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
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    D45(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  lea    D42(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L6
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D42(%rip),%r12
L16:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0xc(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L16
L6:
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <selectionSort@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  mov    %rbx,%rdi
  call   <printArray@plt>
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D45
datatype: string
name: parse_data
arguments: tag: D42
datatype: string
name: parse_data
arguments: tag: D21
datatype: string
name: parse_data
arguments: tag: D17
datatype: string

tool: D45: Enter number of Elements tobe sorted: 

tool: D42: %d

tool: D21: Enter %d Numbers


tool: D17: Sorted array: 

assistant: ```c
int main() {
  int arr[100];
  int n, i;
  printf("Enter number of Elements tobe sorted: \n");
  scanf("%d", &n);
  printf("Enter %d Numbers\n", n);
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  selectionSort(arr, n);
  printf("Sorted array: \n");
  printArray(arr, n);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  lea    D17(%rip),%rdi
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  mov    %r13,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L26
  nopl   0x0(%rax,%rax,1)
L28:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L28
L26:
  mov    %r13,%rdi
  call   <selectionSort@plt>
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  mov    %r13,%rdi
  call   <printArray@plt>
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D17
data_type: string
name: parse_data
arguments: tag: D13
data_type: string
name: parse_data
arguments: tag: D29
data_type: string
name: parse_data
arguments: tag: D35
data_type: string

tool: D17: Enter number of Elements tobe sorted: 

tool: D13: %d

tool: D29: Enter %d Numbers


tool: D35: Sorted array: 

assistant: ```c
int main() {
  int arr[100];
  int n, i;
  printf("Enter number of Elements tobe sorted: \n");
  scanf("%d", &n);
  printf("Enter %d Numbers\n", n);
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  selectionSort(arr, n);
  printf("Sorted array: \n");
  printArray(arr, n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D21(%rip),%rdi
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  mov    %r13,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  test   %esi,%esi
  jle    L14
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L25
L14:
  mov    %r13,%rdi
  call   <selectionSort@plt>
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%esi
  mov    %r13,%rdi
  call   <printArray@plt>
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D21
datatype: string
name: read_data
arguments: data_tag: D28
datatype: string
name: read_data
arguments: data_tag: D3
datatype: string
name: read_data
arguments: data_tag: D35
datatype: string

tool: D21: Enter number of Elements tobe sorted: 

tool: D28: %d

tool: D3: Enter %d Numbers


tool: D35: Sorted array: 

assistant: ```c
int main() {
  int arr[100];
  int n, i;
  printf("Enter number of Elements tobe sorted: \n");
  scanf("%d", &n);
  printf("Enter %d Numbers\n", n);
  for (i = 0; i < n; i++)
    scanf("%d", &arr[i]);
  selectionSort(arr, n);
  printf("Sorted array: \n");
  printArray(arr, n);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Hamming>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L7:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0xc(%rbp),%xmm1
  movsd  D23(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <cos@plt>
  movsd  D9(%rip),%xmm1
  mulsd  %xmm0,%xmm1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  D13(%rip),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D23
datatype: f64
name: read_data
arguments: label: D9
datatype: f64
name: read_data
arguments: label: D13
datatype: f64

tool: D23: 6.283185307179586

tool: D9: 0.46164

tool: D13: 0.53836

assistant: ```c
void Hamming(int sizeWindow, double *pWindow) {
  int i;
  double fSum = 0;
  for (i = 0; i < sizeWindow; i++) {
    fSum += pWindow[i] = 0.53836 - 0.46164 * cos(6.28318530717958647692 * i / (sizeWindow - 1));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Hamming>:
  endbr64
  test   %edi,%edi
  jle    L12
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbp
  mov    %edi,%r12d
  mov    $0x0,%ebx
  sub    $0x1,%edi
  pxor   %xmm3,%xmm3
  cvtsi2sd %edi,%xmm3
  movsd  %xmm3,0x8(%rsp)
L9:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mulsd  D16(%rip),%xmm0
  divsd  0x8(%rsp),%xmm0
  call   <cos@plt>
  mulsd  D23(%rip),%xmm0
  movsd  D26(%rip),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L9
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D16
data_type: f64
name: parse_data
arguments: data_tag: D23
data_type: f64
name: parse_data
arguments: data_tag: D26
data_type: f64

tool: D16: 6.283185307179586

tool: D23: 0.46164

tool: D26: 0.53836

assistant: ```c
void Hamming(int sizeWindow, double *pWindow) {
  int i;
  double fSum = 0;
  for (i = 0; i < sizeWindow; i++) {
    fSum += pWindow[i] = 0.53836 - 0.46164 * cos(6.28318530717958647692 * i / (sizeWindow - 1));
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<Hamming>:
  endbr64
  test   %edi,%edi
  jle    L12
  lea    -0x1(%rdi),%eax
  pxor   %xmm3,%xmm3
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  cvtsi2sd %eax,%xmm3
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  movsd  %xmm3,0x8(%rsp)
  nopl   0x0(%rax)
L17:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mulsd  D15(%rip),%xmm0
  divsd  0x8(%rsp),%xmm0
  call   <cos@plt>
  mulsd  D29(%rip),%xmm0
  movsd  D6(%rip),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  ret
```

tool_calls: name: read_data
arguments: data_tag: D15
type_infered: f64
name: read_data
arguments: data_tag: D29
type_infered: f64
name: read_data
arguments: data_tag: D6
type_infered: f64

tool: D15: 6.283185307179586

tool: D29: 0.46164

tool: D6: 0.53836

assistant: ```c
void Hamming(int sizeWindow, double *pWindow) {
  int i;
  double fSum = 0;
  for (i = 0; i < sizeWindow; i++) {
    fSum += pWindow[i] = 0.53836 - 0.46164 * cos(6.28318530717958647692 * i / (sizeWindow - 1));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Hamming>:
  endbr64
  test   %edi,%edi
  jle    L5
  lea    -0x1(%rdi),%eax
  pxor   %xmm3,%xmm3
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  cvtsi2sd %eax,%xmm3
  movslq %edi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  movsd  %xmm3,0x8(%rsp)
  nopl   0x0(%rax)
L15:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mulsd  D10(%rip),%xmm0
  divsd  0x8(%rsp),%xmm0
  call   <cos@plt>
  mulsd  D5(%rip),%xmm0
  movsd  D22(%rip),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L15
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D10
type: f64
name: disassemble_data
arguments: data_tag: D5
type: f64
name: disassemble_data
arguments: data_tag: D22
type: f64

tool: D10: 6.283185307179586

tool: D5: 0.46164

tool: D22: 0.53836

assistant: ```c
void Hamming(int sizeWindow, double *pWindow) {
  int i;
  double fSum = 0;
  for (i = 0; i < sizeWindow; i++) {
    fSum += pWindow[i] = 0.53836 - 0.46164 * cos(6.28318530717958647692 * i / (sizeWindow - 1));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L23
L17:
  mov    -0x28(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x28(%rbp)
L23:
  mov    -0x28(%rbp),%eax
  cmp    $0x3,%eax
  jbe    L17
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x20(%rbp),%rax
  mov    $0x2,%edx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <rotateLeft@plt>
  mov    $0x2,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x24(%rbp)
  jmp    L46
L40:
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L46:
  mov    -0x24(%rbp),%eax
  cmp    $0x3,%eax
  jbe    L40
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D28
type: string
name: unpack_data
arguments: data_tag: D22
type: string
name: unpack_data
arguments: data_tag: D10
type: string

tool: D28: Original Vector:

tool: D22: %d 

tool: D10: rotated left(%d) Vector:

assistant: ```c
int main() {
  int a[] = {1, 2, 3, 4};
  printf("Original Vector:");
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  rotateLeft(a, sizeof(a) / sizeof(int), 2);
  printf("rotated left(%d) Vector:", 2);
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  lea    0x10(%rsp),%r12
  mov    %rbx,%rbp
  lea    D25(%rip),%r13
L11:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rsp,%rdi
  mov    $0x2,%edx
  mov    $0x4,%esi
  call   <rotateLeft@plt>
  mov    $0x2,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D25(%rip),%rbp
L16:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L16
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D24
type_infered: string
name: read_data
arguments: data_label: D25
type_infered: string
name: read_data
arguments: data_label: D28
type_infered: string

tool: D24: Original Vector:

tool: D25: %d 

tool: D28: rotated left(%d) Vector:

assistant: ```c
int main() {
  int a[] = {1, 2, 3, 4};
  printf("Original Vector:");
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  rotateLeft(a, sizeof(a) / sizeof(int), 2);
  printf("rotated left(%d) Vector:", 2);
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D16(%rip),%rdi
  push   %r13
  lea    D13(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%r14
  lea    0x10(%rsp),%r12
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
L6:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x4,%esi
  mov    %r14,%rdi
  mov    $0x2,%edx
  call   <rotateLeft@plt>
  mov    $0x2,%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L1:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D13
datatype: string
name: unpack_data
arguments: tag: D7
datatype: string

tool: D16: Original Vector:

tool: D13: %d 

tool: D7: rotated left(%d) Vector:

assistant: ```c
int main() {
  int a[] = {1, 2, 3, 4};
  printf("Original Vector:");
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  rotateLeft(a, sizeof(a) / sizeof(int), 2);
  printf("rotated left(%d) Vector:", 2);
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D24(%rip),%rdi
  push   %r13
  lea    D4(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x10(%rsp),%r12
  movaps %xmm0,(%rsp)
  mov    %r14,%rbx
  mov    %r14,%rbp
  call   <printf@plt>
L19:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x4,%esi
  mov    %r14,%rdi
  mov    $0x2,%edx
  call   <rotateLeft@plt>
  mov    $0x2,%esi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L6:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D24
datatype: string
name: read_data
arguments: tag: D4
datatype: string
name: read_data
arguments: tag: D1
datatype: byte[16]
name: read_data
arguments: tag: D3
datatype: string

tool: D24: Original Vector:

tool: D4: %d 

tool: {"D1": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D3: rotated left(%d) Vector:

assistant: ```c
int main() {
  int a[] = {1, 2, 3, 4};
  printf("Original Vector:");
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  rotateLeft(a, sizeof(a) / sizeof(int), 2);
  printf("rotated left(%d) Vector:", 2);
  for (int i = 0; i < sizeof(a) / sizeof(int); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  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
<sum_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L10
L19:
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jb     L19
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

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

```asm
<sum_array>:
  endbr64
  test   %esi,%esi
  je     L24
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rcx
  mov    $0x0,%edx
L0:
  add    (%rax),%rdx
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L0
L18:
  mov    %rdx,%rax
  ret
L24:
  mov    $0x0,%edx
  jmp    L18
```

assistant: ```c
long sum_array(long *array, unsigned int length) {
  long sum = 0;
  unsigned int i;
  for (i = 0; i < length; i++)
    sum += array[i];
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sum_array>:
  endbr64
  test   %esi,%esi
  je     L8
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L0:
  add    (%rdi),%rax
  add    $0x8,%rdi
  cmp    %rdx,%rdi
  jne    L0
  ret
  nopl   (%rax)
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
long sum_array(long *array, unsigned int length) {
  long sum = 0;
  unsigned int i;
  for (i = 0; i < length; i++)
    sum += array[i];
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum_array>:
  endbr64
  test   %esi,%esi
  je     L49
  cmp    $0x1,%esi
  je     L6
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L23:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddq  %xmm2,%xmm0
  cmp    %rax,%rdx
  jne    L23
  movdqa %xmm0,%xmm1
  mov    %esi,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffe,%edx
  and    $0x1,%esi
  paddq  %xmm1,%xmm0
  movq   %xmm0,%rax
  je     L9
L28:
  add    (%rdi,%rdx,8),%rax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  ret
  nopl   0x0(%rax)
L49:
  xor    %eax,%eax
  ret
L6:
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L28
```

assistant: ```c
long sum_array(long *array, unsigned int length) {
  long sum = 0;
  unsigned int i;
  for (i = 0; i < length; i++)
    sum += array[i];
  return sum;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L1
  negl   -0x14(%rbp)
L1:
  movl   $0x0,-0x10(%rbp)
L30:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jg     L30
  jmp    L67
L86:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
L67:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L86
  cmpl   $0x0,-0x8(%rbp)
  jns    L64
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
L64:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0xc(%rbp)
  jmp    L27
L25:
  addl   $0x1,-0xc(%rbp)
L27:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L25
  subl   $0x1,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L10
L35:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
  subl   $0x1,-0xc(%rbp)
L10:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L35
  nop
  pop    %rbp
  ret
```

assistant: ```c
void my_itoa(int n, char s[], int k) {
  int c, i, j, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do {
    s[i++] = n % 10 + '0';
  } while ((n /= 10) > 0);
  for (; i < k;)
    s[i++] = '0';
  if (sign < 0)
    s[i++] = '-';
  s[i] = '\0';
  j = 0;
  while (s[j] != '\0')
    j++;
  --j;
  for (i = 0; i < j; i++, j--) {
    c = s[i];
    s[i] = s[j];
    s[j] = c;
  }
  return;
}
```