ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<max_absf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movss  D9(%rip),%xmm0
  movss  %xmm0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L24
L4:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movd   %xmm0,%eax
  movd   %eax,%xmm1
  comiss -0x4(%rbp),%xmm1
  jbe    L14
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x4(%rbp)
L14:
  addl   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L4
  pxor   %xmm0,%xmm0
  cvtss2sd -0x4(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
datatype: byte[4]

tool: {"D9": [0xca,0xf2,0x49,0xf1]}

assistant: ```c
double max_absf(int n, float *vector) {
  int i;
  float max_val = -1e30f;
  for (i = 0; i < n; i++)
    if (fabs(vector[i]) > max_val)
      max_val = (float)(fabs(vector[i]));
  return max_val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<max_absf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %edi,%edi
  jle    L8
  mov    %rsi,%rbx
  lea    -0x1(%rdi),%eax
  lea    0x4(%rsi,%rax,4),%rbp
  movss  D14(%rip),%xmm2
  movss  %xmm2,0xc(%rsp)
  jmp    L45
L23:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  je     L2
L45:
  movss  (%rbx),%xmm0
  call   <fabs@plt>
  comiss 0xc(%rsp),%xmm0
  jbe    L23
  movss  (%rbx),%xmm0
  call   <fabs@plt>
  movss  %xmm0,0xc(%rsp)
  jmp    L23
L8:
  movss  D14(%rip),%xmm3
  movss  %xmm3,0xc(%rsp)
L2:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type_guessed: byte[4]

tool: {"D14": [0xca,0xf2,0x49,0xf1]}

assistant: ```c
double max_absf(int n, float *vector) {
  int i;
  float max_val = -1e30f;
  for (i = 0; i < n; i++)
    if (fabs(vector[i]) > max_val)
      max_val = (float)(fabs(vector[i]));
  return max_val;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<max_absf>:
  endbr64
  test   %edi,%edi
  jle    L9
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x4(%rsi,%rax,4),%rbp
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movss  D0(%rip),%xmm2
  movss  %xmm2,0xc(%rsp)
  jmp    L44
  nopl   0x0(%rax)
L15:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  je     L41
L44:
  movss  (%rbx),%xmm0
  call   <fabs@plt>
  comiss 0xc(%rsp),%xmm0
  jbe    L15
  movss  (%rbx),%xmm0
  add    $0x4,%rbx
  call   <fabs@plt>
  movss  %xmm0,0xc(%rsp)
  cmp    %rbp,%rbx
  jne    L44
L41:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  movsd  D12(%rip),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
data_type: byte[4]
name: unpack_data
arguments: tag: D12
data_type: byte[8]

tool: {"D0": [0xca,0xf2,0x49,0xf1]}

tool: {"D12": [0x00,0x00,0x00,0x40,0x59,0x3e,0x29,0xc6]}

assistant: ```c
double max_absf(int n, float *vector) {
  int i;
  float max_val = -1e30f;
  for (i = 0; i < n; i++)
    if (fabs(vector[i]) > max_val)
      max_val = (float)(fabs(vector[i]));
  return max_val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<max_absf>:
  endbr64
  test   %edi,%edi
  jle    L8
  push   %rbp
  lea    -0x1(%rdi),%eax
  push   %rbx
  lea    0x4(%rsi,%rax,4),%rbp
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  movss  D17(%rip),%xmm2
  movss  %xmm2,0xc(%rsp)
  jmp    L14
  nopl   0x0(%rax)
L11:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  je     L47
L14:
  movss  (%rbx),%xmm0
  call   <fabs@plt>
  comiss 0xc(%rsp),%xmm0
  jbe    L11
  movss  (%rbx),%xmm0
  add    $0x4,%rbx
  call   <fabs@plt>
  movss  %xmm0,0xc(%rsp)
  cmp    %rbp,%rbx
  jne    L14
L47:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  movsd  D10(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_label: D17
type: byte[4]
name: read_data
arguments: data_label: D10
type: byte[8]

tool: {"D17": [0xca,0xf2,0x49,0xf1]}

tool: {"D10": [0x00,0x00,0x00,0x40,0x59,0x3e,0x29,0xc6]}

assistant: ```c
double max_absf(int n, float *vector) {
  int i;
  float max_val = -1e30f;
  for (i = 0; i < n; i++)
    if (fabs(vector[i]) > max_val)
      max_val = (float)(fabs(vector[i]));
  return max_val;
}
```
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
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <debug_captured@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  debug_captured();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <debug_captured@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  debug_captured();
  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
  xor    %eax,%eax
  call   <debug_captured@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  debug_captured();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <debug_captured@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <add_4@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <print_four@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
type: string

tool: D2: %i


assistant: ```c
int main() {
  int n = get_number();
  int y = add_4(n);
  printf("%i\n", y);
  print_four();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_number@plt>
  mov    %eax,%edi
  call   <add_4@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <print_four@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type_infered: string

tool: D3: %i


assistant: ```c
int main() {
  int n = get_number();
  int y = add_4(n);
  printf("%i\n", y);
  print_four();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_number@plt>
  mov    %eax,%edi
  call   <add_4@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <print_four@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_guessed: string

tool: D4: %i


assistant: ```c
int main() {
  int n = get_number();
  int y = add_4(n);
  printf("%i\n", y);
  print_four();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_number@plt>
  mov    %eax,%edi
  call   <add_4@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <print_four@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string

tool: D8: %i


assistant: ```c
int main() {
  int n = get_number();
  int y = add_4(n);
  printf("%i\n", y);
  print_four();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <dummy@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  extern void dummy();
  dummy();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <dummy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  extern void dummy();
  dummy();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <dummy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  extern void dummy();
  dummy();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <dummy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  extern void dummy();
  dummy();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<iStrNumBaseConvert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %esi,-0x12c(%rbp)
  mov    %rdx,-0x138(%rbp)
  mov    %ecx,-0x130(%rbp)
  mov    %r8d,-0x13c(%rbp)
  mov    %r9d,%eax
  mov    %al,-0x140(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x12c(%rbp),%edx
  lea    -0x110(%rbp),%rsi
  mov    -0x128(%rbp),%rax
  mov    $0x0,%r8d
  mov    %edx,%ecx
  mov    $0x80,%edx
  mov    %rax,%rdi
  call   <iEncode@plt>
  mov    %eax,-0x114(%rbp)
  cmpl   $0x0,-0x114(%rbp)
  je     L2
  mov    -0x114(%rbp),%eax
  jmp    L8
L2:
  mov    -0x13c(%rbp),%ecx
  lea    -0x90(%rbp),%rdx
  mov    -0x12c(%rbp),%esi
  lea    -0x110(%rbp),%rax
  mov    $0x80,%r9d
  mov    %ecx,%r8d
  mov    %rdx,%rcx
  mov    $0x80,%edx
  mov    %rax,%rdi
  call   <iBaseConvert@plt>
  mov    %eax,-0x114(%rbp)
  cmpl   $0x0,-0x114(%rbp)
  je     L43
  mov    -0x114(%rbp),%eax
  jmp    L8
L43:
  movzbl -0x140(%rbp),%edi
  mov    -0x13c(%rbp),%esi
  mov    -0x130(%rbp),%ecx
  mov    -0x138(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    $0x80,%esi
  mov    %rax,%rdi
  call   <iDecode@plt>
  mov    %eax,-0x114(%rbp)
  cmpl   $0x0,-0x114(%rbp)
  je     L30
  mov    -0x114(%rbp),%eax
  jmp    L8
L30:
  mov    $0x0,%eax
L8:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int iStrNumBaseConvert(unsigned char *pszSource, unsigned int uiSourceBase, unsigned char *pszTarget, unsigned int uiTargetSize, unsigned int uiTargetBase, unsigned char ucFillChar) {
  int iRet;
  unsigned char sS[128], sT[128];
  iRet = iEncode(pszSource, sS, 128, uiSourceBase, NULL);
  if (iRet)
    return (iRet);
  iRet = iBaseConvert(sS, uiSourceBase, 128, sT, uiTargetBase, 128);
  if (iRet)
    return (iRet);
  iRet = iDecode(sT, 128, pszTarget, uiTargetSize, uiTargetBase, ucFillChar);
  if (iRet)
    return (iRet);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<iStrNumBaseConvert>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %esi,%ebx
  mov    %rdx,%r12
  mov    %ecx,%r13d
  mov    %r8d,%ebp
  mov    %r9d,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x0,%r8d
  mov    %ebx,%ecx
  mov    $0x80,%edx
  call   <iEncode@plt>
  test   %eax,%eax
  je     L0
L24:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  lea    0x80(%rsp),%rcx
  mov    %rsp,%rdi
  mov    $0x80,%r9d
  mov    %ebp,%r8d
  mov    $0x80,%edx
  mov    %ebx,%esi
  call   <iBaseConvert@plt>
  test   %eax,%eax
  jne    L24
  lea    0x80(%rsp),%rdi
  movzbl %r14b,%r9d
  mov    %ebp,%r8d
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    $0x80,%esi
  call   <iDecode@plt>
  jmp    L24
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iStrNumBaseConvert(unsigned char *pszSource, unsigned int uiSourceBase, unsigned char *pszTarget, unsigned int uiTargetSize, unsigned int uiTargetBase, unsigned char ucFillChar) {
  int iRet;
  unsigned char sS[128], sT[128];
  iRet = iEncode(pszSource, sS, 128, uiSourceBase, NULL);
  if (iRet)
    return (iRet);
  iRet = iBaseConvert(sS, uiSourceBase, 128, sT, uiTargetBase, 128);
  if (iRet)
    return (iRet);
  iRet = iDecode(sT, 128, pszTarget, uiTargetSize, uiTargetBase, ucFillChar);
  if (iRet)
    return (iRet);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<iStrNumBaseConvert>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %ecx,%r13d
  mov    %esi,%ecx
  push   %r12
  mov    %rdx,%r12
  mov    $0x80,%edx
  push   %rbp
  mov    %r9d,%ebp
  push   %rbx
  mov    %r8d,%ebx
  xor    %r8d,%r8d
  sub    $0x128,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  mov    %r15,%rsi
  call   <iEncode@plt>
  test   %eax,%eax
  je     L1
L7:
  mov    0x118(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x128,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    $0x80,%r9d
  mov    %ebx,%r8d
  mov    %r14d,%esi
  mov    %r15,%rdi
  lea    0x90(%rsp),%r10
  mov    $0x80,%edx
  mov    %r10,%rcx
  mov    %r10,0x8(%rsp)
  call   <iBaseConvert@plt>
  test   %eax,%eax
  jne    L7
  mov    0x8(%rsp),%rdi
  movzbl %bpl,%r9d
  mov    %ebx,%r8d
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    $0x80,%esi
  call   <iDecode@plt>
  jmp    L7
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iStrNumBaseConvert(unsigned char *pszSource, unsigned int uiSourceBase, unsigned char *pszTarget, unsigned int uiTargetSize, unsigned int uiTargetBase, unsigned char ucFillChar) {
  int iRet;
  unsigned char sS[128], sT[128];
  iRet = iEncode(pszSource, sS, 128, uiSourceBase, NULL);
  if (iRet)
    return (iRet);
  iRet = iBaseConvert(sS, uiSourceBase, 128, sT, uiTargetBase, 128);
  if (iRet)
    return (iRet);
  iRet = iDecode(sT, 128, pszTarget, uiTargetSize, uiTargetBase, ucFillChar);
  if (iRet)
    return (iRet);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<iStrNumBaseConvert>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %ecx,%r13d
  mov    %esi,%ecx
  push   %r12
  mov    %rdx,%r12
  mov    $0x80,%edx
  push   %rbp
  mov    %r9d,%ebp
  push   %rbx
  mov    %r8d,%ebx
  xor    %r8d,%r8d
  sub    $0x128,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r15
  mov    %r15,%rsi
  call   <iEncode@plt>
  test   %eax,%eax
  je     L25
L19:
  mov    0x118(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x128,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  mov    $0x80,%r9d
  mov    %ebx,%r8d
  mov    %r14d,%esi
  mov    %r15,%rdi
  lea    0x90(%rsp),%r10
  mov    $0x80,%edx
  mov    %r10,%rcx
  mov    %r10,0x8(%rsp)
  call   <iBaseConvert@plt>
  test   %eax,%eax
  jne    L19
  mov    0x8(%rsp),%rdi
  movzbl %bpl,%r9d
  mov    %ebx,%r8d
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    $0x80,%esi
  call   <iDecode@plt>
  jmp    L19
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int iStrNumBaseConvert(unsigned char *pszSource, unsigned int uiSourceBase, unsigned char *pszTarget, unsigned int uiTargetSize, unsigned int uiTargetBase, unsigned char ucFillChar) {
  int iRet;
  unsigned char sS[128], sT[128];
  iRet = iEncode(pszSource, sS, 128, uiSourceBase, NULL);
  if (iRet)
    return (iRet);
  iRet = iBaseConvert(sS, uiSourceBase, 128, sT, uiTargetBase, 128);
  if (iRet)
    return (iRet);
  iRet = iDecode(sT, 128, pszTarget, uiTargetSize, uiTargetBase, ucFillChar);
  if (iRet)
    return (iRet);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<pr_random>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D27(%rip),%eax
  imul   $0x556,%eax,%eax
  lea    0x24d69(%rax),%edx
  movslq %edx,%rax
  imul   $0x5dfc9957,%rax,%rax
  shr    $0x20,%rax
  sar    $0x12,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0xae529,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,D16(%rip)
  mov    D16(%rip),%eax
  pop    %rbp
  ret
```

assistant: ```c
int pr_random(void) {
  static int state = 0;
  return (state = (1366 * state + 150889) % 714025);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<pr_random>:
  endbr64
  imul   $0x556,0x2f1d(%rip),%edx        # 4024 <state.0>
  add    $0x24d69,%edx
  movslq %edx,%rax
  imul   $0x5dfc9957,%rax,%rax
  sar    $0x32,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0xae529,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,D5(%rip)
  ret
```

assistant: ```c
int pr_random(void) {
  static int state = 0;
  return (state = (1366 * state + 150889) % 714025);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pr_random>:
  endbr64
  imul   $0x556,0x2f16(%rip),%edx        # 4024 <state.0>
  add    $0x24d69,%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x5dfc9957,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x32,%rax
  sub    %ecx,%eax
  imul   $0xae529,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,D4(%rip)
  ret
```

assistant: ```c
int pr_random(void) {
  static int state = 0;
  return (state = (1366 * state + 150889) % 714025);
}
```
user: What is the c source code of the assembly code below?

```asm
<pr_random>:
  endbr64
  imul   $0x556,0x2f16(%rip),%edx        # 4024 <state.0>
  add    $0x24d69,%edx
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x5dfc9957,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x32,%rax
  sub    %ecx,%eax
  imul   $0xae529,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,D5(%rip)
  ret
```

assistant: ```c
int pr_random(void) {
  static int state = 0;
  return (state = (1366 * state + 150889) % 714025);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<count_digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
L7:
  addl   $0x1,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movabs $0xcccccccccccccccd,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int count_digit(unsigned long long n) {
  int i = 0;
  do {
    ++i;
    n /= 10;
  } while (n != 0);
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<count_digit>:
  endbr64
  mov    %rdi,%rdx
  mov    $0x0,%ecx
  movabs $0xcccccccccccccccd,%rdi
L6:
  add    $0x1,%ecx
  mov    %rdx,%rsi
  mov    %rdx,%rax
  mul    %rdi
  shr    $0x3,%rdx
  cmp    $0x9,%rsi
  ja     L6
  mov    %ecx,%eax
  ret
```

assistant: ```c
int count_digit(unsigned long long n) {
  int i = 0;
  do {
    ++i;
    n /= 10;
  } while (n != 0);
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<count_digit>:
  endbr64
  movabs $0xcccccccccccccccd,%rsi
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L2:
  mov    %rdx,%rax
  mov    %rdx,%rcx
  add    $0x1,%r8d
  mul    %rsi
  shr    $0x3,%rdx
  cmp    $0x9,%rcx
  ja     L2
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_digit(unsigned long long n) {
  int i = 0;
  do {
    ++i;
    n /= 10;
  } while (n != 0);
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<count_digit>:
  endbr64
  movabs $0xcccccccccccccccd,%rsi
  mov    %rdi,%rdx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L1:
  mov    %rdx,%rax
  mov    %rdx,%rcx
  add    $0x1,%r8d
  mul    %rsi
  shr    $0x3,%rdx
  cmp    $0x9,%rcx
  ja     L1
  mov    %r8d,%eax
  ret
```

assistant: ```c
int count_digit(unsigned long long n) {
  int i = 0;
  do {
    ++i;
    n /= 10;
  } while (n != 0);
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<checaEmpate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <checaVitoria@plt>
  test   %rax,%rax
  jne    L30
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x6,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x7,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    $0x1,%eax
  jmp    L7
L30:
  mov    $0x0,%eax
L7:
  leave
  ret
```

assistant: ```c
int checaEmpate(char grade[]) {
  if ((checaVitoria(grade) == 0) && (grade[0] != ' ') && (grade[1] != ' ') && (grade[2] != ' ') && (grade[3] != ' ') && (grade[4] != ' ') && (grade[5] != ' ') && (grade[6] != ' ') && (grade[7] != ' ') && (grade[8] != ' ')) {
    return 1;
  } else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<checaEmpate>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <checaVitoria@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L71
  cmpb   $0x20,(%rbx)
  je     L71
  cmpb   $0x20,0x1(%rbx)
  je     L71
  cmpb   $0x20,0x2(%rbx)
  je     L71
  cmpb   $0x20,0x3(%rbx)
  je     L71
  cmpb   $0x20,0x4(%rbx)
  je     L71
  cmpb   $0x20,0x5(%rbx)
  je     L71
  cmpb   $0x20,0x6(%rbx)
  je     L71
  cmpb   $0x20,0x7(%rbx)
  je     L71
  cmpb   $0x20,0x8(%rbx)
  setne  %al
  movzbl %al,%eax
L71:
  pop    %rbx
  ret
```

assistant: ```c
int checaEmpate(char grade[]) {
  if ((checaVitoria(grade) == 0) && (grade[0] != ' ') && (grade[1] != ' ') && (grade[2] != ' ') && (grade[3] != ' ') && (grade[4] != ' ') && (grade[5] != ' ') && (grade[6] != ' ') && (grade[7] != ' ') && (grade[8] != ' ')) {
    return 1;
  } else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<checaEmpate>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <checaVitoria@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L83
  cmpb   $0x20,(%rbx)
  je     L83
  cmpb   $0x20,0x1(%rbx)
  je     L83
  cmpb   $0x20,0x2(%rbx)
  je     L83
  cmpb   $0x20,0x3(%rbx)
  je     L83
  cmpb   $0x20,0x4(%rbx)
  je     L83
  cmpb   $0x20,0x5(%rbx)
  je     L83
  cmpb   $0x20,0x6(%rbx)
  je     L83
  cmpb   $0x20,0x7(%rbx)
  je     L83
  xor    %eax,%eax
  cmpb   $0x20,0x8(%rbx)
  setne  %al
L83:
  pop    %rbx
  ret
```

assistant: ```c
int checaEmpate(char grade[]) {
  if ((checaVitoria(grade) == 0) && (grade[0] != ' ') && (grade[1] != ' ') && (grade[2] != ' ') && (grade[3] != ' ') && (grade[4] != ' ') && (grade[5] != ' ') && (grade[6] != ' ') && (grade[7] != ' ') && (grade[8] != ' ')) {
    return 1;
  } else
    return 0;
}
```