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
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  mov    D37(%rip),%rax
  movq   %rax,%xmm1
  mov    D5(%rip),%rax
  movq   %rax,%xmm0
  call   <ang2vec@plt>
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <vec2ang@plt>
  movsd  0x8(%rsp),%xmm3
  mov    D37(%rip),%rax
  movq   %rax,%xmm2
  movsd  (%rsp),%xmm1
  mov    D5(%rip),%rax
  movq   %rax,%xmm0
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D37
type_infered: byte[8]
name: retrieve_data
arguments: label: D5
type_infered: f64
name: retrieve_data
arguments: label: D14
type_infered: string

tool: {"D37": [0x9a,0x99,0x99,0x99,0x99,0x99,0xd9,0xbf]}

tool: D5: 0.5

tool: D14: %f %f, %f %f


assistant: ```c
int main(void) {
  double theta, phi, th1, ph1;
  long nside;
  long ipix, npix, dpix, ip2, ip1, ns1, pp;
  double vec[3];
  nside = 1024;
  dpix = 1;
  theta = 0.5;
  phi = -0.4;
  ang2vec(theta, phi, vec);
  vec2ang(vec, &th1, &ph1);
  printf("%f %f, %f %f\n", theta, th1, phi, ph1);
  theta = -0.1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  movsd  D43(%rip),%xmm2
  movsd  D13(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  movapd %xmm2,%xmm1
  mov    %rbp,%rdi
  call   <ang2vec@plt>
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  call   <vec2ang@plt>
  movsd  0x8(%rsp),%xmm3
  mov    D43(%rip),%rax
  lea    D41(%rip),%rdi
  movsd  (%rsp),%xmm1
  movq   %rax,%xmm2
  mov    D13(%rip),%rax
  movq   %rax,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D43
type_guessed: byte[8]
name: disassemble_data
arguments: data_label: D13
type_guessed: f64
name: disassemble_data
arguments: data_label: D41
type_guessed: string

tool: {"D43": [0x9a,0x99,0x99,0x99,0x99,0x99,0xd9,0xbf]}

tool: D13: 0.5

tool: D41: %f %f, %f %f


assistant: ```c
int main(void) {
  double theta, phi, th1, ph1;
  long nside;
  long ipix, npix, dpix, ip2, ip1, ns1, pp;
  double vec[3];
  nside = 1024;
  dpix = 1;
  theta = 0.5;
  phi = -0.4;
  ang2vec(theta, phi, vec);
  vec2ang(vec, &th1, &ph1);
  printf("%f %f, %f %f\n", theta, th1, phi, ph1);
  theta = -0.1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x30,%rsp
  movsd  D38(%rip),%xmm2
  movsd  D29(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  movapd %xmm2,%xmm1
  mov    %rbp,%rdi
  call   <ang2vec@plt>
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  call   <vec2ang@plt>
  movsd  0x8(%rsp),%xmm3
  mov    D38(%rip),%rax
  lea    D34(%rip),%rdi
  movsd  (%rsp),%xmm1
  movq   %rax,%xmm2
  mov    D29(%rip),%rax
  movq   %rax,%xmm0
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D38
type_infered: byte[8]
name: parse_data
arguments: data_tag: D29
type_infered: f64
name: parse_data
arguments: data_tag: D34
type_infered: string

tool: {"D38": [0x9a,0x99,0x99,0x99,0x99,0x99,0xd9,0xbf]}

tool: D29: 0.5

tool: D34: %f %f, %f %f


assistant: ```c
int main(void) {
  double theta, phi, th1, ph1;
  long nside;
  long ipix, npix, dpix, ip2, ip1, ns1, pp;
  double vec[3];
  nside = 1024;
  dpix = 1;
  theta = 0.5;
  phi = -0.4;
  ang2vec(theta, phi, vec);
  vec2ang(vec, &th1, &ph1);
  printf("%f %f, %f %f\n", theta, th1, phi, ph1);
  theta = -0.1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <print_binary@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_guessed: string

tool: D9: %i

assistant: ```c
int main(int argc, const char *argv[]) {
  int x;
  scanf("%i", &x);
  print_binary(x);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_binary@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D4
type: string

tool: D4: %i

assistant: ```c
int main(int argc, const char *argv[]) {
  int x;
  scanf("%i", &x);
  print_binary(x);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_binary@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D5
type: string

tool: D5: %i

assistant: ```c
int main(int argc, const char *argv[]) {
  int x;
  scanf("%i", &x);
  print_binary(x);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D6(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <print_binary@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D6: %i

assistant: ```c
int main(int argc, const char *argv[]) {
  int x;
  scanf("%i", &x);
  print_binary(x);
  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
<test2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x5,%edi
  call   <xyzzy2@plt>
  pop    %rbp
  ret
```

assistant: ```c
int test2(void) {
  extern int xyzzy2(int);
  return xyzzy2(5);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test2>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x5,%edi
  call   <xyzzy2@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int test2(void) {
  extern int xyzzy2(int);
  return xyzzy2(5);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test2>:
  endbr64
  mov    $0x5,%edi
  jmp    1050 <xyzzy2@plt>
```

assistant: ```c
int test2(void) {
  extern int xyzzy2(int);
  return xyzzy2(5);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test2>:
  endbr64
  mov    $0x5,%edi
  jmp    1050 <xyzzy2@plt>
```

assistant: ```c
int test2(void) {
  extern int xyzzy2(int);
  return xyzzy2(5);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<md5_encode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L15:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl %al,%eax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    $0x8,%eax
  cltq
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl %al,%eax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    $0x10,%eax
  cltq
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  add    $0x2,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl %al,%eax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sar    $0x18,%eax
  cltq
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  add    $0x3,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl %al,%eax
  mov    %rax,(%rdx)
  addl   $0x1,-0x8(%rbp)
  addl   $0x4,-0x4(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L15
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void md5_encode(uint8_t *output, const uint32_t *input, uint32_t len) {
  uint32_t i, j;
  for (i = 0, j = 0; i < len; i++, j += 4) {
    output[j] = (uint8_t)(input[i] & 0xff);
    output[j + 1] = (uint8_t)((input[i] >> 8) & 0xff);
    output[j + 2] = (uint8_t)((input[i] >> 16) & 0xff);
    output[j + 3] = (uint8_t)((input[i] >> 24) & 0xff);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<md5_encode>:
  endbr64
  test   %edx,%edx
  jle    L3
  mov    %rsi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rsi,%rdx,4),%rcx
L15:
  movzbl (%rax),%edx
  mov    %rdx,(%rdi)
  movzbl 0x1(%rax),%edx
  mov    %rdx,0x8(%rdi)
  movzbl 0x2(%rax),%edx
  mov    %rdx,0x10(%rdi)
  movzbl 0x3(%rax),%edx
  mov    %rdx,0x18(%rdi)
  add    $0x4,%rax
  add    $0x20,%rdi
  cmp    %rcx,%rax
  jne    L15
L3:
  ret
```

assistant: ```c
void md5_encode(uint8_t *output, const uint32_t *input, uint32_t len) {
  uint32_t i, j;
  for (i = 0, j = 0; i < len; i++, j += 4) {
    output[j] = (uint8_t)(input[i] & 0xff);
    output[j + 1] = (uint8_t)((input[i] >> 8) & 0xff);
    output[j + 2] = (uint8_t)((input[i] >> 16) & 0xff);
    output[j + 3] = (uint8_t)((input[i] >> 24) & 0xff);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<md5_encode>:
  endbr64
  test   %edx,%edx
  jle    L0
  lea    -0x1(%rdx),%eax
  lea    0x4(%rsi,%rax,4),%rcx
L14:
  mov    (%rsi),%eax
  add    $0x4,%rsi
  add    $0x20,%rdi
  movzbl %al,%edx
  mov    %rdx,-0x20(%rdi)
  movzbl %ah,%edx
  movslq %edx,%rdx
  mov    %rdx,-0x18(%rdi)
  mov    %eax,%edx
  shr    $0x18,%eax
  sar    $0x10,%edx
  mov    %rax,-0x8(%rdi)
  movzbl %dl,%edx
  mov    %rdx,-0x10(%rdi)
  cmp    %rsi,%rcx
  jne    L14
L0:
  ret
```

assistant: ```c
void md5_encode(uint8_t *output, const uint32_t *input, uint32_t len) {
  uint32_t i, j;
  for (i = 0, j = 0; i < len; i++, j += 4) {
    output[j] = (uint8_t)(input[i] & 0xff);
    output[j + 1] = (uint8_t)((input[i] >> 8) & 0xff);
    output[j + 2] = (uint8_t)((input[i] >> 16) & 0xff);
    output[j + 3] = (uint8_t)((input[i] >> 24) & 0xff);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<quick_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  ja     L38
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D5(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D5(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x0,%eax
  jmp    L60
  mov    $0x1,%eax
  jmp    L60
  mov    $0x9,%eax
  jmp    L60
  mov    $0x49,%eax
  jmp    L60
  mov    $0x249,%eax
  jmp    L60
  mov    $0x1249,%eax
  jmp    L60
  mov    $0x9249,%eax
  jmp    L60
  mov    $0x49249,%eax
  jmp    L60
  mov    $0x249249,%eax
  jmp    L60
  mov    $0x1249249,%eax
  jmp    L60
  mov    $0x9249249,%eax
  jmp    L60
L38:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  mov    $0x8,%edi
  call   <pow@plt>
  sub    $0x1,%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L60:
  leave
  ret
```

assistant: ```c
int quick_base(int depth) {
  switch (depth) {
  case 0:
    return 0;
  case 1:
    return 1;
  case 2:
    return 9;
  case 3:
    return 73;
  case 4:
    return 585;
  case 5:
    return 4681;
  case 6:
    return 37449;
  case 7:
    return 299593;
  case 8:
    return 2396745;
  case 9:
    return 19173961;
  case 10:
    return 153391689;
  default:
    return (pow(8, depth) - 1) / (8 - 1);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<quick_base>:
  endbr64
  cmp    $0xa,%edi
  ja     L5
  mov    %edi,%eax
  lea    D3(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x9,%eax
  ret
  mov    $0x49,%eax
  ret
  mov    $0x249,%eax
  ret
  mov    $0x1249,%eax
  ret
  mov    $0x9249,%eax
  ret
  mov    $0x49249,%eax
  ret
  mov    $0x249249,%eax
  ret
  mov    $0x1249249,%eax
  ret
  mov    $0x9249249,%eax
  ret
L5:
  sub    $0x8,%rsp
  mov    %edi,%esi
  mov    $0x8,%edi
  call   <pow@plt>
  sub    $0x1,%eax
  movslq %eax,%rdi
  imul   $0xffffffff92492493,%rdi,%rdi
  shr    $0x20,%rdi
  add    %eax,%edi
  sar    $0x2,%edi
  cltd
  mov    %edi,%eax
  sub    %edx,%eax
  add    $0x8,%rsp
  ret
  mov    %edi,%eax
  ret
```

assistant: ```c
int quick_base(int depth) {
  switch (depth) {
  case 0:
    return 0;
  case 1:
    return 1;
  case 2:
    return 9;
  case 3:
    return 73;
  case 4:
    return 585;
  case 5:
    return 4681;
  case 6:
    return 37449;
  case 7:
    return 299593;
  case 8:
    return 2396745;
  case 9:
    return 19173961;
  case 10:
    return 153391689;
  default:
    return (pow(8, depth) - 1) / (8 - 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quick_base>:
  endbr64
  cmp    $0xa,%edi
  ja     L1
  mov    %edi,%esi
  lea    D9(%rip),%rax
  mov    (%rax,%rsi,4),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  sub    $0x8,%rsp
  mov    %edi,%esi
  mov    $0x8,%edi
  call   <pow@plt>
  add    $0x8,%rsp
  lea    -0x1(%rax),%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x1f,%edx
  sar    $0x2,%eax
  sub    %edx,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
datatype: string

tool: 

assistant: ```c
int quick_base(int depth) {
  switch (depth) {
  case 0:
    return 0;
  case 1:
    return 1;
  case 2:
    return 9;
  case 3:
    return 73;
  case 4:
    return 585;
  case 5:
    return 4681;
  case 6:
    return 37449;
  case 7:
    return 299593;
  case 8:
    return 2396745;
  case 9:
    return 19173961;
  case 10:
    return 153391689;
  default:
    return (pow(8, depth) - 1) / (8 - 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quick_base>:
  endbr64
  cmp    $0xa,%edi
  ja     L0
  mov    %edi,%esi
  lea    D2(%rip),%rax
  mov    (%rax,%rsi,4),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  sub    $0x8,%rsp
  mov    %edi,%esi
  mov    $0x8,%edi
  call   <pow@plt>
  add    $0x8,%rsp
  lea    -0x1(%rax),%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x1f,%edx
  sar    $0x2,%eax
  sub    %edx,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type: string

tool: 

assistant: ```c
int quick_base(int depth) {
  switch (depth) {
  case 0:
    return 0;
  case 1:
    return 1;
  case 2:
    return 9;
  case 3:
    return 73;
  case 4:
    return 585;
  case 5:
    return 4681;
  case 6:
    return 37449;
  case 7:
    return 299593;
  case 8:
    return 2396745;
  case 9:
    return 19173961;
  case 10:
    return 153391689;
  default:
    return (pow(8, depth) - 1) / (8 - 1);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<aes_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0xd0(%rbp),%rdx
  lea    -0xc0(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <expand_key@plt>
  lea    -0xc0(%rbp),%rdx
  mov    -0xc8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <aes_encr@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void aes_encrypt(unsigned char *state, unsigned char *key) {
  unsigned char expanded_key[176];
  expand_key(expanded_key, key);
  aes_encr(state, expanded_key);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aes_encrypt>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xc8,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <expand_key@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <aes_encr@plt>
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0xc8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aes_encrypt(unsigned char *state, unsigned char *key) {
  unsigned char expanded_key[176];
  expand_key(expanded_key, key);
  aes_encr(state, expanded_key);
}
```
user: What is the c source code of the assembly code below:

```asm
<aes_encrypt>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0xc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <expand_key@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <aes_encr@plt>
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xc8,%rsp
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aes_encrypt(unsigned char *state, unsigned char *key) {
  unsigned char expanded_key[176];
  expand_key(expanded_key, key);
  aes_encr(state, expanded_key);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<aes_encrypt>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0xc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <expand_key@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <aes_encr@plt>
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xc8,%rsp
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aes_encrypt(unsigned char *state, unsigned char *key) {
  unsigned char expanded_key[176];
  expand_key(expanded_key, key);
  aes_encr(state, expanded_key);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Hash_lookup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x1505,-0x8(%rbp)
  jmp    L0
L2:
  mov    -0x8(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int Hash_lookup(const char *key) {
  int c;
  unsigned int hash = 5381;
  while ((c = *key++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Hash_lookup>:
  endbr64
  lea    0x1(%rdi),%rcx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L5
  mov    $0x1505,%eax
L13:
  mov    %eax,%esi
  shl    $0x5,%esi
  add    %esi,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L13
  ret
L5:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int Hash_lookup(const char *key) {
  int c;
  unsigned int hash = 5381;
  while ((c = *key++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Hash_lookup>:
  endbr64
  movsbl (%rdi),%edx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L16
  nopl   0x0(%rax)
L17:
  mov    %eax,%esi
  add    $0x1,%rcx
  shl    $0x5,%esi
  add    %esi,%eax
  add    %edx,%eax
  movsbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L17
  ret
  xchg   %ax,%ax
L16:
  ret
```

assistant: ```c
unsigned int Hash_lookup(const char *key) {
  int c;
  unsigned int hash = 5381;
  while ((c = *key++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Hash_lookup>:
  endbr64
  movsbl (%rdi),%edx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L11
  nopl   0x0(%rax)
L18:
  mov    %eax,%esi
  add    $0x1,%rcx
  shl    $0x5,%esi
  add    %esi,%eax
  add    %edx,%eax
  movsbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L18
  ret
  xchg   %ax,%ax
L11:
  ret
```

assistant: ```c
unsigned int Hash_lookup(const char *key) {
  int c;
  unsigned int hash = 5381;
  while ((c = *key++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dv2o_mult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  mov    -0x8(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm1
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dv2o_mult(double *output, double *a, double *b) {
  output[0] = a[0] * b[0];
  output[1] = a[1] * b[1];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<dv2o_mult>:
  endbr64
  movsd  (%rsi),%xmm0
  mulsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movsd  0x8(%rsi),%xmm0
  mulsd  0x8(%rdx),%xmm0
  movsd  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void dv2o_mult(double *output, double *a, double *b) {
  output[0] = a[0] * b[0];
  output[1] = a[1] * b[1];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dv2o_mult>:
  endbr64
  movsd  (%rsi),%xmm0
  mulsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movsd  0x8(%rsi),%xmm0
  mulsd  0x8(%rdx),%xmm0
  movsd  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void dv2o_mult(double *output, double *a, double *b) {
  output[0] = a[0] * b[0];
  output[1] = a[1] * b[1];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<dv2o_mult>:
  endbr64
  movsd  (%rsi),%xmm0
  mulsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movsd  0x8(%rsi),%xmm0
  mulsd  0x8(%rdx),%xmm0
  movsd  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void dv2o_mult(double *output, double *a, double *b) {
  output[0] = a[0] * b[0];
  output[1] = a[1] * b[1];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movb   $0x61,-0x4(%rbp)
  movb   $0x62,-0x3(%rbp)
  movb   $0x63,-0x2(%rbp)
  movb   $0x65,-0x1(%rbp)
  movsbl -0x1(%rbp),%esi
  movsbl -0x2(%rbp),%ecx
  movsbl -0x3(%rbp),%edx
  movsbl -0x4(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D6: Chars: %c %c %c %c

assistant: ```c
int main() {
  char a = 'a';
  char b = 'b';
  char c = 'c';
  char d = 'e';
  printf("Chars: %c %c %c %c", a, b, c, d);
  return 0;
}
```