r13
  lea    -0x1(%rsi),%r14d
  xor    %r12d,%r12d
  lea    0x0(%rbp,%r8,1),%rax
  mov    %rax,-0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L63:
  mov    -0x1c(%rsp),%eax
  test   %eax,%eax
  jle    L23
  mov    -0x8(%rsp),%rsi
  movslq %r15d,%rax
  mov    -0x10(%rsp),%r11
  xor    %r10d,%r10d
  lea    (%rsi,%rax,8),%rsi
  nopl   0x0(%rax)
L7:
  movsd  (%rbx,%r12,8),%xmm1
  mov    %r11,%rcx
  mov    %r13,%rdx
  mov    $0x1,%eax
  mulsd  0x0(%rbp,%r10,8),%xmm1
  movsd  %xmm1,(%rsi)
  cmp    $0x1,%edi
  jle    L42
  nopl   0x0(%rax)
L1:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx),%xmm0
  add    $0x1,%eax
  add    %r9,%rdx
  add    %r8,%rcx
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsi)
  cmp    %eax,%edi
  jne    L1
L42:
  lea    0x1(%r10),%rax
  add    $0x8,%rsi
  add    $0x8,%r11
  cmp    %r14,%r10
  je     L23
  mov    %rax,%r10
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L23:
  add    $0x1,%r12
  add    $0x8,%r13
  add    -0x1c(%rsp),%r15d
  cmp    -0x18(%rsp),%r12
  jne    L63
L33:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void crossprod(double *x, double *y, double *z, int m, int n, int p) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      z[i * p + j] = x[k * n + i] * y[k * p + j];
      for (k = 1; k < m; k++)
        z[i * p + j] += x[k * n + i] * y[k * p + j];
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<crossprod>:
  endbr64
  push   %r15
  movslq %r8d,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,-0x10(%rsp)
  test   %eax,%eax
  jle    L47
  movslq %r9d,%r14
  test   %r14d,%r14d
  jle    L47
  mov    %rax,-0x8(%rsp)
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %ecx,%edi
  lea    0x0(,%r14,8),%r8
  lea    0x0(,%rax,8),%r9
  xor    %r15d,%r15d
  xor    %r12d,%r12d
  lea    (%rsi,%r8,1),%rax
  lea    (%rbx,%r9,1),%r13
  mov    %rax,-0x18(%rsp)
  nopw   0x0(%rax,%rax,1)
L58:
  mov    -0x10(%rsp),%rax
  mov    -0x18(%rsp),%r11
  xor    %r10d,%r10d
  lea    (%rax,%r15,8),%rsi
  nopl   0x0(%rax)
L27:
  movsd  (%rbx,%r12,8),%xmm1
  mov    %r11,%rcx
  mov    %r13,%rdx
  mov    $0x1,%eax
  mulsd  0x0(%rbp,%r10,8),%xmm1
  movsd  %xmm1,(%rsi)
  cmp    $0x1,%edi
  jle    L48
  nopl   0x0(%rax)
L9:
  movsd  (%rdx),%xmm0
  mulsd  (%rcx),%xmm0
  add    $0x1,%eax
  add    %r9,%rdx
  add    %r8,%rcx
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rsi)
  cmp    %eax,%edi
  jne    L9
L48:
  add    $0x1,%r10
  add    $0x8,%rsi
  add    $0x8,%r11
  cmp    %r10,%r14
  jne    L27
  add    $0x1,%r12
  add    $0x8,%r13
  add    %r14,%r15
  cmp    -0x8(%rsp),%r12
  jne    L58
L47:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void crossprod(double *x, double *y, double *z, int m, int n, int p) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < p; j++) {
      k = 0;
      z[i * p + j] = x[k * n + i] * y[k * p + j];
      for (k = 1; k < m; k++)
        z[i * p + j] += x[k * n + i] * y[k * p + j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0x50(%rbp)
  movl   $0x8,-0x4c(%rbp)
  movl   $0x2,-0x48(%rbp)
  movl   $0x5,-0x44(%rbp)
  movl   $0x1,-0x40(%rbp)
  movl   $0x6,-0x3c(%rbp)
  movl   $0x3,-0x38(%rbp)
  movl   $0x7,-0x34(%rbp)
  movl   $0x8,-0x30(%rbp)
  movl   $0x4,-0x2c(%rbp)
  movl   $0x5,-0x28(%rbp)
  movl   $0x2,-0x24(%rbp)
  movl   $0x6,-0x20(%rbp)
  movl   $0x7,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  lea    -0x30(%rbp),%rsi
  lea    -0x50(%rbp),%rax
  mov    $0x7,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <BuildTree@plt>
  mov    %rax,-0x58(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <preOrder@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int main(void) {
  int in[] = {4, 8, 2, 5, 1, 6, 3, 7};
  int post[] = {8, 4, 5, 2, 6, 7, 3, 1};
  struct node *root = BuildTree(in, post, 0, 7);
  preOrder(root);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x4,(%rsp)
  movl   $0x8,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x5,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x7,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  movl   $0x4,0x24(%rsp)
  movl   $0x5,0x28(%rsp)
  movl   $0x2,0x2c(%rsp)
  movl   $0x6,0x30(%rsp)
  movl   $0x7,0x34(%rsp)
  movl   $0x3,0x38(%rsp)
  movl   $0x1,0x3c(%rsp)
  lea    0x20(%rsp),%rsi
  mov    %rsp,%rdi
  mov    $0x7,%ecx
  mov    $0x0,%edx
  call   <BuildTree@plt>
  mov    %rax,%rdi
  call   <preOrder@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int in[] = {4, 8, 2, 5, 1, 6, 3, 7};
  int post[] = {8, 4, 5, 2, 6, 7, 3, 1};
  struct node *root = BuildTree(in, post, 0, 7);
  preOrder(root);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  xor    %edx,%edx
  mov    $0x7,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x800000004,%rax
  mov    %rsp,%rdi
  lea    0x20(%rsp),%rsi
  mov    %rax,(%rsp)
  movabs $0x500000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x600000001,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x700000003,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x400000008,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x200000005,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x700000006,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x100000003,%rax
  mov    %rax,0x38(%rsp)
  call   <BuildTree@plt>
  mov    %rax,%rdi
  call   <preOrder@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int in[] = {4, 8, 2, 5, 1, 6, 3, 7};
  int post[] = {8, 4, 5, 2, 6, 7, 3, 1};
  struct node *root = BuildTree(in, post, 0, 7);
  preOrder(root);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  movdqa D1(%rip),%xmm0
  xor    %edx,%edx
  mov    $0x7,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x20(%rsp),%rsi
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movdqa D13(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D26(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  call   <BuildTree@plt>
  mov    %rax,%rdi
  call   <preOrder@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: byte[16]
name: unpack_data
arguments: data_label: D20
type_infered: byte[16]
name: unpack_data
arguments: data_label: D13
type_infered: byte[16]
name: unpack_data
arguments: data_label: D26
type_infered: byte[16]

tool: {"D1": [0x04,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D20": [0x01,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D13": [0x08,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: {"D26": [0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int in[] = {4, 8, 2, 5, 1, 6, 3, 7};
  int post[] = {8, 4, 5, 2, 6, 7, 3, 1};
  struct node *root = BuildTree(in, post, 0, 7);
  preOrder(root);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<funcAngCor_alpha1_2plus_2plus_15097_L0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm1
  movsd  %xmm1,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <EvalAngCor_alpha1_2plus_2plus_15097_L0@plt>
  mulsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double funcAngCor_alpha1_2plus_2plus_15097_L0(double *x, double *par) {
  double result;
  result = par[0] * EvalAngCor_alpha1_2plus_2plus_15097_L0(x[0]);
  return result;
}
```