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

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L10:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L10
  sub    $0x258,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rbp
  jmp    L57
L43:
  cmp    $0x31,%bl
  mov    $0x2,%eax
  cmovne %eax,%esi
L3:
  mov    %rbp,%rdi
  call   <printf@plt>
L57:
  mov    %rsp,%rdi
  call   <gets@plt>
  movzbl (%rsp),%ebx
  cmp    $0x45,%bl
  je     L42
  mov    %rsp,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  cmp    $0x1,%eax
  je     L43
  cmp    $0x9,%eax
  setg   %sil
  movzbl %sil,%esi
  add    $0x3,%esi
  jmp    L3
L42:
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0xf4258,%rsp
  pop    %rbx
  pop    %rbp
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D4
data_type: string

tool: D4: %d


assistant: ```c
int main(void) {
  char number[1000000 + 1];
  int len, result;
  while (1) {
    gets(number);
    if (number[0] == 'E')
      break;
    len = (int)strlen(number);
    if (len == 1) {
      if (number[0] == '1')
        result = 1;
      else
        result = 2;
    } else {
      if (len < 10)
        result = 3;
      else
        result = 4;
    }
    printf("%d\n", result);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L9:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L9
  sub    $0x250,%rsp
  lea    D5(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L39:
  xor    %esi,%esi
  cmp    $0x9,%rax
  setg   %sil
  add    $0x3,%esi
L11:
  mov    %r12,%rdi
  call   <printf@plt>
L14:
  mov    %rbp,%rdi
  call   <gets@plt>
  movzbl (%rsp),%ebx
  cmp    $0x45,%bl
  je     L51
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x1,%rax
  jne    L39
  xor    %esi,%esi
  cmp    $0x31,%bl
  setne  %sil
  add    $0x1,%esi
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L51:
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0xf4250,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string

tool: D5: %d


assistant: ```c
int main(void) {
  char number[1000000 + 1];
  int len, result;
  while (1) {
    gets(number);
    if (number[0] == 'E')
      break;
    len = (int)strlen(number);
    if (len == 1) {
      if (number[0] == '1')
        result = 1;
      else
        result = 2;
    } else {
      if (len < 10)
        result = 3;
      else
        result = 4;
    }
    printf("%d\n", result);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L6:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L6
  sub    $0x250,%rsp
  lea    D6(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %esi,%esi
  cmp    $0x9,%rax
  setg   %sil
  add    $0x3,%esi
L12:
  mov    %r12,%rdi
  call   <printf@plt>
L1:
  mov    %rbp,%rdi
  call   <gets@plt>
  movzbl (%rsp),%ebx
  cmp    $0x45,%bl
  je     L40
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x1,%rax
  jne    L8
  xor    %esi,%esi
  cmp    $0x31,%bl
  setne  %sil
  add    $0x1,%esi
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L40:
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0xf4250,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: %d


assistant: ```c
int main(void) {
  char number[1000000 + 1];
  int len, result;
  while (1) {
    gets(number);
    if (number[0] == 'E')
      break;
    len = (int)strlen(number);
    if (len == 1) {
      if (number[0] == '1')
        result = 1;
      else
        result = 2;
    } else {
      if (len < 10)
        result = 3;
      else
        result = 4;
    }
    printf("%d\n", result);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<padcon_get_standard_pole>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    %r8,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  lea    -0x20(%rbp),%rcx
  lea    -0x3c(%rbp),%rdx
  lea    -0x2c(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x367,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x30(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x48(%rbp),%rax
  movsd  %xmm0,(%rax)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x50(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x30(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int padcon_get_standard_pole(int leadidx, int poleidx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[3];
  double nbr[2];
  inpars[0] = 0;
  inpars[1] = leadidx;
  inpars[2] = poleidx;
  fail = _ada_use_c2phc4c(871, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: What is the c source code of the assembly code below:

```asm
<padcon_get_standard_pole>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  mov    %rcx,%rbp
  mov    %r8,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  mov    %edi,0x18(%rsp)
  mov    %esi,0x1c(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  lea    0x14(%rsp),%rsi
  mov    $0x0,%r8d
  mov    $0x367,%edi
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int padcon_get_standard_pole(int leadidx, int poleidx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[3];
  double nbr[2];
  inpars[0] = 0;
  inpars[1] = leadidx;
  inpars[2] = poleidx;
  fail = _ada_use_c2phc4c(871, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<padcon_get_standard_pole>:
  endbr64
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  xor    %r8d,%r8d
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %edi,0x18(%rsp)
  mov    $0x367,%edi
  mov    %esi,0x1c(%rsp)
  lea    0x14(%rsp),%rsi
  movl   $0x0,0x14(%rsp)
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int padcon_get_standard_pole(int leadidx, int poleidx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[3];
  double nbr[2];
  inpars[0] = 0;
  inpars[1] = leadidx;
  inpars[2] = poleidx;
  fail = _ada_use_c2phc4c(871, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<padcon_get_standard_pole>:
  endbr64
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  xor    %r8d,%r8d
  sub    $0x48,%rsp
  mov    %edx,0xc(%rsp)
  lea    0x20(%rsp),%rcx
  lea    0xc(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %edi,0x18(%rsp)
  mov    $0x367,%edi
  mov    %esi,0x1c(%rsp)
  lea    0x14(%rsp),%rsi
  movl   $0x0,0x14(%rsp)
  call   <_ada_use_c2phc4c@plt>
  movsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x0(%rbp)
  movsd  0x28(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int padcon_get_standard_pole(int leadidx, int poleidx, int verbose, double *cre, double *cim) {
  int fail;
  int inpars[3];
  double nbr[2];
  inpars[0] = 0;
  inpars[1] = leadidx;
  inpars[2] = poleidx;
  fail = _ada_use_c2phc4c(871, inpars, &verbose, nbr, 0);
  *cre = nbr[0];
  *cim = nbr[1];
  return fail;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<generateHashValue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L46
L28:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,-0xd(%rbp)
  cmpb   $0x2e,-0xd(%rbp)
  je     L44
  cmpb   $0x5c,-0xd(%rbp)
  jne    L12
  movb   $0x2f,-0xd(%rbp)
L12:
  movsbq -0xd(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  add    $0x77,%eax
  cltq
  imul   %rdx,%rax
  add    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L46:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  jmp    L2
L44:
  nop
L2:
  andq   $0x3ff,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long generateHashValue(const char *fname) {
  int i;
  long hash;
  char letter;
  hash = 0;
  i = 0;
  while (fname[i] != '\0') {
    letter = tolower(fname[i]);
    if (letter == '.') {
      break;
    }
    if (letter == '\\') {
      letter = '/';
    }
    hash += (long)(letter) * (i + 119);
    i++;
  }
  hash &= (1024 - 1);
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<generateHashValue>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L14
  mov    $0x77,%ebx
  mov    $0x0,%ebp
  mov    $0x2f,%r13d
L16:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %eax,%edx
  cmp    $0x2e,%al
  je     L35
  cmp    $0x5c,%al
  cmove  %r13d,%edx
  movsbq %dl,%rdx
  imul   %rbx,%rdx
  add    %rdx,%rbp
  add    $0x1,%rbx
  movzbl -0x77(%r12,%rbx,1),%edi
  test   %dil,%dil
  jne    L16
L35:
  mov    %rbp,%rax
  and    $0x3ff,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  mov    $0x0,%ebp
  jmp    L35
```

assistant: ```c
long generateHashValue(const char *fname) {
  int i;
  long hash;
  char letter;
  hash = 0;
  i = 0;
  while (fname[i] != '\0') {
    letter = tolower(fname[i]);
    if (letter == '.') {
      break;
    }
    if (letter == '\\') {
      letter = '/';
    }
    hash += (long)(letter) * (i + 119);
    i++;
  }
  hash &= (1024 - 1);
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<generateHashValue>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L31
  mov    $0x77,%r13d
  xor    %ebx,%ebx
  mov    $0x2f,%r12d
  jmp    L25
  nopl   0x0(%rax)
L48:
  cmp    $0x5c,%al
  cmove  %r12,%rdx
  imul   %r13,%rdx
  add    $0x1,%r13
  movsbl -0x77(%rbp,%r13,1),%edi
  add    %rdx,%rbx
  test   %dil,%dil
  je     L42
L25:
  call   <tolower@plt>
  movsbq %al,%rdx
  cmp    $0x2e,%al
  jne    L48
L42:
  mov    %rbx,%rax
  and    $0x3ff,%eax
L46:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  xor    %eax,%eax
  jmp    L46
```

assistant: ```c
long generateHashValue(const char *fname) {
  int i;
  long hash;
  char letter;
  hash = 0;
  i = 0;
  while (fname[i] != '\0') {
    letter = tolower(fname[i]);
    if (letter == '.') {
      break;
    }
    if (letter == '\\') {
      letter = '/';
    }
    hash += (long)(letter) * (i + 119);
    i++;
  }
  hash &= (1024 - 1);
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<generateHashValue>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L15
  mov    $0x77,%r12d
  xor    %ebx,%ebx
  jmp    L35
  nopl   (%rax)
L4:
  movsbq %al,%rax
  imul   %r12,%rax
L45:
  add    $0x1,%r12
  add    %rax,%rbx
  movsbl -0x77(%rbp,%r12,1),%edi
  test   %dil,%dil
  je     L29
L35:
  call   <tolower@plt>
  cmp    $0x2e,%al
  je     L29
  cmp    $0x5c,%al
  jne    L4
  lea    (%r12,%r12,2),%rax
  shl    $0x4,%rax
  sub    %r12,%rax
  jmp    L45
  nopl   0x0(%rax)
L29:
  mov    %rbx,%rax
  and    $0x3ff,%eax
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  xor    %eax,%eax
  jmp    L5
```

assistant: ```c
long generateHashValue(const char *fname) {
  int i;
  long hash;
  char letter;
  hash = 0;
  i = 0;
  while (fname[i] != '\0') {
    letter = tolower(fname[i]);
    if (letter == '.') {
      break;
    }
    if (letter == '\\') {
      letter = '/';
    }
    hash += (long)(letter) * (i + 119);
    i++;
  }
  hash &= (1024 - 1);
  return hash;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<diffTime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssq -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssq -0x20(%rbp),%xmm1
  subss  %xmm1,%xmm0
  movss  D9(%rip),%xmm2
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  movss  D5(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_infered: byte[4]
name: extract_symbolic_data
arguments: tag: D5
type_infered: f32

tool: {"D9": [0x00,0x24,0x74,0x49]}

tool: D5: 1000

assistant: ```c
float diffTime(clock_t t2, clock_t t1) {
  float diff = (((float)t2 - (float)t1) / 1000000.0F) * 1000;
  return diff;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<diffTime>:
  endbr64
  pxor   %xmm0,%xmm0
  cvtsi2ss %rdi,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %rsi,%xmm1
  subss  %xmm1,%xmm0
  divss  D17(%rip),%xmm0
  mulss  D13(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D17
type_guessed: byte[4]
name: retrieve_data
arguments: data_label: D13
type_guessed: f32

tool: {"D17": [0x00,0x24,0x74,0x49]}

tool: D13: 1000

assistant: ```c
float diffTime(clock_t t2, clock_t t1) {
  float diff = (((float)t2 - (float)t1) / 1000000.0F) * 1000;
  return diff;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<diffTime>:
  endbr64
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %rdi,%xmm0
  cvtsi2ss %rsi,%xmm1
  subss  %xmm1,%xmm0
  divss  D7(%rip),%xmm0
  mulss  D2(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
data_type: byte[4]
name: extract_symbolic_data
arguments: data_label: D2
data_type: f32

tool: {"D7": [0x00,0x24,0x74,0x49]}

tool: D2: 1000

assistant: ```c
float diffTime(clock_t t2, clock_t t1) {
  float diff = (((float)t2 - (float)t1) / 1000000.0F) * 1000;
  return diff;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<diffTime>:
  endbr64
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %rdi,%xmm0
  cvtsi2ss %rsi,%xmm1
  subss  %xmm1,%xmm0
  divss  D5(%rip),%xmm0
  mulss  D12(%rip),%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_infered: byte[4]
name: disassemble_data
arguments: data_tag: D12
type_infered: f32

tool: {"D5": [0x00,0x24,0x74,0x49]}

tool: D12: 1000

assistant: ```c
float diffTime(clock_t t2, clock_t t1) {
  float diff = (((float)t2 - (float)t1) / 1000000.0F) * 1000;
  return diff;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vector_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <clear@plt>
  movl   $0x0,-0x18(%rbp)
  jmp    L26
L25:
  movl   $0x0,-0x14(%rbp)
  jmp    L19
L22:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <place@plt>
  mov    %eax,(%rbx)
  addl   $0x1,-0x14(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L22
  addl   $0x1,-0x18(%rbp)
L26:
  cmpl   $0x0,-0x18(%rbp)
  jle    L25
  mov    -0x2c(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <check@plt>
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int vector_1(int *ary, int size) {
  clear(ary, size);
#pragma acc parallel num_workers(32) vector_length(32) copy(ary [0:size]) firstprivate(size)
  {
#pragma acc loop gang
    for (int jx = 0; jx < 1; jx++)
#pragma acc loop auto
      for (int ix = 0; ix < size; ix++)
        ary[ix] = place();
  }
  return check(ary, size, 0, 1, 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vector_1>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%r12d
  call   <clear@plt>
  test   %r12d,%r12d
  jle    L4
  mov    %r13,%rbx
  lea    -0x1(%r12),%eax
  lea    0x4(%r13,%rax,4),%rbp
L13:
  mov    $0x0,%eax
  call   <place@plt>
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L13
L4:
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x0,%edx
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <check@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int vector_1(int *ary, int size) {
  clear(ary, size);
#pragma acc parallel num_workers(32) vector_length(32) copy(ary [0:size]) firstprivate(size)
  {
#pragma acc loop gang
    for (int jx = 0; jx < 1; jx++)
#pragma acc loop auto
      for (int ix = 0; ix < size; ix++)
        ary[ix] = place();
  }
  return check(ary, size, 0, 1, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<vector_1>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <clear@plt>
  test   %r12d,%r12d
  jle    L17
  lea    -0x1(%r12),%eax
  mov    %r13,%rbx
  lea    0x4(%r13,%rax,4),%rbp
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <place@plt>
  mov    %eax,-0x4(%rbx)
  cmp    %rbp,%rbx
  jne    L3
L17:
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    $0x1,%r8d
  pop    %rbx
  mov    $0x1,%ecx
  pop    %rbp
  xor    %edx,%edx
  pop    %r12
  pop    %r13
  jmp    1080 <check@plt>
```

assistant: ```c
int vector_1(int *ary, int size) {
  clear(ary, size);
#pragma acc parallel num_workers(32) vector_length(32) copy(ary [0:size]) firstprivate(size)
  {
#pragma acc loop gang
    for (int jx = 0; jx < 1; jx++)
#pragma acc loop auto
      for (int ix = 0; ix < size; ix++)
        ary[ix] = place();
  }
  return check(ary, size, 0, 1, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<vector_1>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <clear@plt>
  test   %r12d,%r12d
  jle    L8
  lea    -0x1(%r12),%eax
  mov    %r13,%rbx
  lea    0x4(%r13,%rax,4),%rbp
  nopl   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <place@plt>
  mov    %eax,-0x4(%rbx)
  cmp    %rbp,%rbx
  jne    L13
L8:
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %r13,%rdi
  mov    $0x1,%r8d
  pop    %rbx
  mov    $0x1,%ecx
  pop    %rbp
  xor    %edx,%edx
  pop    %r12
  pop    %r13
  jmp    1080 <check@plt>
```

assistant: ```c
int vector_1(int *ary, int size) {
  clear(ary, size);
#pragma acc parallel num_workers(32) vector_length(32) copy(ary [0:size]) firstprivate(size)
  {
#pragma acc loop gang
    for (int jx = 0; jx < 1; jx++)
#pragma acc loop auto
      for (int ix = 0; ix < size; ix++)
        ary[ix] = place();
  }
  return check(ary, size, 0, 1, 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<uniphier_board_reset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void uniphier_board_reset(void) {
  do {
  } while (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<uniphier_board_reset>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_reset(void) {
  do {
  } while (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<uniphier_board_reset>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_reset(void) {
  do {
  } while (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<uniphier_board_reset>:
  endbr64
  ret
```

assistant: ```c
void uniphier_board_reset(void) {
  do {
  } while (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L7
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *fill(int *a, int size) {
  a = malloc(size * sizeof(int));
  int i;
  for (i = 0; i < size; i++) {
    a[i] = i + 1;
  }
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fill>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L5
  lea    0x1(%rbx),%ecx
  mov    $0x1,%edx
L11:
  mov    %edx,-0x4(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L11
L5:
  pop    %rbx
  ret
```

assistant: ```c
int *fill(int *a, int size) {
  a = malloc(size * sizeof(int));
  int i;
  for (i = 0; i < size; i++) {
    a[i] = i + 1;
  }
  return a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fill>:
  endbr64
  push   %rbx
  lea    0x0(,%rsi,4),%edi
  mov    %esi,%ebx
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L0
  lea    0x1(%rbx),%ecx
  mov    $0x1,%edx
  nop
L15:
  mov    %edx,-0x4(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L15
L0:
  pop    %rbx
  ret
```

assistant: ```c
int *fill(int *a, int size) {
  a = malloc(size * sizeof(int));
  int i;
  for (i = 0; i < size; i++) {
    a[i] = i + 1;
  }
  return a;
}
```