ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<tls1_check_curve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <tls1_get_curvelist@plt>
  movq   $0x0,-0x10(%rbp)
  jmp    L11
L21:
  mov    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  cmp    %rax,-0x30(%rbp)
  jne    L20
  mov    $0x1,%eax
  jmp    L25
L20:
  addq   $0x1,-0x10(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jb     L21
  mov    $0x0,%eax
L25:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

assistant: ```c
int tls1_check_curve(SSL *s, const uint16_t curve_id) {
  const uint16_t *curves;
  size_t curveslen, i;
  tls1_get_curvelist(s, 0, &curves, &curveslen);
  for (i = 0; i < curveslen; i++) {
    if (curves[i] == curve_id)
      return (1);
  }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tls1_check_curve>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    $0x0,%esi
  call   <tls1_get_curvelist@plt>
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L20
  mov    0x8(%rsp),%rcx
  mov    $0x0,%eax
L15:
  cmp    %rbx,(%rcx,%rax,8)
  je     L39
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L15
  mov    $0x0,%eax
  jmp    L58
L20:
  mov    $0x0,%eax
  jmp    L58
L39:
  mov    $0x1,%eax
L58:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L59
  add    $0x20,%rsp
  pop    %rbx
  ret
L59:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int tls1_check_curve(SSL *s, const uint16_t curve_id) {
  const uint16_t *curves;
  size_t curveslen, i;
  tls1_get_curvelist(s, 0, &curves, &curveslen);
  for (i = 0; i < curveslen; i++) {
    if (curves[i] == curve_id)
      return (1);
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tls1_check_curve>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rcx
  call   <tls1_get_curvelist@plt>
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L28
  mov    0x8(%rsp),%rcx
  xor    %eax,%eax
  jmp    L17
L27:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L28
L17:
  cmp    %rbx,(%rcx,%rax,8)
  jne    L27
  mov    $0x1,%eax
  jmp    L30
  cs nopw 0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
L30:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x20,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int tls1_check_curve(SSL *s, const uint16_t curve_id) {
  const uint16_t *curves;
  size_t curveslen, i;
  tls1_get_curvelist(s, 0, &curves, &curveslen);
  for (i = 0; i < curveslen; i++) {
    if (curves[i] == curve_id)
      return (1);
  }
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tls1_check_curve>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rcx
  call   <tls1_get_curvelist@plt>
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L40
  mov    0x8(%rsp),%rcx
  xor    %eax,%eax
  jmp    L37
L14:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L40
L37:
  cmp    %rbx,(%rcx,%rax,8)
  jne    L14
  mov    $0x1,%eax
  jmp    L38
  cs nopw 0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
L38:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L58
  add    $0x20,%rsp
  pop    %rbx
  ret
L58:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int tls1_check_curve(SSL *s, const uint16_t curve_id) {
  const uint16_t *curves;
  size_t curveslen, i;
  tls1_get_curvelist(s, 0, &curves, &curveslen);
  for (i = 0; i < curveslen; i++) {
    if (curves[i] == curve_id)
      return (1);
  }
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sumOfDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L22
  mov    $0x1,%eax
  jmp    L24
L22:
  cmpl   $0x0,-0x14(%rbp)
  jle    L5
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ebx
  mov    %ebx,%eax
  shl    $0x2,%eax
  add    %ebx,%eax
  add    %eax,%eax
  mov    %edx,%ebx
  sub    %eax,%ebx
  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,%edi
  call   <sumOfDigits>
  add    %ebx,%eax
  jmp    L24
L5:
  mov    $0x0,%eax
L24:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int sumOfDigits(int num) {
  if (num == 1)
    return 1;
  if (num > 0) {
    return num % 10 + sumOfDigits(num / 10);
  } else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sumOfDigits>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%edi
  je     L1
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x0,%eax
  test   %edi,%edi
  jg     L13
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  movslq %edi,%rbp
  imul   $0x66666667,%rbp,%rbp
  sar    $0x22,%rbp
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebp
  mov    %ebp,%edi
  call   <sumOfDigits>
  lea    0x0(%rbp,%rbp,4),%edx
  add    %edx,%edx
  sub    %edx,%ebx
  add    %ebx,%eax
  jmp    L7
L1:
  ret
```

assistant: ```c
int sumOfDigits(int num) {
  if (num == 1)
    return 1;
  if (num > 0) {
    return num % 10 + sumOfDigits(num / 10);
  } else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sumOfDigits>:
  endbr64
  cmp    $0x1,%edi
  je     L55
  xor    %r8d,%r8d
  mov    $0xcccccccd,%esi
  test   %edi,%edi
  jg     L19
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L48:
  cmp    $0x9,%ecx
  jle    L1
L19:
  mov    %edi,%eax
  imul   %rsi,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %edi,%eax
  add    %ecx,%ecx
  sub    %ecx,%eax
  mov    %edi,%ecx
  mov    %edx,%edi
  add    %eax,%r8d
  cmp    $0x1,%edx
  jne    L48
  add    $0x1,%r8d
L1:
  mov    %r8d,%eax
  ret
L55:
  mov    $0x1,%r8d
  jmp    L1
```

assistant: ```c
int sumOfDigits(int num) {
  if (num == 1)
    return 1;
  if (num > 0) {
    return num % 10 + sumOfDigits(num / 10);
  } else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sumOfDigits>:
  endbr64
  cmp    $0x1,%edi
  je     L13
  xor    %r8d,%r8d
  mov    $0xcccccccd,%esi
  test   %edi,%edi
  jg     L54
  jmp    L57
  nopw   0x0(%rax,%rax,1)
L8:
  cmp    $0x9,%ecx
  jle    L57
L54:
  mov    %edi,%eax
  imul   %rsi,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  mov    %eax,%edx
  mov    %edi,%eax
  add    %ecx,%ecx
  sub    %ecx,%eax
  mov    %edi,%ecx
  mov    %edx,%edi
  add    %eax,%r8d
  cmp    $0x1,%edx
  jne    L8
  add    $0x1,%r8d
L57:
  mov    %r8d,%eax
  ret
L13:
  mov    $0x1,%r8d
  jmp    L57
```

assistant: ```c
int sumOfDigits(int num) {
  if (num == 1)
    return 1;
  if (num > 0) {
    return num % 10 + sumOfDigits(num / 10);
  } else
    return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L65
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L14
L65:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L14:
  movl   $0x1,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L67
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L77
L67:
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L77:
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L16
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L40
L16:
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L40:
  movq   $0x1,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L29
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L2
L29:
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L2:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D70
data_type: string
name: read_data
arguments: data_tag: D1
data_type: string
name: read_data
arguments: data_tag: D21
data_type: string
name: read_data
arguments: data_tag: D27
data_type: string
name: read_data
arguments: data_tag: D37
data_type: string
name: read_data
arguments: data_tag: D56
data_type: string
name: read_data
arguments: data_tag: D13
data_type: string
name: read_data
arguments: data_tag: D45
data_type: string

tool: D70: A


tool: D1: B


tool: D21: C


tool: D27: D


tool: D37: >A


tool: D56: >B


tool: D13: >C


tool: D45: >D


assistant: ```c
int main(void) {
  int a;
  long long e;
  a = 0;
  if (a)
    printf("A\n");
  else
    printf("B\n");
  a = 1;
  if (a)
    printf("C\n");
  else
    printf("D\n");
  e = 0;
  if (e)
    printf(">A\n");
  else
    printf(">B\n");
  e = 1;
  if (e)
    printf(">C\n");
  else
    printf(">D\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
  sub    $0x8,%rsp
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D21
datatype: string
name: parse_data
arguments: label: D13
datatype: string
name: parse_data
arguments: label: D15
datatype: string
name: parse_data
arguments: label: D3
datatype: string

tool: D21: B


tool: D13: C


tool: D15: >B


tool: D3: >C


assistant: ```c
int main(void) {
  int a;
  long long e;
  a = 0;
  if (a)
    printf("A\n");
  else
    printf("B\n");
  a = 1;
  if (a)
    printf("C\n");
  else
    printf("D\n");
  e = 0;
  if (e)
    printf(">A\n");
  else
    printf(">B\n");
  e = 1;
  if (e)
    printf(">C\n");
  else
    printf(">D\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
data_type: string
name: parse_data
arguments: data_tag: D21
data_type: string
name: parse_data
arguments: data_tag: D13
data_type: string
name: parse_data
arguments: data_tag: D12
data_type: string

tool: D1: B


tool: D21: C


tool: D13: >B


tool: D12: >C


assistant: ```c
int main(void) {
  int a;
  long long e;
  a = 0;
  if (a)
    printf("A\n");
  else
    printf("B\n");
  a = 1;
  if (a)
    printf("C\n");
  else
    printf("D\n");
  e = 0;
  if (e)
    printf(">A\n");
  else
    printf(">B\n");
  e = 1;
  if (e)
    printf(">C\n");
  else
    printf(">D\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D27
type_guessed: string
name: retrieve_data
arguments: label: D17
type_guessed: string
name: retrieve_data
arguments: label: D13
type_guessed: string
name: retrieve_data
arguments: label: D30
type_guessed: string

tool: D27: B


tool: D17: C


tool: D13: >B


tool: D30: >C


assistant: ```c
int main(void) {
  int a;
  long long e;
  a = 0;
  if (a)
    printf("A\n");
  else
    printf("B\n");
  a = 1;
  if (a)
    printf("C\n");
  else
    printf("D\n");
  e = 0;
  if (e)
    printf(">A\n");
  else
    printf(">B\n");
  e = 1;
  if (e)
    printf(">C\n");
  else
    printf(">D\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<predict_8x8c_dc>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  movslq %esi,%r11
  lea    0x0(,%r11,8),%rdi
  lea    -0x8(%rax),%rcx
  neg    %esi
  movslq %esi,%rsi
  lea    (%rax,%rsi,8),%rbp
  mov    %r11,%rdx
  neg    %rdx
  lea    (%rax,%rdx,8),%rbx
  shl    $0x5,%r11
  mov    $0x0,%edx
  mov    $0x0,%r8d
  mov    $0x0,%esi
  mov    $0x0,%r9d
  mov    $0x0,%r10d
L2:
  add    0x0(%rbp,%rdx,1),%r10d
  add    0x20(%rbx,%rdx,1),%r9d
  add    (%rcx),%esi
  add    (%rcx,%r11,1),%r8d
  add    %rdi,%rcx
  add    $0x8,%rdx
  cmp    $0x20,%rdx
  jne    L2
  lea    0x4(%r10,%rsi,1),%ebx
  sar    $0x3,%ebx
  imul   $0x1010101,%ebx,%ebx
  movslq %ebx,%rbx
  lea    0x2(%r9),%r10d
  sar    $0x2,%r10d
  imul   $0x1010101,%r10d,%r10d
  movslq %r10d,%r10
  lea    0x2(%r8),%esi
  sar    $0x2,%esi
  imul   $0x1010101,%esi,%esi
  movslq %esi,%rsi
  lea    0x4(%r9,%r8,1),%r8d
  sar    $0x3,%r8d
  imul   $0x1010101,%r8d,%r8d
  movslq %r8d,%r8
  mov    %rax,%rdx
  mov    $0x4,%ecx
L6:
  mov    %rbx,(%rdx)
  mov    %r10,0x8(%rdx)
  add    %rdi,%rdx
  sub    $0x1,%ecx
  jne    L6
  add    %r11,%rax
  mov    $0x4,%edx
L15:
  mov    %rsi,(%rax)
  mov    %r8,0x8(%rax)
  add    %rdi,%rax
  sub    $0x1,%edx
  jne    L15
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void predict_8x8c_dc(uint8_t *src, int i_stride) {
  int y;
  int s0 = 0, s1 = 0, s2 = 0, s3 = 0;
  uint32_t dc0, dc1, dc2, dc3;
  int i;
  for (i = 0; i < 4; i++) {
    s0 += src[i - i_stride];
    s1 += src[i + 4 - i_stride];
    s2 += src[-1 + i * i_stride];
    s3 += src[-1 + (i + 4) * i_stride];
  }
  dc0 = ((s0 + s2 + 4) >> 3) * 0x01010101;
  dc1 = ((s1 + 2) >> 2) * 0x01010101;
  dc2 = ((s3 + 2) >> 2) * 0x01010101;
  dc3 = ((s1 + s3 + 4) >> 3) * 0x01010101;
  for (y = 0; y < 4; y++) {
    uint32_t *p = (uint32_t *)src;
    *p++ = dc0;
    *p++ = dc1;
    src += i_stride;
  }
  for (y = 0; y < 4; y++) {
    uint32_t *p = (uint32_t *)src;
    *p++ = dc2;
    *p++ = dc3;
    src += i_stride;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<predict_8x8c_dc>:
  endbr64
  mov    %rdi,%rax
  movslq %esi,%rsi
  push   %rbx
  xor    %edx,%edx
  mov    %rax,%rbx
  lea    -0x8(%rax),%rcx
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  lea    0x0(,%rsi,8),%rdi
  xor    %r10d,%r10d
  shl    $0x5,%rsi
  xor    %r11d,%r11d
  sub    %rdi,%rbx
L21:
  add    (%rbx,%rdx,1),%r11d
  add    0x20(%rbx,%rdx,1),%r10d
  add    $0x8,%rdx
  add    (%rcx),%r8d
  add    (%rcx,%rsi,1),%r9d
  add    %rdi,%rcx
  cmp    $0x20,%rdx
  jne    L21
  lea    0x4(%r11,%r8,1),%ebx
  lea    0x2(%r9),%r8d
  mov    %rax,%rdx
  mov    $0x4,%ecx
  lea    0x2(%r10),%r11d
  lea    0x4(%r10,%r9,1),%r9d
  sar    $0x3,%ebx
  sar    $0x2,%r8d
  sar    $0x2,%r11d
  sar    $0x3,%r9d
  imul   $0x1010101,%ebx,%ebx
  imul   $0x1010101,%r11d,%r11d
  imul   $0x1010101,%r8d,%r8d
  imul   $0x1010101,%r9d,%r9d
  movslq %ebx,%rbx
  movslq %r11d,%r11
  movslq %r8d,%r8
  movslq %r9d,%r9
L8:
  mov    %rbx,(%rdx)
  mov    %r11,0x8(%rdx)
  add    %rdi,%rdx
  sub    $0x1,%ecx
  jne    L8
  add    %rsi,%rax
  mov    $0x4,%edx
L28:
  mov    %r8,(%rax)
  mov    %r9,0x8(%rax)
  add    %rdi,%rax
  sub    $0x1,%edx
  jne    L28
  pop    %rbx
  ret
```

assistant: ```c
void predict_8x8c_dc(uint8_t *src, int i_stride) {
  int y;
  int s0 = 0, s1 = 0, s2 = 0, s3 = 0;
  uint32_t dc0, dc1, dc2, dc3;
  int i;
  for (i = 0; i < 4; i++) {
    s0 += src[i - i_stride];
    s1 += src[i + 4 - i_stride];
    s2 += src[-1 + i * i_stride];
    s3 += src[-1 + (i + 4) * i_stride];
  }
  dc0 = ((s0 + s2 + 4) >> 3) * 0x01010101;
  dc1 = ((s1 + 2) >> 2) * 0x01010101;
  dc2 = ((s3 + 2) >> 2) * 0x01010101;
  dc3 = ((s1 + s3 + 4) >> 3) * 0x01010101;
  for (y = 0; y < 4; y++) {
    uint32_t *p = (uint32_t *)src;
    *p++ = dc0;
    *p++ = dc1;
    src += i_stride;
  }
  for (y = 0; y < 4; y++) {
    uint32_t *p = (uint32_t *)src;
    *p++ = dc2;
    *p++ = dc3;
    src += i_stride;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<intraprocdpd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    -0x10(%rbp),%rax
  movl   $0x2a,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void intraprocdpd() {
  int *y;
  int *z;
  *y = 42;
  z = y;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<intraprocdpd>:
  endbr64
  movl   $0x2a,0x0
  ret
```

assistant: ```c
void intraprocdpd() {
  int *y;
  int *z;
  *y = 42;
  z = y;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<intraprocdpd>:
  endbr64
  movl   $0x2a,0x0
  ret
```

assistant: ```c
void intraprocdpd() {
  int *y;
  int *z;
  *y = 42;
  z = y;
}
```
user: What is the c source code of the assembly code below?

```asm
<ExecPressCommand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    D66(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L6
  mov    -0x8(%rbp),%rax
  lea    D56(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ProcessCommand@plt>
  jmp    L4
L6:
  mov    -0x10(%rbp),%rax
  lea    D34(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L2
  mov    -0x8(%rbp),%rax
  lea    D82(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ProcessCommand@plt>
  jmp    L4
L2:
  mov    -0x10(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L43
  mov    -0x8(%rbp),%rax
  lea    D19(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ProcessCommand@plt>
  jmp    L4
L43:
  mov    -0x10(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L19
  mov    -0x8(%rbp),%rax
  lea    D35(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ProcessCommand@plt>
  jmp    L4
L19:
  mov    -0x10(%rbp),%rax
  lea    D80(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L34
  mov    -0x8(%rbp),%rax
  lea    D84(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ProcessCommand@plt>
  jmp    L4
L34:
  mov    $0x2,%eax
L4:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D66
data_type: string
name: disassemble_data
arguments: data_tag: D56
data_type: string
name: disassemble_data
arguments: data_tag: D34
data_type: string
name: disassemble_data
arguments: data_tag: D82
data_type: string
name: disassemble_data
arguments: data_tag: D6
data_type: string
name: disassemble_data
arguments: data_tag: D19
data_type: string
name: disassemble_data
arguments: data_tag: D12
data_type: string
name: disassemble_data
arguments: data_tag: D35
data_type: string
name: disassemble_data
arguments: data_tag: D80
data_type: string
name: disassemble_data
arguments: data_tag: D84
data_type: string

tool: D66: right

tool: D56: C3A

tool: D34: left

tool: D82: C3B

tool: D6: up

tool: D19: C3C

tool: D12: down

tool: D35: C3D

tool: D80: enter

tool: D84: C3F

assistant: ```c
int ExecPressCommand(char ret[], const char *arg) {
  if (!strcmp(arg, "right"))
    return (ProcessCommand(ret, "C3A\r"));
  if (!strcmp(arg, "left"))
    return (ProcessCommand(ret, "C3B\r"));
  if (!strcmp(arg, "up"))
    return (ProcessCommand(ret, "C3C\r"));
  if (!strcmp(arg, "down"))
    return (ProcessCommand(ret, "C3D\r"));
  if (!strcmp(arg, "enter"))
    return (ProcessCommand(ret, "C3F\r"));
  return (2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ExecPressCommand>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  lea    D77(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L68
  lea    D83(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L61
  lea    D70(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L31
  lea    D31(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L1
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %eax,%edx
  mov    $0x2,%eax
  test   %edx,%edx
  je     L53
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L68:
  lea    D66(%rip),%rsi
  mov    %rbp,%rdi
  call   <ProcessCommand@plt>
  jmp    L7
L61:
  lea    D55(%rip),%rsi
  mov    %rbp,%rdi
  call   <ProcessCommand@plt>
  jmp    L7
L31:
  lea    D43(%rip),%rsi
  mov    %rbp,%rdi
  call   <ProcessCommand@plt>
  jmp    L7
L1:
  lea    D57(%rip),%rsi
  mov    %rbp,%rdi
  call   <ProcessCommand@plt>
  jmp    L7
L53:
  lea    D59(%rip),%rsi
  mov    %rbp,%rdi
  call   <ProcessCommand@plt>
  jmp    L7
```

tool_calls: name: extract_symbolic_data
arguments: tag: D77
datatype: string
name: extract_symbolic_data
arguments: tag: D83
datatype: string
name: extract_symbolic_data
arguments: tag: D70
datatype: string
name: extract_symbolic_data
arguments: tag: D31
datatype: string
name: extract_symbolic_data
arguments: tag: D5
datatype: string
name: extract_symbolic_data
arguments: tag: D66
datatype: string
name: extract_symbolic_data
arguments: tag: D55
datatype: string
name: extract_symbolic_data
arguments: tag: D43
datatype: string
name: extract_symbolic_data
arguments: tag: D57
datatype: string
name: extract_symbolic_data
arguments: tag: D59
datatype: string

tool: D77: right

tool: D83: left

tool: D70: up

tool: D31: down

tool: D5: enter

tool: D66: C3A

tool: D55: C3B

tool: D43: C3C

tool: D57: C3D

tool: D59: C3F

assistant: ```c
int ExecPressCommand(char ret[], const char *arg) {
  if (!strcmp(arg, "right"))
    return (ProcessCommand(ret, "C3A\r"));
  if (!strcmp(arg, "left"))
    return (ProcessCommand(ret, "C3B\r"));
  if (!strcmp(arg, "up"))
    return (ProcessCommand(ret, "C3C\r"));
  if (!strcmp(arg, "down"))
    return (ProcessCommand(ret, "C3D\r"));
  if (!strcmp(arg, "enter"))
    return (ProcessCommand(ret, "C3F\r"));
  return (2);
}
```