ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cocktailSort>:
  endbr64
  cmp    %edx,%esi
  jge    L7
  mov    %rdi,%rcx
  lea    -0x2(%rdx),%r11d
  movslq %edx,%rax
  push   %rbp
  mov    %esi,%r10d
  push   %rbx
  mov    %edx,%edi
  movslq %esi,%rbx
  sub    %esi,%r11d
  lea    (%rcx,%rax,4),%rbp
  nopw   0x0(%rax,%rax,1)
L38:
  mov    %rbx,%rax
  mov    $0x1,%r9d
  nopl   0x0(%rax)
L25:
  mov    (%rcx,%rax,4),%edx
  mov    0x4(%rcx,%rax,4),%esi
  cmp    %esi,%edx
  jle    L21
  mov    %esi,(%rcx,%rax,4)
  xor    %r9d,%r9d
  mov    %edx,0x4(%rcx,%rax,4)
L21:
  add    $0x1,%rax
  cmp    %eax,%edi
  jg     L25
  sub    $0x1,%edi
  cmp    %r10d,%edi
  jle    L18
  mov    %r11d,%edx
  movslq %edi,%rax
  lea    -0x8(%rbp),%r8
  shl    $0x2,%rdx
  lea    (%rcx,%rax,4),%rax
  sub    %rdx,%r8
  nopw   0x0(%rax,%rax,1)
L29:
  mov    (%rax),%edx
  mov    -0x4(%rax),%esi
  cmp    %esi,%edx
  jge    L74
  mov    %esi,(%rax)
  xor    %r9d,%r9d
  mov    %edx,-0x4(%rax)
L74:
  sub    $0x4,%rax
  cmp    %rax,%r8
  jne    L29
L18:
  add    $0x1,%r10d
  add    $0x1,%rbx
  sub    $0x2,%r11d
  sub    $0x4,%rbp
  test   %r9d,%r9d
  jne    L28
  cmp    %r10d,%edi
  jg     L38
L28:
  pop    %rbx
  pop    %rbp
  ret
L7:
  ret
```

assistant: ```c
void cocktailSort(int list[10], int bottom, int top) {
  int swapped = 0, i = 0, aux = 0;
  while (swapped == 0 && bottom < top) {
    swapped = 1;
    for (i = bottom; i < top; i++) {
      if (list[i] > list[i + 1]) {
        aux = list[i];
        list[i] = list[i + 1];
        list[i + 1] = aux;
        swapped = 0;
      }
    }
    top = top - 1;
    for (i = top; i > bottom; i--) {
      if (list[i] < list[i - 1]) {
        aux = list[i];
        list[i] = list[i - 1];
        list[i - 1] = aux;
        swapped = 0;
      }
    }
    bottom = bottom + 1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cocktailSort>:
  endbr64
  mov    %esi,%r9d
  cmp    %edx,%esi
  jge    L20
  lea    -0x2(%rdx),%r10d
  movslq %edx,%rax
  push   %rbp
  mov    %edx,%esi
  push   %rbx
  movslq %r9d,%r11
  sub    %r9d,%r10d
  lea    (%rdi,%rax,4),%rbx
  nopl   0x0(%rax)
L45:
  mov    %r11,%rdx
  mov    $0x1,%r8d
  nopl   0x0(%rax)
L63:
  movq   (%rdi,%rdx,4),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L72
  movq   %xmm1,(%rdi,%rdx,4)
  xor    %r8d,%r8d
L72:
  add    $0x1,%rdx
  cmp    %edx,%esi
  jg     L63
  sub    $0x1,%esi
  cmp    %r9d,%esi
  jle    L17
  movslq %esi,%rax
  lea    -0xc(%rbx),%rbp
  lea    -0x4(%rdi,%rax,4),%rdx
  mov    %r10d,%eax
  shl    $0x2,%rax
  sub    %rax,%rbp
  nopl   0x0(%rax)
L24:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm3
  movd   %xmm0,%ecx
  movd   %xmm3,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L32
  movq   %xmm1,(%rdx)
  xor    %r8d,%r8d
L32:
  sub    $0x4,%rdx
  cmp    %rdx,%rbp
  jne    L24
L17:
  add    $0x1,%r9d
  add    $0x1,%r11
  sub    $0x2,%r10d
  sub    $0x4,%rbx
  test   %r8d,%r8d
  jne    L30
  cmp    %r9d,%esi
  jg     L45
L30:
  pop    %rbx
  pop    %rbp
  ret
L20:
  ret
```

assistant: ```c
void cocktailSort(int list[10], int bottom, int top) {
  int swapped = 0, i = 0, aux = 0;
  while (swapped == 0 && bottom < top) {
    swapped = 1;
    for (i = bottom; i < top; i++) {
      if (list[i] > list[i + 1]) {
        aux = list[i];
        list[i] = list[i + 1];
        list[i + 1] = aux;
        swapped = 0;
      }
    }
    top = top - 1;
    for (i = top; i > bottom; i--) {
      if (list[i] < list[i - 1]) {
        aux = list[i];
        list[i] = list[i - 1];
        list[i - 1] = aux;
        swapped = 0;
      }
    }
    bottom = bottom + 1;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <setLastError@plt>
  mov    $0x3e7,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
type_guessed: string

tool: D8: TestFunction

assistant: ```c
int Test() {
  setLastError("TestFunction");
  return 999;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Test>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <setLastError@plt>
  mov    $0x3e7,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: TestFunction

assistant: ```c
int Test() {
  setLastError("TestFunction");
  return 999;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Test>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <setLastError@plt>
  mov    $0x3e7,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type_infered: string

tool: D3: TestFunction

assistant: ```c
int Test() {
  setLastError("TestFunction");
  return 999;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Test>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <setLastError@plt>
  mov    $0x3e7,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
datatype: string

tool: D6: TestFunction

assistant: ```c
int Test() {
  setLastError("TestFunction");
  return 999;
}
```
user: What is the c source code of the assembly code below:

```asm
<set_tag>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x7e,%eax
  jle    L1
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <fatal_error@plt>
L1:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_infered: string

tool: D9: %s too long

assistant: ```c
const char *set_tag(const char *tag, const char *s) {
  if (strlen(s) >= 128 - 1)
    fatal_error("%s too long", tag);
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<set_tag>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rsi,%rdi
  call   <strlen@plt>
  cmp    $0x7e,%eax
  jg     L16
L8:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    %rbp,%rsi
  lea    D6(%rip),%rdi
  call   <fatal_error@plt>
  jmp    L8
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D6: %s too long

assistant: ```c
const char *set_tag(const char *tag, const char *s) {
  if (strlen(s) >= 128 - 1)
    fatal_error("%s too long", tag);
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<set_tag>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x7e,%eax
  jg     L0
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <fatal_error@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
datatype: string

tool: D4: %s too long

assistant: ```c
const char *set_tag(const char *tag, const char *s) {
  if (strlen(s) >= 128 - 1)
    fatal_error("%s too long", tag);
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<set_tag>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    $0x7e,%eax
  jg     L5
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L5:
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <fatal_error@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: %s too long

assistant: ```c
const char *set_tag(const char *tag, const char *s) {
  if (strlen(s) >= 128 - 1)
    fatal_error("%s too long", tag);
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<createl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L32
L42:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L35
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L2
L35:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  jmp    L25
L2:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
L25:
  addl   $0x1,-0x10(%rbp)
L32:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L42
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *createl(int size) {
  int i;
  int length = size * size;
  int *arr = malloc(length * sizeof(int));
  for (i = 0; i < length; i++) {
    if (i < size || i % size == 0) {
      arr[i] = 1;
    } else {
      arr[i] = 0;
    }
  }
  return arr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<createl>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %edi,%ebp
  imul   %edi,%ebp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %ebp,%ebp
  jle    L27
  mov    %ebp,%esi
  mov    $0x0,%ecx
  jmp    L14
L5:
  movl   $0x1,(%rdi,%rcx,4)
L41:
  add    $0x1,%rcx
  cmp    %rsi,%rcx
  je     L27
L14:
  mov    %ecx,%eax
  cmp    %ecx,%ebx
  jg     L5
  cltd
  idiv   %ebx
  test   %edx,%edx
  je     L5
  movl   $0x0,(%rdi,%rcx,4)
  jmp    L41
L27:
  mov    %rdi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *createl(int size) {
  int i;
  int length = size * size;
  int *arr = malloc(length * sizeof(int));
  for (i = 0; i < length; i++) {
    if (i < size || i % size == 0) {
      arr[i] = 1;
    } else {
      arr[i] = 0;
    }
  }
  return arr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<createl>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  imul   %edi,%ebp
  mov    %edi,%ebx
  sub    $0x8,%rsp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  lea    -0x1(%rbp),%esi
  xor    %ecx,%ecx
  mov    %rax,%r8
  test   %ebp,%ebp
  jne    L13
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %rax,%rcx
L13:
  mov    %ecx,%eax
  mov    $0x1,%edx
  cmp    %ecx,%ebx
  jg     L3
  cltd
  idiv   %ebx
  test   %edx,%edx
  sete   %dl
  movzbl %dl,%edx
L3:
  mov    %edx,(%r8,%rcx,4)
  lea    0x1(%rcx),%rax
  cmp    %rsi,%rcx
  jne    L22
L16:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *createl(int size) {
  int i;
  int length = size * size;
  int *arr = malloc(length * sizeof(int));
  for (i = 0; i < length; i++) {
    if (i < size || i % size == 0) {
      arr[i] = 1;
    } else {
      arr[i] = 0;
    }
  }
  return arr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<createl>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  imul   %edi,%ebp
  mov    %edi,%ebx
  sub    $0x8,%rsp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %ebp,%ebp
  je     L65
  test   %ebx,%ebx
  jle    L16
  cmp    %ebp,%ebx
  mov    %ebp,%edi
  mov    $0x1,%esi
  cmovle %ebx,%edi
  test   %edi,%edi
  cmovg  %edi,%esi
  cmp    $0x3,%edi
  jle    L79
  mov    %esi,%edx
  movdqa D3(%rip),%xmm1
  movdqa D11(%rip),%xmm3
  shr    $0x2,%edx
  movdqa D10(%rip),%xmm2
  shl    $0x4,%rdx
  add    %rax,%rdx
  xchg   %ax,%ax
L55:
  movdqa %xmm1,%xmm0
  movups %xmm2,(%rax)
  add    $0x10,%rax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %rax,%rdx
  jne    L55
  mov    %esi,%eax
  pshufd $0xff,%xmm0,%xmm0
  and    $0xfffffffc,%eax
  and    $0x3,%esi
  movd   %xmm0,%ecx
  je     L73
L52:
  movslq %eax,%rdx
  lea    0x1(%rax),%ecx
  movl   $0x1,(%r8,%rdx,4)
  lea    0x0(,%rdx,4),%rsi
  cmp    %ecx,%edi
  jle    L73
  movl   $0x1,0x4(%r8,%rsi,1)
  lea    0x2(%rax),%ecx
  cmp    %ecx,%edi
  jle    L73
  movl   $0x1,0x8(%r8,%rsi,1)
  lea    0x3(%rax),%ecx
L73:
  cmp    %ecx,%ebp
  jle    L65
L119:
  movslq %ecx,%rcx
  nopl   0x0(%rax,%rax,1)
L71:
  mov    %ecx,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L31
  movl   $0x1,(%r8,%rcx,4)
  add    $0x1,%rcx
  cmp    %ecx,%ebp
  jg     L71
L65:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L31:
  movl   $0x0,(%r8,%rcx,4)
  add    $0x1,%rcx
  cmp    %ecx,%ebp
  jg     L71
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  xor    %ecx,%ecx
  jmp    L119
L79:
  xor    %eax,%eax
  jmp    L52
```

tool_calls: name: parse_data
arguments: tag: D3
type: byte[16]
name: parse_data
arguments: tag: D11
type: byte[16]
name: parse_data
arguments: tag: D10
type: byte[16]

tool: {"D3": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D11": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D10": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int *createl(int size) {
  int i;
  int length = size * size;
  int *arr = malloc(length * sizeof(int));
  for (i = 0; i < length; i++) {
    if (i < size || i % size == 0) {
      arr[i] = 1;
    } else {
      arr[i] = 0;
    }
  }
  return arr;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  lea    D33(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    $0xc8,%edi
  call   <malloc@plt>
  mov    %rax,-0x78(%rbp)
  cmpq   $0x0,-0x78(%rbp)
  jne    L19
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L11
L19:
  mov    -0x78(%rbp),%rax
  lea    D48(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L11:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D33
type_guessed: string
name: unpack_data
arguments: label: D3
type_guessed: string
name: unpack_data
arguments: label: D48
type_guessed: string
name: unpack_data
arguments: label: D15
type_guessed: string
name: unpack_data
arguments: label: D22
type_guessed: string

tool: D33: Short string of text.


tool: D3: Unable to allocate.

tool: D48: Long string of text.


tool: D15: 1st: %s


tool: D22: 2nd: %s


assistant: ```c
int main() {
  char staticMemory[100];
  char *dynamicMemory;
  strcpy(staticMemory, "Short string of text.\n");
  dynamicMemory = malloc(200 * sizeof(char));
  if (dynamicMemory == NULL) {
    printf("Unable to allocate.\n");
  } else {
    strcpy(dynamicMemory, "Long string of text.\n");
  }
  printf("1st: %s\n", staticMemory);
  printf("2nd: %s\n", dynamicMemory);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D40(%rip),%rsi
  call   <strcpy@plt>
  mov    $0xc8,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L4
  lea    D43(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L14:
  mov    %rsp,%rsi
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L4:
  lea    D18(%rip),%rdi
  call   <puts@plt>
  jmp    L14
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D40
type_guessed: string
name: retrieve_data
arguments: data_tag: D43
type_guessed: string
name: retrieve_data
arguments: data_tag: D48
type_guessed: string
name: retrieve_data
arguments: data_tag: D24
type_guessed: string
name: retrieve_data
arguments: data_tag: D18
type_guessed: string

tool: D40: Short string of text.


tool: D43: Long string of text.


tool: D48: 1st: %s


tool: D24: 2nd: %s


tool: D18: Unable to allocate.

assistant: ```c
int main() {
  char staticMemory[100];
  char *dynamicMemory;
  strcpy(staticMemory, "Short string of text.\n");
  dynamicMemory = malloc(200 * sizeof(char));
  if (dynamicMemory == NULL) {
    printf("Unable to allocate.\n");
  } else {
    strcpy(dynamicMemory, "Long string of text.\n");
  }
  printf("1st: %s\n", staticMemory);
  printf("2nd: %s\n", dynamicMemory);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  lea    D29(%rip),%rsi
  push   %rbp
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    $0xc8,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L22
  lea    D22(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L2:
  mov    %r12,%rsi
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D37(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L22:
  lea    D49(%rip),%rdi
  call   <puts@plt>
  jmp    L2
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D29
type: string
name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D25
type: string
name: unpack_data
arguments: tag: D37
type: string
name: unpack_data
arguments: tag: D49
type: string

tool: D29: Short string of text.


tool: D22: Long string of text.


tool: D25: 1st: %s


tool: D37: 2nd: %s


tool: D49: Unable to allocate.

assistant: ```c
int main() {
  char staticMemory[100];
  char *dynamicMemory;
  strcpy(staticMemory, "Short string of text.\n");
  dynamicMemory = malloc(200 * sizeof(char));
  if (dynamicMemory == NULL) {
    printf("Unable to allocate.\n");
  } else {
    strcpy(dynamicMemory, "Long string of text.\n");
  }
  printf("1st: %s\n", staticMemory);
  printf("2nd: %s\n", dynamicMemory);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D20(%rip),%rsi
  push   %rbp
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    $0xc8,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L26
  lea    D23(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L1:
  mov    %r12,%rsi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L26:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  jmp    L1
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D20
datatype: string
name: retrieve_data
arguments: label: D23
datatype: string
name: retrieve_data
arguments: label: D3
datatype: string
name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D35
datatype: string

tool: D20: Short string of text.


tool: D23: Long string of text.


tool: D3: 1st: %s


tool: D12: 2nd: %s


tool: D35: Unable to allocate.

assistant: ```c
int main() {
  char staticMemory[100];
  char *dynamicMemory;
  strcpy(staticMemory, "Short string of text.\n");
  dynamicMemory = malloc(200 * sizeof(char));
  if (dynamicMemory == NULL) {
    printf("Unable to allocate.\n");
  } else {
    strcpy(dynamicMemory, "Long string of text.\n");
  }
  printf("1st: %s\n", staticMemory);
  printf("2nd: %s\n", dynamicMemory);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_bitonic>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x30(%rbp)
  movl   $0xb,-0x2c(%rbp)
  movl   $0x2,-0x28(%rbp)
  movl   $0xa,-0x24(%rbp)
  movl   $0x4,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <bitonic@plt>
  cmp    $0x6,%eax
  sete   %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int test_bitonic() {
  int input[] = {1, 11, 2, 10, 4, 5, 2, 1};
  return 6 == bitonic(input, 8);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_bitonic>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0xb,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0xa,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  movl   $0x1,0x1c(%rsp)
  mov    %rsp,%rdi
  mov    $0x8,%esi
  call   <bitonic@plt>
  cmp    $0x6,%eax
  sete   %al
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  movzbl %al,%eax
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_bitonic() {
  int input[] = {1, 11, 2, 10, 4, 5, 2, 1};
  return 6 == bitonic(input, 8);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_bitonic>:
  endbr64
  sub    $0x38,%rsp
  mov    $0x8,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xb00000001,%rax
  mov    %rsp,%rdi
  mov    %rax,(%rsp)
  movabs $0xa00000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x500000004,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x100000002,%rax
  mov    %rax,0x18(%rsp)
  call   <bitonic@plt>
  cmp    $0x6,%eax
  sete   %al
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  movzbl %al,%eax
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_bitonic() {
  int input[] = {1, 11, 2, 10, 4, 5, 2, 1};
  return 6 == bitonic(input, 8);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_bitonic>:
  endbr64
  sub    $0x38,%rsp
  movdqa D6(%rip),%xmm0
  mov    $0x8,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movaps %xmm0,(%rsp)
  movdqa D12(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  call   <bitonic@plt>
  cmp    $0x6,%eax
  sete   %al
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  movzbl %al,%eax
  add    $0x38,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: byte[16]
name: retrieve_data
arguments: label: D12
type_infered: byte[16]

tool: {"D6": [0x01,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

tool: {"D12": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int test_bitonic() {
  int input[] = {1, 11, 2, 10, 4, 5, 2, 1};
  return 6 == bitonic(input, 8);
}
```