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

```asm
<collatzConjectureIterative>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L22
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L18
L12:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L45
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L22
L45:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
L22:
  cmpl   $0x1,-0x4(%rbp)
  jg     L12
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L18:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D17
data_type: string
name: extract_symbolic_data
arguments: data_label: D10
data_type: string

tool: D17: Invalid input for the Collatz Conjecture.

tool: D10: %d


assistant: ```c
void collatzConjectureIterative(int x) {
  if (x < 1) {
    printf("Invalid input for the Collatz Conjecture.\n");
    return;
  }
  while (x > 1) {
    printf("%d\n", x);
    if ((x % 2) == 0) {
      x = x / 2;
    } else {
      x = (3 * x) + 1;
    }
  }
  printf("%d\n", x);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<collatzConjectureIterative>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L46
  mov    %edi,%ebx
  lea    D7(%rip),%rbp
  cmp    $0x1,%edi
  jg     L54
L25:
  mov    %ebx,%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L56:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L46:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  jmp    L56
L11:
  lea    0x1(%rbx,%rbx,2),%ebx
L6:
  cmp    $0x1,%ebx
  jle    L25
L54:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   $0x1,%bl
  jne    L11
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %eax,%ebx
  sar    %ebx
  jmp    L6
```

tool_calls: name: unpack_data
arguments: label: D7
type_guessed: string
name: unpack_data
arguments: label: D11
type_guessed: string

tool: D7: %d


tool: D11: Invalid input for the Collatz Conjecture.

assistant: ```c
void collatzConjectureIterative(int x) {
  if (x < 1) {
    printf("Invalid input for the Collatz Conjecture.\n");
    return;
  }
  while (x > 1) {
    printf("%d\n", x);
    if ((x % 2) == 0) {
      x = x / 2;
    } else {
      x = (3 * x) + 1;
    }
  }
  printf("%d\n", x);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<collatzConjectureIterative>:
  endbr64
  test   %edi,%edi
  jle    L38
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L47
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  test   $0x1,%bl
  jne    L46
  sar    %ebx
  cmp    $0x1,%ebx
  jne    L17
L47:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  jmp    1070 <printf@plt>
  nopl   (%rax)
L46:
  lea    0x1(%rbx,%rbx,2),%ebx
  jmp    L17
  cs nopw 0x0(%rax,%rax,1)
L38:
  lea    D0(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: label: D6
datatype: string
name: disassemble_data
arguments: label: D0
datatype: string

tool: D6: %d


tool: D0: Invalid input for the Collatz Conjecture.

assistant: ```c
void collatzConjectureIterative(int x) {
  if (x < 1) {
    printf("Invalid input for the Collatz Conjecture.\n");
    return;
  }
  while (x > 1) {
    printf("%d\n", x);
    if ((x % 2) == 0) {
      x = x / 2;
    } else {
      x = (3 * x) + 1;
    }
  }
  printf("%d\n", x);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<collatzConjectureIterative>:
  endbr64
  test   %edi,%edi
  jle    L24
  push   %rbp
  lea    D3(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L33
  nopl   0x0(%rax)
L40:
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  test   $0x1,%bl
  jne    L19
  sar    %ebx
  cmp    $0x1,%ebx
  jne    L40
L33:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  jmp    1070 <printf@plt>
  nopl   (%rax)
L19:
  lea    0x1(%rbx,%rbx,2),%ebx
  jmp    L40
  cs nopw 0x0(%rax,%rax,1)
L24:
  lea    D4(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: label: D3
type: string
name: disassemble_data
arguments: label: D4
type: string

tool: D3: %d


tool: D4: Invalid input for the Collatz Conjecture.

assistant: ```c
void collatzConjectureIterative(int x) {
  if (x < 1) {
    printf("Invalid input for the Collatz Conjecture.\n");
    return;
  }
  while (x > 1) {
    printf("%d\n", x);
    if ((x % 2) == 0) {
      x = x / 2;
    } else {
      x = (3 * x) + 1;
    }
  }
  printf("%d\n", x);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<f2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x3,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x3,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x3,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <f3@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D18
type_infered: string

tool: D18: %d


assistant: ```c
void f2(int i) {
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  f3(i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<f2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %edi,%esi
  lea    D7(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    0x3(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    0x6(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x9,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <f3@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
datatype: string

tool: D7: %d


assistant: ```c
void f2(int i) {
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  f3(i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<f2>:
  endbr64
  push   %r12
  lea    D8(%rip),%r12
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  lea    0x3(%rbp),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  lea    0x6(%rbp),%esi
  mov    %r12,%rdi
  add    $0x9,%ebp
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %ebp,%esi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <f3@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
data_type: string

tool: D8: %d


assistant: ```c
void f2(int i) {
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  f3(i);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f2>:
  endbr64
  push   %r12
  lea    D1(%rip),%r12
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <printf@plt>
  lea    0x3(%rbp),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  lea    0x6(%rbp),%esi
  mov    %r12,%rdi
  add    $0x9,%ebp
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %ebp,%esi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1070 <f3@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D1: %d


assistant: ```c
void f2(int i) {
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  i = i + 3;
  printf("%d\n", i);
  f3(i);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<stnputs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <makestrspace@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <mempcpy@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *stnputs(const char *s, size_t n, char *p) {
  p = makestrspace(n, p);
  p = mempcpy(p, s, n);
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<stnputs>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  mov    %rbx,%rdi
  call   <makestrspace@plt>
  mov    %rax,%rdi
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  call   <mempcpy@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stnputs(const char *s, size_t n, char *p) {
  p = makestrspace(n, p);
  p = mempcpy(p, s, n);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<stnputs>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    %rdx,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <makestrspace@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <mempcpy@plt>
```

assistant: ```c
char *stnputs(const char *s, size_t n, char *p) {
  p = makestrspace(n, p);
  p = mempcpy(p, s, n);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<stnputs>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    %rdx,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <makestrspace@plt>
  add    $0x8,%rsp
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <mempcpy@plt>
```

assistant: ```c
char *stnputs(const char *s, size_t n, char *p) {
  p = makestrspace(n, p);
  p = mempcpy(p, s, n);
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  movq   $0x8000,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L1
L21:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x24(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x24(%rbp),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x24(%rbp)
L1:
  mov    -0x20(%rbp),%rax
  cmp    %eax,-0x24(%rbp)
  jl     L21
  movl   $0x0,-0x24(%rbp)
  jmp    L25
L28:
  mov    -0x24(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  lea    0x1(%rax),%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %edx,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x24(%rbp)
L25:
  cmpl   $0x1f,-0x24(%rbp)
  jle    L28
  mov    -0x8(%rbp),%rdi
  mov    -0x8(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <kernel@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <dump@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  float *g1, *g2, *k;
  size_t nelms = 32 * 32 * 32;
  g1 = (float *)malloc(sizeof(float) * nelms);
  g2 = (float *)malloc(sizeof(float) * nelms);
  k = (float *)malloc(sizeof(float) * 32);
  int i;
  for (i = 0; i < (int)nelms; i++) {
    g1[i] = i;
    g2[i] = i;
  }
  for (i = 0; i < 32; ++i) {
    k[i] = 1 + (i % 2);
  }
  kernel(g1, g2, k, k, k);
  dump(g2);
  free(g1);
  free(g2);
  free(k);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x20000,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x20000,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r12,%rdx
  mov    $0x0,%eax
  mov    %rbp,%rcx
  sub    %r12,%rcx
L6:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,(%rdx)
  movss  %xmm0,(%rcx,%rdx,1)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    $0x8000,%eax
  jne    L6
  mov    %rbx,%rcx
  mov    $0x0,%edx
L19:
  mov    %edx,%esi
  shr    $0x1f,%esi
  lea    (%rdx,%rsi,1),%eax
  and    $0x1,%eax
  sub    %esi,%eax
  add    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,(%rcx)
  add    $0x1,%edx
  add    $0x4,%rcx
  cmp    $0x20,%edx
  jne    L19
  mov    %rbx,%r8
  mov    %rbx,%rcx
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <kernel@plt>
  mov    %rbp,%rdi
  call   <dump@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  float *g1, *g2, *k;
  size_t nelms = 32 * 32 * 32;
  g1 = (float *)malloc(sizeof(float) * nelms);
  g2 = (float *)malloc(sizeof(float) * nelms);
  k = (float *)malloc(sizeof(float) * 32);
  int i;
  for (i = 0; i < (int)nelms; i++) {
    g1[i] = i;
    g2[i] = i;
  }
  for (i = 0; i < 32; ++i) {
    k[i] = 1 + (i % 2);
  }
  kernel(g1, g2, k, k, k);
  dump(g2);
  free(g1);
  free(g2);
  free(k);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x20000,%edi
  push   %r12
  push   %rbp
  call   <malloc@plt>
  mov    $0x20000,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    $0x80,%edi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %rax,%r13
  sub    %r12,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L6:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  add    $0x1,%eax
  movss  %xmm0,(%rdx)
  movss  %xmm0,(%rcx,%rdx,1)
  add    $0x4,%rdx
  cmp    $0x8000,%eax
  jne    L6
  mov    %r13,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L11:
  mov    %eax,%edx
  pxor   %xmm0,%xmm0
  add    $0x1,%eax
  add    $0x4,%rcx
  and    $0x1,%edx
  add    $0x1,%edx
  cvtsi2ss %edx,%xmm0
  movss  %xmm0,-0x4(%rcx)
  cmp    $0x20,%eax
  jne    L11
  mov    %r13,%r8
  mov    %r13,%rcx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <kernel@plt>
  mov    %rbp,%rdi
  call   <dump@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  float *g1, *g2, *k;
  size_t nelms = 32 * 32 * 32;
  g1 = (float *)malloc(sizeof(float) * nelms);
  g2 = (float *)malloc(sizeof(float) * nelms);
  k = (float *)malloc(sizeof(float) * 32);
  int i;
  for (i = 0; i < (int)nelms; i++) {
    g1[i] = i;
    g2[i] = i;
  }
  for (i = 0; i < 32; ++i) {
    k[i] = 1 + (i % 2);
  }
  kernel(g1, g2, k, k, k);
  dump(g2);
  free(g1);
  free(g2);
  free(k);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x20000,%edi
  push   %r12
  push   %rbp
  call   <malloc@plt>
  mov    $0x20000,%edi
  mov    %rax,%r12
  call   <malloc@plt>
  mov    $0x80,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  lea    0x4(%r12),%rdx
  mov    %rax,%rbp
  mov    %r13,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L14
  mov    %r13,%rdx
  movdqa D11(%rip),%xmm1
  mov    %r12,%rax
  movdqa D20(%rip),%xmm2
  lea    0x20000(%r12),%rcx
  sub    %r12,%rdx
  cs nopw 0x0(%rax,%rax,1)
L30:
  movdqa %xmm1,%xmm0
  paddd  %xmm2,%xmm1
  cvtdq2ps %xmm0,%xmm0
  movups %xmm0,(%rax)
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rcx,%rax
  jne    L30
L34:
  movaps D17(%rip),%xmm0
  mov    %rbp,%r8
  mov    %rbp,%rcx
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  movups %xmm0,0x0(%rbp)
  movups %xmm0,0x10(%rbp)
  movups %xmm0,0x20(%rbp)
  movups %xmm0,0x30(%rbp)
  movups %xmm0,0x40(%rbp)
  movups %xmm0,0x50(%rbp)
  movups %xmm0,0x60(%rbp)
  movups %xmm0,0x70(%rbp)
  call   <kernel@plt>
  mov    %r13,%rdi
  call   <dump@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
L14:
  mov    %r13,%rcx
  mov    %r12,%rdx
  xor    %eax,%eax
  sub    %r12,%rcx
  nopl   0x0(%rax)
L6:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  add    $0x1,%eax
  movss  %xmm0,(%rdx)
  movss  %xmm0,(%rcx,%rdx,1)
  add    $0x4,%rdx
  cmp    $0x8000,%eax
  jne    L6
  jmp    L34
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type: byte[16]
name: extract_symbolic_data
arguments: label: D20
type: byte[16]
name: extract_symbolic_data
arguments: label: D17
type: byte[16]

tool: {"D11": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D20": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D17": [0x00,0x00,0x80,0x3f,0x00,0x00,0x00,0x40,0x00,0x00,0x80,0x3f,0x00,0x00,0x00,0x40]}

assistant: ```c
int main(int argc, char *argv[]) {
  float *g1, *g2, *k;
  size_t nelms = 32 * 32 * 32;
  g1 = (float *)malloc(sizeof(float) * nelms);
  g2 = (float *)malloc(sizeof(float) * nelms);
  k = (float *)malloc(sizeof(float) * 32);
  int i;
  for (i = 0; i < (int)nelms; i++) {
    g1[i] = i;
    g2[i] = i;
  }
  for (i = 0; i < 32; ++i) {
    k[i] = 1 + (i % 2);
  }
  kernel(g1, g2, k, k, k);
  dump(g2);
  free(g1);
  free(g2);
  free(k);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<aes_round>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    -0x10(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    -0x28(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    -0x30(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x10(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void aes_round(__m128i key, __m128i *x0, __m128i *x1, __m128i *x2, __m128i *x3, __m128i *x4, __m128i *x5, __m128i *x6, __m128i *x7) {
  *x0 = _mm_aesenc_si128(*x0, key);
  *x1 = _mm_aesenc_si128(*x1, key);
  *x2 = _mm_aesenc_si128(*x2, key);
  *x3 = _mm_aesenc_si128(*x3, key);
  *x4 = _mm_aesenc_si128(*x4, key);
  *x5 = _mm_aesenc_si128(*x5, key);
  *x6 = _mm_aesenc_si128(*x6, key);
  *x7 = _mm_aesenc_si128(*x7, key);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<aes_round>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r15
  mov    %rdx,%r14
  mov    %rcx,%r13
  mov    %r8,%r12
  mov    %r9,%rbp
  mov    %edi,%esi
  mov    (%r15),%edi
  call   <_mm_aesenc_si128@plt>
  mov    %eax,(%r15)
  mov    %ebx,%esi
  mov    (%r14),%edi
  call   <_mm_aesenc_si128@plt>
  mov    %eax,(%r14)
  mov    %ebx,%esi
  mov    0x0(%r13),%edi
  call   <_mm_aesenc_si128@plt>
  mov    %eax,0x0(%r13)
  mov    %ebx,%esi
  mov    (%r12),%edi
  call   <_mm_aesenc_si128@plt>
  mov    %eax,(%r12)
  mov    %ebx,%esi
  mov    0x0(%rbp),%edi
  call   <_mm_aesenc_si128@plt>
  mov    %eax,0x0(%rbp)
  mov    %ebx,%esi
  mov    0x40(%rsp),%rax
  mov    (%rax),%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x40(%rsp),%rdx
  mov    %eax,(%rdx)
  mov    %ebx,%esi
  mov    0x48(%rsp),%rax
  mov    (%rax),%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x48(%rsp),%rcx
  mov    %eax,(%rcx)
  mov    %ebx,%esi
  mov    0x50(%rsp),%rax
  mov    (%rax),%edi
  call   <_mm_aesenc_si128@plt>
  mov    0x50(%rsp),%rdx
  mov    %eax,(%rdx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void aes_round(__m128i key, __m128i *x0, __m128i *x1, __m128i *x2, __m128i *x3, __m128i *x4, __m128i *x5, __m128i *x6, __m128i *x7) {
  *x0 = _mm_aesenc_si128(*x0, key);
  *x1 = _mm_aesenc_si128(*x1, key);
  *x2 = _mm_aesenc_si128(*x2, key);
  *x3 = _mm_aesenc_si128(*x3, key);
  *x4 = _mm_aesenc_si128(*x4, key);
  *x5 = _mm_aesenc_si128(*x5, key);
  *x6 = _mm_aesenc_si128(*x6, key);
  *x7 = _mm_aesenc_si128(*x7, key);
}
```