ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61,0x4(%rsp)
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  lea    D9(%rip),%r13
  mov    %r13,%rdi
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  movsbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%esi
  movsbl 0x7(%rsp),%r8d
  lea    D2(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D29(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x4c,0x4(%rsp)
  movb   $0x65,0x5(%rsp)
  movb   $0x6f,0x6(%rsp)
  movb   $0xa,0x7(%rsp)
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  movsbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%esi
  movsbl 0x7(%rsp),%r8d
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D12(%rip),%rsi
  lea    D47(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x6c,%r8d
  mov    $0x75,%ecx
  mov    $0x61,%edx
  mov    $0x50,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
type_guessed: string
name: read_data
arguments: data_tag: D2
type_guessed: string
name: read_data
arguments: data_tag: D29
type_guessed: string
name: read_data
arguments: data_tag: D12
type_guessed: string
name: read_data
arguments: data_tag: D47
type_guessed: string
name: read_data
arguments: data_tag: D1
type_guessed: string

tool: D9: numbers: %d %d %d %d


tool: D2: name each: %c %c %c %c


tool: D29: name: %s


tool: D12: Paul

tool: D47: another: %s


tool: D1: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'L';
  name[1] = 'e';
  name[2] = 'o';
  name[3] = '\n';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Paul";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D3(%rip),%r14
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   %r13
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  push   %r12
  lea    D0(%rip),%r13
  push   %rbp
  lea    D12(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61,0x4(%rsp)
  lea    0x4(%rsp),%r12
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%r8d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  movl   $0xa6f654c,0x4(%rsp)
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D16(%rip),%rsi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    $0x6c,%r8d
  mov    $0x75,%ecx
  mov    $0x61,%edx
  mov    $0x50,%esi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D3
type: string
name: read_data
arguments: data_tag: D0
type: string
name: read_data
arguments: data_tag: D12
type: string
name: read_data
arguments: data_tag: D16
type: string
name: read_data
arguments: data_tag: D27
type: string
name: read_data
arguments: data_tag: D40
type: string

tool: D3: numbers: %d %d %d %d


tool: D0: name each: %c %c %c %c


tool: D12: name: %s


tool: D16: Paul

tool: D27: another: %s


tool: D40: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'L';
  name[1] = 'e';
  name[2] = 'o';
  name[3] = '\n';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Paul";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D38(%rip),%r14
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  push   %r13
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  push   %r12
  lea    D3(%rip),%r13
  push   %rbp
  lea    D36(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x61,0x4(%rsp)
  lea    0x4(%rsp),%r12
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x4,%r8d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  movl   $0xa6f654c,0x4(%rsp)
  call   <printf@plt>
  movsbl 0x6(%rsp),%ecx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsbl 0x5(%rsp),%edx
  movsbl 0x7(%rsp),%r8d
  movsbl 0x4(%rsp),%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D30(%rip),%rsi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    $0x6c,%r8d
  mov    $0x75,%ecx
  mov    $0x61,%edx
  mov    $0x50,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D38
datatype: string
name: unpack_data
arguments: label: D3
datatype: string
name: unpack_data
arguments: label: D36
datatype: string
name: unpack_data
arguments: label: D30
datatype: string
name: unpack_data
arguments: label: D24
datatype: string
name: unpack_data
arguments: label: D0
datatype: string

tool: D38: numbers: %d %d %d %d


tool: D3: name each: %c %c %c %c


tool: D36: name: %s


tool: D30: Paul

tool: D24: another: %s


tool: D0: another each: %c %c %c %c


assistant: ```c
int main(int argc, char *argv[]) {
  int numbers[4] = {0};
  char name[4] = {'a'};
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  numbers[0] = 1;
  numbers[1] = 2;
  numbers[2] = 3;
  numbers[3] = 4;
  name[0] = 'L';
  name[1] = 'e';
  name[2] = 'o';
  name[3] = '\n';
  printf("numbers: %d %d %d %d\n", numbers[0], numbers[1], numbers[2], numbers[3]);
  printf("name each: %c %c %c %c\n", name[0], name[1], name[2], name[3]);
  printf("name: %s\n", name);
  char *another = "Paul";
  printf("another: %s\n", another);
  printf("another each: %c %c %c %c\n", another[0], another[1], another[2], another[3]);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%eax
  imul   -0x34(%rbp),%eax
  add    $0x5,%eax
  mov    %eax,%edi
  call   <f>
  mov    -0x34(%rbp),%edx
  add    %edx,%eax
  mov    %eax,-0x30(%rbp)
  mov    -0x28(%rbp),%eax
  imul   -0x34(%rbp),%eax
  add    $0x5,%eax
  mov    %eax,%edi
  call   <f>
  mov    -0x34(%rbp),%edx
  add    %edx,%eax
  mov    %eax,-0x30(%rbp)
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int f(int b) {
  int y;
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <f>
  mov    $0x5,%edi
  call   <f>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int f(int b) {
  int y;
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f>:
  endbr64
  push   %rax
  pop    %rax
  sub    $0x8,%rsp
L4:
  mov    $0x5,%edi
  call   <f>
  jmp    L4
```

assistant: ```c
int f(int b) {
  int y;
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f>:
  endbr64
  push   %rax
  pop    %rax
  sub    $0x8,%rsp
L9:
  mov    $0x5,%edi
  call   <f>
  jmp    L9
```

assistant: ```c
int f(int b) {
  int y;
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
  {
    int x[10];
    x[2 + 3 - 5] = b + f(5 + x[2] * b);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<matrixTrace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L8:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double matrixTrace(double **A, int n) {
  double sum = 0;
  int i;
  for (i = 0; i < n; i++)
    sum += A[i][i];
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<matrixTrace>:
  endbr64
  test   %esi,%esi
  jle    L4
  mov    %esi,%esi
  lea    0x0(,%rsi,8),%rcx
  mov    $0x0,%eax
  pxor   %xmm0,%xmm0
L3:
  mov    (%rdi,%rax,1),%rdx
  addsd  (%rdx,%rax,1),%xmm0
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L3
  ret
L4:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double matrixTrace(double **A, int n) {
  double sum = 0;
  int i;
  for (i = 0; i < n; i++)
    sum += A[i][i];
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<matrixTrace>:
  endbr64
  test   %esi,%esi
  jle    L12
  movslq %esi,%rsi
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  lea    0x0(,%rsi,8),%rcx
  nopl   0x0(%rax)
L15:
  mov    (%rdi,%rax,1),%rdx
  addsd  (%rdx,%rax,1),%xmm0
  add    $0x8,%rax
  cmp    %rax,%rcx
  jne    L15
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double matrixTrace(double **A, int n) {
  double sum = 0;
  int i;
  for (i = 0; i < n; i++)
    sum += A[i][i];
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<matrixTrace>:
  endbr64
  test   %esi,%esi
  jle    L6
  movslq %esi,%rsi
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  lea    0x0(,%rsi,8),%rcx
  nopl   0x0(%rax)
L14:
  mov    (%rdi,%rax,1),%rdx
  addsd  (%rdx,%rax,1),%xmm0
  add    $0x8,%rax
  cmp    %rax,%rcx
  jne    L14
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double matrixTrace(double **A, int n) {
  double sum = 0;
  int i;
  for (i = 0; i < n; i++)
    sum += A[i][i];
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isxdigit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L13
  mov    $0x1,%eax
  jmp    L5
L13:
  mov    $0x0,%eax
L5:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
data_type: string

tool: D0: 0123456789abcdefABCDEF

assistant: ```c
int isxdigit(int c) {
  const char *hex = "0123456789abcdefABCDEF";
  if (strchr(hex, c))
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isxdigit>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D4(%rip),%rdi
  call   <strchr@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
data_type: string

tool: D4: 0123456789abcdefABCDEF

assistant: ```c
int isxdigit(int c) {
  const char *hex = "0123456789abcdefABCDEF";
  if (strchr(hex, c))
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isxdigit>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D0(%rip),%rdi
  call   <strchr@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type_infered: string

tool: D0: 0123456789abcdefABCDEF

assistant: ```c
int isxdigit(int c) {
  const char *hex = "0123456789abcdefABCDEF";
  if (strchr(hex, c))
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isxdigit>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D7(%rip),%rdi
  call   <strchr@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

tool_calls: name: read_data
arguments: label: D7
type: string

tool: D7: 0123456789abcdefABCDEF

assistant: ```c
int isxdigit(int c) {
  const char *hex = "0123456789abcdefABCDEF";
  if (strchr(hex, c))
    return 1;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_2x2_quantization>:
  endbr64
  push   %rbx
  lea    D21(%rip),%rdi
  call   <puts@plt>
  mov    $0xfffffd03,%esi
  lea    D1(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xffffffeb,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0xfffffd03,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xffffffeb,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D21
datatype: string
name: disassemble_data
arguments: tag: D1
datatype: string
name: disassemble_data
arguments: tag: D10
datatype: string

tool: D21: before

tool: D1: %5d   

tool: D10: after

assistant: ```c
void test_2x2_quantization() {
  int qp = 6;
  int coeff[2][2] = {{-765, -19}, {-21, 1}};
  int dclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int dcrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int aclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int acrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int leveloffset[4][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}};
  int coeff_cost = 0;
  printf("before\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  printf("after\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test_2x2_quantization>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rdi
  lea    D16(%rip),%rbp
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0xfffffd03,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  mov    $0xffffffeb,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D25(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0xfffffd03,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  mov    $0xffffffeb,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type: string
name: read_data
arguments: tag: D16
type: string
name: read_data
arguments: tag: D25
type: string

tool: D9: before

tool: D16: %5d   

tool: D25: after

assistant: ```c
void test_2x2_quantization() {
  int qp = 6;
  int coeff[2][2] = {{-765, -19}, {-21, 1}};
  int dclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int dcrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int aclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int acrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int leveloffset[4][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}};
  int coeff_cost = 0;
  printf("before\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  printf("after\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<test_2x2_quantization>:
  endbr64
  push   %rbp
  lea    D29(%rip),%rdi
  lea    D5(%rip),%rbp
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0xfffffd03,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  mov    $0xffffffeb,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0xfffffd03,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xffffffed,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rdi
  mov    $0xffffffeb,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  pop    %rbp
  jmp    1070 <putchar@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D29
data_type: string
name: disassemble_data
arguments: tag: D5
data_type: string
name: disassemble_data
arguments: tag: D11
data_type: string

tool: D29: before

tool: D5: %5d   

tool: D11: after

assistant: ```c
void test_2x2_quantization() {
  int qp = 6;
  int coeff[2][2] = {{-765, -19}, {-21, 1}};
  int dclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int dcrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int aclevel[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int acrun[16] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int leveloffset[4][4] = {{0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}};
  int coeff_cost = 0;
  printf("before\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  printf("after\n");
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      printf("%5d   ", coeff[i][j]);
    }
    printf("\n");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<count_greater>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L7
  mov    $0x0,%eax
  jmp    L14
L7:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <binarySearch@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  js     L35
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
  jmp    L14
L35:
  mov    -0x8(%rbp),%rax
  not    %rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
L14:
  leave
  ret
```

assistant: ```c
int32_t count_greater(const uint16_t *array, int32_t lenarray, uint16_t ikey) {
  if (lenarray == 0)
    return 0;
  int32_t pos = binarySearch(array, lenarray, ikey);
  if (pos >= 0) {
    return lenarray - (pos + 1);
  } else {
    return lenarray - (-pos - 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_greater>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  test   %rsi,%rsi
  jne    L2
L18:
  mov    %rbx,%rax
  pop    %rbx
  ret
L2:
  call   <binarySearch@plt>
  test   %rax,%rax
  js     L24
  add    $0x1,%rax
  sub    %rax,%rbx
  jmp    L18
L24:
  not    %rax
  sub    %rax,%rbx
  jmp    L18
```

assistant: ```c
int32_t count_greater(const uint16_t *array, int32_t lenarray, uint16_t ikey) {
  if (lenarray == 0)
    return 0;
  int32_t pos = binarySearch(array, lenarray, ikey);
  if (pos >= 0) {
    return lenarray - (pos + 1);
  } else {
    return lenarray - (-pos - 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<count_greater>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  test   %rsi,%rsi
  jne    L13
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L13:
  call   <binarySearch@plt>
  test   %rax,%rax
  js     L1
  add    $0x1,%rax
  sub    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
  nop
L1:
  not    %rax
  sub    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
int32_t count_greater(const uint16_t *array, int32_t lenarray, uint16_t ikey) {
  if (lenarray == 0)
    return 0;
  int32_t pos = binarySearch(array, lenarray, ikey);
  if (pos >= 0) {
    return lenarray - (pos + 1);
  } else {
    return lenarray - (-pos - 1);
  }
}
```