ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<xtoi>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  jne    L63
  jmp    L69
  nop
L10:
  movsbl %cl,%ecx
L6:
  movzbl 0x1(%rdi),%edx
  shl    $0x4,%eax
  add    $0x1,%rdi
  add    %ecx,%eax
  test   %dl,%dl
  je     L71
L63:
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L10
  lea    -0x41(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L62
  sub    $0x37,%edx
  add    $0x1,%rdi
  shl    $0x4,%eax
  movsbl %dl,%ecx
  movzbl (%rdi),%edx
  add    %ecx,%eax
  test   %dl,%dl
  jne    L63
L71:
  ret
  nopl   0x0(%rax,%rax,1)
L62:
  lea    -0x61(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L69
  sub    $0x57,%edx
  movsbl %dl,%ecx
  jmp    L6
L69:
  ret
```

assistant: ```c
int xtoi(char *p) {
  int curr;
  int res = 0;
  while (*p) {
    curr = ((*p) >= '0' && (*p) <= '9' ? (*p) - '0' : (*p) >= 'A' && (*p) <= 'F' ? (*p) - 'A' + 10 : (*p) >= 'a' && (*p) <= 'f' ? (*p) - 'a' + 10 : -1);
    if (curr == -1)
      break;
    res = res * 0x10 + curr;
    p++;
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<finitpa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  lea    -0x1c(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <finitpa_@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  sub    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L16
  nop
  nop
  leave
  ret
```

assistant: ```c
void finitpa(int a[], int na, int iord[], int nb) {
  int i;
  finitpa_(a, &na, iord, &nb);
  for (i = 0; i < nb; i++)
    iord[i]--;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<finitpa>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,0xc(%rsp)
  mov    %rdx,%rbx
  mov    %ecx,0x8(%rsp)
  lea    0x8(%rsp),%rcx
  lea    0xc(%rsp),%rsi
  call   <finitpa_@plt>
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L7
  mov    %rbx,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rbx,%rdx,4),%rdx
L19:
  subl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L19
L7:
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
void finitpa(int a[], int na, int iord[], int nb) {
  int i;
  finitpa_(a, &na, iord, &nb);
  for (i = 0; i < nb; i++)
    iord[i]--;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<finitpa>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x10,%rsp
  mov    %esi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    %ecx,0x8(%rsp)
  lea    0x8(%rsp),%rcx
  call   <finitpa_@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  jle    L13
  sub    $0x1,%eax
  mov    %rbx,%rdx
  lea    0x4(%rbx,%rax,4),%rax
  cs nopw 0x0(%rax,%rax,1)
L18:
  subl   $0x1,(%rdx)
  add    $0x4,%rdx
  cmp    %rax,%rdx
  jne    L18
L13:
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
void finitpa(int a[], int na, int iord[], int nb) {
  int i;
  finitpa_(a, &na, iord, &nb);
  for (i = 0; i < nb; i++)
    iord[i]--;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<finitpa>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x10,%rsp
  mov    %esi,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  mov    %ecx,0x8(%rsp)
  lea    0x8(%rsp),%rcx
  call   <finitpa_@plt>
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L14
  lea    -0x1(%rdx),%eax
  cmp    $0x2,%eax
  jbe    L48
  mov    %edx,%ecx
  mov    %rbx,%rax
  pcmpeqd %xmm1,%xmm1
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rbx,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L4:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  paddd  %xmm1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L4
  mov    %edx,%ecx
  and    $0xfffffffc,%ecx
  test   $0x3,%dl
  je     L14
L30:
  movslq %ecx,%rax
  lea    0x1(%rcx),%esi
  shl    $0x2,%rax
  subl   $0x1,(%rbx,%rax,1)
  cmp    %esi,%edx
  jle    L14
  add    $0x2,%ecx
  subl   $0x1,0x4(%rbx,%rax,1)
  cmp    %ecx,%edx
  jle    L14
  subl   $0x1,0x8(%rbx,%rax,1)
L14:
  add    $0x10,%rsp
  pop    %rbx
  ret
L48:
  xor    %ecx,%ecx
  jmp    L30
```

assistant: ```c
void finitpa(int a[], int na, int iord[], int nb) {
  int i;
  finitpa_(a, &na, iord, &nb);
  for (i = 0; i < nb; i++)
    iord[i]--;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<vectorFPMultiply>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L7
L12:
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movsd  (%rax),%xmm1
  mov    -0x20(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movsd  (%rax),%xmm0
  mov    -0x28(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  subl   $0x1,-0x4(%rbp)
L7:
  cmpl   $0x0,-0x4(%rbp)
  jg     L12
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vectorFPMultiply(double *a, double *b, double *c, int nsize) {
  int j;
  for (j = nsize; j > 0; j--)
    *(c++) = *(a++) * *(b++);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vectorFPMultiply>:
  endbr64
  test   %ecx,%ecx
  jle    L12
  mov    %ecx,%ecx
  mov    $0x0,%eax
L11:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L11
L12:
  ret
```

assistant: ```c
void vectorFPMultiply(double *a, double *b, double *c, int nsize) {
  int j;
  for (j = nsize; j > 0; j--)
    *(c++) = *(a++) * *(b++);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vectorFPMultiply>:
  endbr64
  test   %ecx,%ecx
  jle    L19
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   (%rax)
L17:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdx,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L17
L19:
  ret
```

assistant: ```c
void vectorFPMultiply(double *a, double *b, double *c, int nsize) {
  int j;
  for (j = nsize; j > 0; j--)
    *(c++) = *(a++) * *(b++);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vectorFPMultiply>:
  endbr64
  mov    %rdx,%r8
  movslq %ecx,%rdx
  test   %edx,%edx
  jle    L12
  lea    0x8(%rsi),%rax
  cmp    %rax,%r8
  lea    0x8(%rdi),%rax
  setne  %cl
  cmp    %rax,%r8
  setne  %al
  test   %al,%cl
  je     L5
  lea    -0x1(%rdx),%eax
  cmp    $0x1,%eax
  jbe    L5
  mov    %edx,%ecx
  xor    %eax,%eax
  shr    %ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax)
L15:
  movupd (%rdi,%rax,1),%xmm0
  movupd (%rsi,%rax,1),%xmm1
  mulpd  %xmm1,%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L15
  mov    %edx,%ecx
  and    $0xfffffffe,%ecx
  mov    %ecx,%eax
  shl    $0x3,%rax
  add    %rax,%rdi
  add    %rax,%rsi
  add    %r8,%rax
  cmp    %ecx,%edx
  je     L12
  movsd  (%rdi),%xmm0
  mulsd  (%rsi),%xmm0
  movsd  %xmm0,(%rax)
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L11:
  movsd  (%rdi,%rax,8),%xmm0
  mulsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%r8,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L11
L12:
  ret
```

assistant: ```c
void vectorFPMultiply(double *a, double *b, double *c, int nsize) {
  int j;
  for (j = nsize; j > 0; j--)
    *(c++) = *(a++) * *(b++);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ascii>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x9(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movzbl -0x9(%rbp),%eax
  movsbl %al,%edx
  movzbl -0x9(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type: string
name: retrieve_data
arguments: tag: D28
type: string
name: retrieve_data
arguments: tag: D4
type: string

tool: D3: Enter a character

tool: D28: %c

tool: D4: The ASCII value of %C = %d

assistant: ```c
void ascii() {
  char c;
  printf("Enter a character");
  scanf("%c", &c);
  printf("\n");
  printf("The ASCII value of %C = %d", c, c);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ascii>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbl 0x7(%rsp),%esi
  mov    %esi,%edx
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D23
type: string
name: extract_symbolic_data
arguments: data_label: D8
type: string
name: extract_symbolic_data
arguments: data_label: D24
type: string

tool: D23: Enter a character

tool: D8: %c

tool: D24: The ASCII value of %C = %d

assistant: ```c
void ascii() {
  char c;
  printf("Enter a character");
  scanf("%c", &c);
  printf("\n");
  printf("The ASCII value of %C = %d", c, c);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ascii>:
  endbr64
  sub    $0x18,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D27(%rip),%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbl 0x7(%rsp),%esi
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D15
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string
name: read_data
arguments: data_tag: D4
datatype: string

tool: D15: Enter a character

tool: D27: %c

tool: D4: The ASCII value of %C = %d

assistant: ```c
void ascii() {
  char c;
  printf("Enter a character");
  scanf("%c", &c);
  printf("\n");
  printf("The ASCII value of %C = %d", c, c);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ascii>:
  endbr64
  sub    $0x18,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x7(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbl 0x7(%rsp),%esi
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  mov    %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D3
data_type: string
name: unpack_data
arguments: label: D4
data_type: string
name: unpack_data
arguments: label: D2
data_type: string

tool: D3: Enter a character

tool: D4: %c

tool: D2: The ASCII value of %C = %d

assistant: ```c
void ascii() {
  char c;
  printf("Enter a character");
  scanf("%c", &c);
  printf("\n");
  printf("The ASCII value of %C = %d", c, c);
}
```
user: What is the c source code of the assembly code below?

```asm
<gf_log>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movb   $0x1,-0x3(%rbp)
  movb   $0x0,-0x2(%rbp)
L27:
  movzbl -0x3(%rbp),%eax
  cmp    -0x14(%rbp),%al
  je     L24
  movzbl -0x3(%rbp),%eax
  mov    %al,-0x1(%rbp)
  shlb   -0x3(%rbp)
  movzbl -0x1(%rbp),%eax
  test   %al,%al
  jns    L22
  xorb   $0x1b,-0x3(%rbp)
L22:
  movzbl -0x1(%rbp),%eax
  xor    %al,-0x3(%rbp)
  addb   $0x1,-0x2(%rbp)
  cmpb   $0x0,-0x2(%rbp)
  jne    L27
  jmp    L10
L24:
  nop
L10:
  movzbl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char gf_log(unsigned char x) {
  unsigned char atb = 1, i = 0, z;
  do {
    if (atb == x)
      break;
    z = atb;
    atb <<= 1;
    if (z & 0x80)
      atb ^= 0x1b;
    atb ^= z;
  } while (++i > 0);
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gf_log>:
  endbr64
  mov    $0x0,%ecx
  mov    $0x1,%eax
L5:
  mov    %ecx,%edx
  cmp    %dil,%al
  je     L15
  lea    (%rax,%rax,1),%edx
  mov    %edx,%esi
  xor    $0x1b,%esi
  test   %al,%al
  cmovs  %esi,%edx
  xor    %edx,%eax
  add    $0x1,%ecx
  cmp    $0x100,%ecx
  jne    L5
  mov    $0x0,%edx
L15:
  mov    %edx,%eax
  ret
```

assistant: ```c
unsigned char gf_log(unsigned char x) {
  unsigned char atb = 1, i = 0, z;
  do {
    if (atb == x)
      break;
    z = atb;
    atb <<= 1;
    if (z & 0x80)
      atb ^= 0x1b;
    atb ^= z;
  } while (++i > 0);
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gf_log>:
  endbr64
  xor    %ecx,%ecx
  mov    $0x1,%eax
  jmp    L19
  nopl   (%rax)
L9:
  lea    (%rax,%rax,1),%edx
  mov    %edx,%esi
  xor    $0x1b,%esi
  test   %al,%al
  cmovs  %esi,%edx
  add    $0x1,%ecx
  xor    %edx,%eax
  cmp    $0x100,%ecx
  je     L28
L19:
  mov    %ecx,%r8d
  cmp    %dil,%al
  jne    L9
  mov    %r8d,%eax
  ret
L28:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned char gf_log(unsigned char x) {
  unsigned char atb = 1, i = 0, z;
  do {
    if (atb == x)
      break;
    z = atb;
    atb <<= 1;
    if (z & 0x80)
      atb ^= 0x1b;
    atb ^= z;
  } while (++i > 0);
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<gf_log>:
  endbr64
  xor    %ecx,%ecx
  mov    $0x1,%eax
  jmp    L8
  nopl   (%rax)
L22:
  lea    (%rax,%rax,1),%edx
  mov    %edx,%esi
  xor    $0x1b,%esi
  test   %al,%al
  cmovs  %esi,%edx
  add    $0x1,%ecx
  xor    %edx,%eax
  cmp    $0x100,%ecx
  je     L15
L8:
  mov    %ecx,%r8d
  cmp    %dil,%al
  jne    L22
  mov    %r8d,%eax
  ret
L15:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned char gf_log(unsigned char x) {
  unsigned char atb = 1, i = 0, z;
  do {
    if (atb == x)
      break;
    z = atb;
    atb <<= 1;
    if (z & 0x80)
      atb ^= 0x1b;
    atb ^= z;
  } while (++i > 0);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    -0xa4(%rbp),%rsi
  lea    D27(%rip),%rdi
  call   <scanf@plt>
  mov    -0xa4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  je     L35
  lea    -0xa0(%rbp),%rax
  mov    %rax,-0xb8(%rbp)
  lea    D10(%rip),%r13
  mov    $0x10,%r14d
  jmp    L53
L56:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L95
  orq    $0x0,-0x8(%rsp,%rdx,1)
L95:
  mov    %rsp,%r12
  test   %rcx,%rcx
  jle    L112
  mov    $0x0,%eax
L87:
  movq   $0x0,(%r12,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L87
  mov    $0x0,%ebx
L8:
  lea    -0x98(%rbp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x98(%rbp),%rax
  addq   $0x1,(%r12,%rax,8)
  add    $0x1,%rbx
  cmp    %rbx,-0xa0(%rbp)
  jg     L8
L68:
  mov    $0x0,%eax
  jmp    L31
L112:
  mov    $0x0,%ebx
  jmp    L68
L132:
  add    $0x1,%rbx
  cmp    $0x2,%eax
  je     L21
L31:
  cmpq   $0x0,(%r12,%rbx,8)
  je     L132
  movslq %eax,%rdx
  mov    %rbx,-0x90(%rbp,%rdx,8)
  add    $0x1,%eax
  jmp    L132
L21:
  mov    -0x88(%rbp),%rsi
  sub    -0x90(%rbp),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %r15,%rsp
  mov    -0xa4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  je     L35
L53:
  mov    %rsp,%r15
  mov    -0xb8(%rbp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xa0(%rbp),%rcx
  lea    0xf(,%rcx,8),%rax
  mov    $0x0,%edx
  div    %r14
  shl    $0x4,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffff000,%rsi
  mov    %rsp,%rdx
  sub    %rsi,%rdx
L55:
  cmp    %rdx,%rsp
  je     L56
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L55
L35:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L101
  mov    $0x0,%eax
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L101:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D27
type_infered: string
name: disassemble_data
arguments: tag: D10
type_infered: string
name: disassemble_data
arguments: tag: D1
type_infered: string

tool: D27: %d

tool: D10: %lld

tool: D1: %lld


assistant: ```c
int main() {
  int t;
  long long int n, i, no;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    long long int diff[n];
    for (i = 0; i < n; i++) {
      diff[i] = 0;
    }
    for (i = 0; i < n; i++) {
      scanf("%lld", &no);
      diff[no]++;
    }
    int count = 0;
    long long int ar[10];
    while (count != 2) {
      if (diff[i] != 0) {
        ar[count] = i;
        count++;
      }
      i++;
    }
    printf("%lld\n", ar[1] - ar[0]);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0xa4(%rbp),%rsi
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    -0xa4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  je     L27
  lea    -0xa0(%rbp),%rax
  lea    D6(%rip),%rbx
  mov    %rax,-0xb8(%rbp)
  lea    -0x98(%rbp),%r12
  nopl   (%rax)
L52:
  mov    -0xb8(%rbp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <scanf@plt>
  mov    -0xa0(%rbp),%rsi
  mov    %rsp,%rdi
  lea    0x0(,%rsi,8),%rdx
  lea    0xf(%rdx),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rdi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rdi,%rsp
  je     L74
L47:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L47
L74:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L58
  orq    $0x0,-0x8(%rsp,%rcx,1)
L58:
  mov    %rsp,%r15
  test   %rsi,%rsi
  jle    L104
  xor    %esi,%esi
  mov    %r15,%rdi
  xor    %r14d,%r14d
  call   <memset@plt>
  nopl   0x0(%rax)
L30:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <scanf@plt>
  mov    -0x98(%rbp),%rax
  addq   $0x1,(%r15,%rax,8)
  cmp    %r14,-0xa0(%rbp)
  jg     L30
L22:
  xor    %eax,%eax
L35:
  cmpq   $0x0,(%r15,%r14,8)
  je     L80
  movslq %eax,%rsi
  add    $0x1,%eax
  mov    %r14,-0x90(%rbp,%rsi,8)
  add    $0x1,%r14
  cmp    $0x2,%eax
  jne    L35
  mov    -0x88(%rbp),%rsi
  lea    D2(%rip),%rdi
  sub    -0x90(%rbp),%rsi
  call   <printf@plt>
  mov    -0xa4(%rbp),%eax
  mov    %r13,%rsp
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  jne    L52
L27:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L84
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L80:
  add    $0x1,%r14
  jmp    L35
L104:
  xor    %r14d,%r14d
  jmp    L22
L84:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
type_guessed: string
name: parse_data
arguments: tag: D6
type_guessed: string
name: parse_data
arguments: tag: D2
type_guessed: string

tool: D3: %d

tool: D6: %lld

tool: D2: %lld


assistant: ```c
int main() {
  int t;
  long long int n, i, no;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    long long int diff[n];
    for (i = 0; i < n; i++) {
      diff[i] = 0;
    }
    for (i = 0; i < n; i++) {
      scanf("%lld", &no);
      diff[no]++;
    }
    int count = 0;
    long long int ar[10];
    while (count != 2) {
      if (diff[i] != 0) {
        ar[count] = i;
        count++;
      }
      i++;
    }
    printf("%lld\n", ar[1] - ar[0]);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D19(%rip),%rdi
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0xa4(%rbp),%rsi
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  mov    -0xa4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  je     L16
  lea    -0xa0(%rbp),%rax
  lea    D20(%rip),%rbx
  mov    %rax,-0xb8(%rbp)
  cs nopw 0x0(%rax,%rax,1)
L50:
  mov    -0xb8(%rbp),%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <scanf@plt>
  mov    -0xa0(%rbp),%rsi
  mov    %rsp,%rdi
  lea    0x0(,%rsi,8),%rdx
  lea    0xf(%rdx),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rdi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rdi,%rsp
  je     L128
L27:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L27
L128:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L100
  orq    $0x0,-0x8(%rsp,%rcx,1)
L100:
  mov    %rsp,%r15
  test   %rsi,%rsi
  jle    L47
  xor    %esi,%esi
  mov    %r15,%rdi
  lea    -0x98(%rbp),%r12
  xor    %r14d,%r14d
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L121:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <scanf@plt>
  mov    -0x98(%rbp),%rax
  addq   $0x1,(%r15,%rax,8)
  cmp    %r14,-0xa0(%rbp)
  jg     L121
L17:
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L35:
  cmpq   $0x0,(%r15,%r14,8)
  je     L6
L112:
  movslq %eax,%rsi
  mov    %r14,-0x90(%rbp,%rsi,8)
  add    $0x1,%r14
  cmp    $0x1,%eax
  je     L86
  cmpq   $0x0,(%r15,%r14,8)
  mov    $0x1,%eax
  jne    L112
L6:
  add    $0x1,%r14
  jmp    L35
  nopl   (%rax)
L86:
  mov    -0x88(%rbp),%rsi
  lea    D0(%rip),%rdi
  sub    -0x90(%rbp),%rsi
  call   <printf@plt>
  mov    -0xa4(%rbp),%eax
  mov    %r13,%rsp
  lea    -0x1(%rax),%edx
  mov    %edx,-0xa4(%rbp)
  test   %eax,%eax
  jne    L50
L16:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L117
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L47:
  xor    %r14d,%r14d
  jmp    L17
L117:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D19
datatype: string
name: retrieve_data
arguments: data_tag: D20
datatype: string
name: retrieve_data
arguments: data_tag: D0
datatype: string

tool: D19: %d

tool: D20: %lld

tool: D0: %lld


assistant: ```c
int main() {
  int t;
  long long int n, i, no;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    long long int diff[n];
    for (i = 0; i < n; i++) {
      diff[i] = 0;
    }
    for (i = 0; i < n; i++) {
      scanf("%lld", &no);
      diff[no]++;
    }
    int count = 0;
    long long int ar[10];
    while (count != 2) {
      if (diff[i] != 0) {
        ar[count] = i;
        count++;
      }
      i++;
    }
    printf("%lld\n", ar[1] - ar[0]);
  }
  return 0;
}
```