ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<RevertJulian>:
  endbr64
  mov    %rdi,%rax
  movsd  D25(%rip),%xmm2
  mov    %rdx,%rdi
  mov    %rcx,%r8
  cmp    $0x231518,%rax
  jle    L16
  fnstcw -0xa(%rsp)
  lea    -0x1c7dd0(%rax),%rdx
  pxor   %xmm0,%xmm0
  flds   0xefe(%rip)        # 2030 <_fini+0xdd8>
  cvtsi2ss %rdx,%xmm0
  movzwl -0xa(%rsp),%edx
  or     $0xc,%dh
  cvtss2sd %xmm0,%xmm0
  subsd  %xmm2,%xmm0
  mov    %dx,-0xc(%rsp)
  movsd  %xmm0,-0x18(%rsp)
  fdivrl -0x18(%rsp)
  pxor   %xmm0,%xmm0
  fldcw  -0xc(%rsp)
  fistpll -0x18(%rsp)
  fldcw  -0xa(%rsp)
  mov    -0x18(%rsp),%rdx
  cvtsi2sd %rdx,%xmm0
  mulsd  D24(%rip),%xmm0
  lea    0x1(%rdx,%rax,1),%rax
  cvttsd2si %xmm0,%rdx
  sub    %rdx,%rax
L16:
  lea    0x5f4(%rax),%rdx
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  sub    $0x2534ca,%rax
  cvtsi2ss %rax,%xmm0
  cvtss2sd %xmm0,%xmm0
  subsd  D3(%rip),%xmm0
  divsd  D39(%rip),%xmm0
  addsd  D21(%rip),%xmm0
  cvttsd2si %xmm0,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm1
  lea    (%rax,%rax,8),%rcx
  lea    (%rax,%rcx,8),%rcx
  lea    (%rcx,%rcx,4),%rcx
  cvtsi2sd %rcx,%xmm0
  mov    %rdx,%rcx
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  movsd  D54(%rip),%xmm1
  cvttsd2si %xmm0,%r9
  pxor   %xmm0,%xmm0
  sub    %r9,%rcx
  sub    %r9d,%edx
  cvtsi2sd %rcx,%xmm0
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rcx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rcx,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%r9d
  sub    %r9d,%edx
  mov    %edx,(%rsi)
  lea    -0x1(%rcx),%edx
  sub    $0xd,%ecx
  cmp    $0xd,%edx
  cmovge %ecx,%edx
  mov    %edx,(%rdi)
  lea    -0x126b(%rax),%edx
  mov    %edx,(%r8)
  cmpl   $0x2,(%rdi)
  jle    L7
  lea    -0x126c(%rax),%edx
  mov    %edx,(%r8)
L7:
  test   %edx,%edx
  jle    L5
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  sub    $0x1,%edx
  mov    %edx,(%r8)
  ret
```

tool_calls: name: read_data
arguments: data_label: D25
datatype: f64
name: read_data
arguments: data_label: D24
datatype: f64
name: read_data
arguments: data_label: D3
datatype: f64
name: read_data
arguments: data_label: D39
datatype: f64
name: read_data
arguments: data_label: D21
datatype: f64
name: read_data
arguments: data_label: D54
datatype: f64

tool: D25: 0.25

tool: D24: 0.24

tool: D3: 122.1

tool: D39: 365.25

tool: D21: 6680

tool: D54: 30.6001

assistant: ```c
void RevertJulian(long Julian, int *Day, int *Month, int *Year) {
  long ja, jalpha, jb, jc, jd, je;
  if (Julian >= 2299161L) {
    jalpha = ((float)(Julian - 1867216L) - 0.25) / 36524.25L;
    ja = Julian + 1 + jalpha - (long)(0.24 * jalpha);
  } else
    ja = Julian;
  jb = ja + 1524;
  jc = 6680.0 + ((float)(jb - 2439870L) - 122.1) / 365.25;
  jd = 365 * jc + (0.25 * jc);
  je = (jb - jd) / 30.6001;
  *Day = (int)jb - jd - ((int)(30.6001 * je));
  *Month = (int)je - 1;
  if (*Month > 12)
    *Month -= 12;
  *Year = (int)jc - 4715;
  if (*Month > 2)
    --(*Year);
  if (*Year <= 0)
    --(*Year);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RevertJulian>:
  endbr64
  mov    %rdi,%rax
  movsd  D21(%rip),%xmm2
  mov    %rdx,%rdi
  mov    %rcx,%r8
  cmp    $0x231518,%rax
  jle    L17
  fnstcw -0xa(%rsp)
  lea    -0x1c7dd0(%rax),%rdx
  pxor   %xmm0,%xmm0
  flds   0xefe(%rip)        # 2030 <_fini+0xdd8>
  cvtsi2ss %rdx,%xmm0
  movzwl -0xa(%rsp),%edx
  or     $0xc,%dh
  cvtss2sd %xmm0,%xmm0
  subsd  %xmm2,%xmm0
  mov    %dx,-0xc(%rsp)
  movsd  %xmm0,-0x18(%rsp)
  fdivrl -0x18(%rsp)
  pxor   %xmm0,%xmm0
  fldcw  -0xc(%rsp)
  fistpll -0x18(%rsp)
  fldcw  -0xa(%rsp)
  mov    -0x18(%rsp),%rdx
  cvtsi2sd %rdx,%xmm0
  mulsd  D45(%rip),%xmm0
  lea    0x1(%rdx,%rax,1),%rax
  cvttsd2si %xmm0,%rdx
  sub    %rdx,%rax
L17:
  lea    0x5f4(%rax),%rdx
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  sub    $0x2534ca,%rax
  cvtsi2ss %rax,%xmm0
  cvtss2sd %xmm0,%xmm0
  subsd  D13(%rip),%xmm0
  divsd  D44(%rip),%xmm0
  addsd  D27(%rip),%xmm0
  cvttsd2si %xmm0,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm1
  lea    (%rax,%rax,8),%rcx
  lea    (%rax,%rcx,8),%rcx
  lea    (%rcx,%rcx,4),%rcx
  cvtsi2sd %rcx,%xmm0
  mov    %rdx,%rcx
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  movsd  D54(%rip),%xmm1
  cvttsd2si %xmm0,%r9
  pxor   %xmm0,%xmm0
  sub    %r9,%rcx
  sub    %r9d,%edx
  cvtsi2sd %rcx,%xmm0
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rcx
  pxor   %xmm0,%xmm0
  cvtsi2sd %rcx,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%r9d
  sub    %r9d,%edx
  mov    %edx,(%rsi)
  lea    -0x1(%rcx),%edx
  sub    $0xd,%ecx
  cmp    $0xd,%edx
  cmovge %ecx,%edx
  mov    %edx,(%rdi)
  lea    -0x126b(%rax),%edx
  mov    %edx,(%r8)
  cmpl   $0x2,(%rdi)
  jle    L29
  lea    -0x126c(%rax),%edx
  mov    %edx,(%r8)
L29:
  test   %edx,%edx
  jle    L28
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  sub    $0x1,%edx
  mov    %edx,(%r8)
  ret
```

tool_calls: name: unpack_data
arguments: label: D21
type_infered: f64
name: unpack_data
arguments: label: D45
type_infered: f64
name: unpack_data
arguments: label: D13
type_infered: f64
name: unpack_data
arguments: label: D44
type_infered: f64
name: unpack_data
arguments: label: D27
type_infered: f64
name: unpack_data
arguments: label: D54
type_infered: f64

tool: D21: 0.25

tool: D45: 0.24

tool: D13: 122.1

tool: D44: 365.25

tool: D27: 6680

tool: D54: 30.6001

assistant: ```c
void RevertJulian(long Julian, int *Day, int *Month, int *Year) {
  long ja, jalpha, jb, jc, jd, je;
  if (Julian >= 2299161L) {
    jalpha = ((float)(Julian - 1867216L) - 0.25) / 36524.25L;
    ja = Julian + 1 + jalpha - (long)(0.24 * jalpha);
  } else
    ja = Julian;
  jb = ja + 1524;
  jc = 6680.0 + ((float)(jb - 2439870L) - 122.1) / 365.25;
  jd = 365 * jc + (0.25 * jc);
  je = (jb - jd) / 30.6001;
  *Day = (int)jb - jd - ((int)(30.6001 * je));
  *Month = (int)je - 1;
  if (*Month > 12)
    *Month -= 12;
  *Year = (int)jc - 4715;
  if (*Month > 2)
    --(*Year);
  if (*Year <= 0)
    --(*Year);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RSHash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x5c6b7,-0x4(%rbp)
  movl   $0xf8c9,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L1
L4:
  mov    -0x8(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L1:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  mov    -0x8(%rbp),%eax
  and    $0x7fffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int RSHash(char *str) {
  unsigned int b = 378551;
  unsigned int a = 63689;
  unsigned int hash = 0;
  while (*str) {
    hash = hash * a + (*str++);
    a *= b;
  }
  return (hash & 0x7FFFFFFF);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RSHash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L6
  mov    $0x0,%eax
  mov    $0xf8c9,%ecx
L3:
  imul   %ecx,%eax
  add    $0x1,%rdi
  movsbl %dl,%edx
  add    %edx,%eax
  imul   $0x5c6b7,%ecx,%ecx
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L3
L13:
  and    $0x7fffffff,%eax
  ret
L6:
  mov    $0x0,%eax
  jmp    L13
```

assistant: ```c
unsigned int RSHash(char *str) {
  unsigned int b = 378551;
  unsigned int a = 63689;
  unsigned int hash = 0;
  while (*str) {
    hash = hash * a + (*str++);
    a *= b;
  }
  return (hash & 0x7FFFFFFF);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<RSHash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L15
  mov    $0xf8c9,%ecx
  nopw   0x0(%rax,%rax,1)
L5:
  imul   %ecx,%eax
  add    $0x1,%rdi
  imul   $0x5c6b7,%ecx,%ecx
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L5
  and    $0x7fffffff,%eax
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
unsigned int RSHash(char *str) {
  unsigned int b = 378551;
  unsigned int a = 63689;
  unsigned int hash = 0;
  while (*str) {
    hash = hash * a + (*str++);
    a *= b;
  }
  return (hash & 0x7FFFFFFF);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RSHash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L15
  mov    $0xf8c9,%ecx
  nopw   0x0(%rax,%rax,1)
L1:
  imul   %ecx,%eax
  add    $0x1,%rdi
  imul   $0x5c6b7,%ecx,%ecx
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L1
  and    $0x7fffffff,%eax
  ret
  nopl   0x0(%rax)
L15:
  ret
```

assistant: ```c
unsigned int RSHash(char *str) {
  unsigned int b = 378551;
  unsigned int a = 63689;
  unsigned int hash = 0;
  while (*str) {
    hash = hash * a + (*str++);
    a *= b;
  }
  return (hash & 0x7FFFFFFF);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<slen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L8:
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int slen(char *ins) {
  int index = 0;
  while (ins[index] != 0) {
    index++;
  }
  return index;
}
```
user: What is the c source code of the assembly code below:

```asm
<slen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L11
  mov    $0x1,%eax
L20:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L20
L7:
  mov    %edx,%eax
  ret
L11:
  mov    $0x0,%edx
  jmp    L7
```

assistant: ```c
int slen(char *ins) {
  int index = 0;
  while (ins[index] != 0) {
    index++;
  }
  return index;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<slen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L11
  mov    $0x1,%eax
  xchg   %ax,%ax
L8:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L8
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int slen(char *ins) {
  int index = 0;
  while (ins[index] != 0) {
    index++;
  }
  return index;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<slen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L2
  mov    $0x1,%eax
  xchg   %ax,%ax
L7:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L7
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int slen(char *ins) {
  int index = 0;
  while (ins[index] != 0) {
    index++;
  }
  return index;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  jmp    L26
L20:
  movl   $0x2,-0x18(%rbp)
  addl   $0x1,-0x14(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L26:
  cmpl   $0x9,-0x14(%rbp)
  jle    L20
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  int *p;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  *p;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  int *p;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  *p;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  int *p;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  *p;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  int *p;
  while (i < 10) {
    int x = 2;
    i++;
    p = &x;
  }
  *p;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_sleep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0xc350,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%r8
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <select@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void _sleep() {
  struct timeval tv = {0, 50000};
  select(0, NULL, NULL, NULL, &tv);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_sleep>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0xc350,0x4(%rsp)
  mov    %rsp,%r8
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <select@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _sleep() {
  struct timeval tv = {0, 50000};
  select(0, NULL, NULL, NULL, &tv);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_sleep>:
  endbr64
  sub    $0x18,%rsp
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  movabs $0xc35000000000,%rax
  xor    %edi,%edi
  mov    %rsp,%r8
  mov    %rax,(%rsp)
  call   <select@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _sleep() {
  struct timeval tv = {0, 50000};
  select(0, NULL, NULL, NULL, &tv);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_sleep>:
  endbr64
  sub    $0x18,%rsp
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  xor    %edi,%edi
  mov    %rsp,%r8
  mov    %rax,(%rsp)
  call   <select@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_infered: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x50,0xc3,0x00,0x00]}

assistant: ```c
void _sleep() {
  struct timeval tv = {0, 50000};
  select(0, NULL, NULL, NULL, &tv);
}
```
user: What is the c source code of the assembly code below:

```asm

:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  je     L84
  cmpl   $0x4,-0x4(%rbp)
  jg     L142
  cmpl   $0x3,-0x4(%rbp)
  je     L65
  cmpl   $0x3,-0x4(%rbp)
  jg     L142
  cmpl   $0x1,-0x4(%rbp)
  je     L32
  cmpl   $0x2,-0x4(%rbp)
  je     L31
  jmp    L142
L32:
  mov    $0x3,%eax
  jmp    L133
L31:
  mov    $0xc,%eax
  jmp    L133
L65:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  jmp    L133
L84:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  cmp    $0x1,%eax
  je     L36
  cmp    $0x2,%eax
  je     L67
  jmp    L40
L36:
  mov    $0x12,%eax
  jmp    L133
L67:
  addl   $0x1,-0x4(%rbp)
L40:
  mov    -0x4(%rbp),%eax
  jmp    L133
L142:
  mov    $0xc,%eax
L133:
  pop    %rbp
  ret
```

assistant: ```c
int s(int z) {
  switch (z) {
  case 1:
    return 3;
  case 2:
    return 12;
  case 3:
    z++;
    return z;
    z--;
  case 4:
    switch (z + 1) {
    case 1:
      return 18;
    case 2:
      z++;
    }
    return z;
  default:
    return 12;
  }
  z++;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm

:
  endbr64
  lea    -0x3(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L7
  cmp    $0x1,%edi
  setne  %al
  movzbl %al,%eax
  lea    0x3(%rax,%rax,8),%eax
  ret
L7:
  mov    $0x4,%eax
  ret
```

assistant: ```c
int s(int z) {
  switch (z) {
  case 1:
    return 3;
  case 2:
    return 12;
  case 3:
    z++;
    return z;
    z--;
  case 4:
    switch (z + 1) {
    case 1:
      return 18;
    case 2:
      z++;
    }
    return z;
  default:
    return 12;
  }
  z++;
}
```
user: Translate the below assembly code into C programming language:

```asm

:
  endbr64
  sub    $0x1,%edi
  mov    $0xc,%eax
  cmp    $0x3,%edi
  ja     L8
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L8:
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string

tool: D9: 

assistant: ```c
int s(int z) {
  switch (z) {
  case 1:
    return 3;
  case 2:
    return 12;
  case 3:
    z++;
    return z;
    z--;
  case 4:
    switch (z + 1) {
    case 1:
      return 18;
    case 2:
      z++;
    }
    return z;
  default:
    return 12;
  }
  z++;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm

:
  endbr64
  sub    $0x1,%edi
  mov    $0xc,%eax
  cmp    $0x3,%edi
  ja     L1
  lea    D9(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L1:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type: string

tool: D9: 

assistant: ```c
int s(int z) {
  switch (z) {
  case 1:
    return 3;
  case 2:
    return 12;
  case 3:
    z++;
    return z;
    z--;
  case 4:
    switch (z + 1) {
    case 1:
      return 18;
    case 2:
      z++;
    }
    return z;
  default:
    return 12;
  }
  z++;
}
```
user: Translate the below assembly code into C programming language:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x14(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  nop
  leave
  ret
```

assistant: ```c
void swap(void *arr, unsigned int entry_size, int i, int j, void *swap_area) {
  memcpy(swap_area, (arr + (j * entry_size)), entry_size);
  memcpy((arr + (j * entry_size)), (arr + (i * entry_size)), entry_size);
  memcpy((arr + (i * entry_size)), swap_area, entry_size);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %edx,%r13d
  mov    %r8,%r14
  imul   %esi,%ecx
  lea    (%rdi,%rcx,1),%r12
  mov    %esi,%edx
  mov    %r12,%rsi
  mov    %r8,%rdi
  call   <memcpy@plt>
  imul   %ebp,%r13d
  add    %r13,%rbx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void swap(void *arr, unsigned int entry_size, int i, int j, void *swap_area) {
  memcpy(swap_area, (arr + (j * entry_size)), entry_size);
  memcpy((arr + (j * entry_size)), (arr + (i * entry_size)), entry_size);
  memcpy((arr + (i * entry_size)), swap_area, entry_size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<swap>:
  endbr64
  push   %r14
  imul   %esi,%ecx
  mov    %r8,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  imul   %r13d,%r12d
  push   %rbp
  lea    (%rdi,%rcx,1),%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbp,%rsi
  mov    %r8,%rdi
  add    %rbx,%r12
  call   <memcpy@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  pop    %rbx
  mov    %r13d,%edx
  pop    %rbp
  mov    %r14,%rsi
  mov    %r12,%rdi
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <memcpy@plt>
```

assistant: ```c
void swap(void *arr, unsigned int entry_size, int i, int j, void *swap_area) {
  memcpy(swap_area, (arr + (j * entry_size)), entry_size);
  memcpy((arr + (j * entry_size)), (arr + (i * entry_size)), entry_size);
  memcpy((arr + (i * entry_size)), swap_area, entry_size);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<swap>:
  endbr64
  push   %r14
  imul   %esi,%ecx
  mov    %r8,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  imul   %r13d,%r12d
  push   %rbp
  lea    (%rdi,%rcx,1),%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rbp,%rsi
  mov    %r8,%rdi
  add    %rbx,%r12
  call   <memcpy@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  pop    %rbx
  mov    %r13d,%edx
  pop    %rbp
  mov    %r14,%rsi
  mov    %r12,%rdi
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <memcpy@plt>
```

assistant: ```c
void swap(void *arr, unsigned int entry_size, int i, int j, void *swap_area) {
  memcpy(swap_area, (arr + (j * entry_size)), entry_size);
  memcpy((arr + (j * entry_size)), (arr + (i * entry_size)), entry_size);
  memcpy((arr + (i * entry_size)), swap_area, entry_size);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<process_buffer>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %rsi,%r12
  mov    %edx,%r14d
  mov    %rcx,%r13
  mov    %r8,0x18(%rsp)
  mov    %r9,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x20(%rsp)
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  jmp    L29
L31:
  mov    %r14d,%edx
  sub    %ebp,%edx
  movslq %ebp,%rsi
  add    %r12,%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
  mov    0x0(%r13),%eax
  sub    %ebp,%eax
  mov    %eax,0x0(%r13)
  cltq
  movb   $0x0,(%r12,%rax,1)
L29:
  mov    $0xa,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L32
  mov    0x8(%rsp),%rdx
  lea    D14(%rip),%rsi
  mov    %r12,%rdi
  call   <strtok_r@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%ebp
  mov    $0x4,%edx
  lea    D20(%rip),%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L31
  mov    $0x5,%edx
  lea    D21(%rip),%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  je     L31
  mov    0x18(%rsp),%rdx
  mov    %rbx,%rsi
  mov    0x10(%rsp),%rdi
  call   <add_parsed_line@plt>
  mov    %r14d,%edx
  sub    %ebp,%edx
  movslq %ebp,%rsi
  add    %r12,%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
  mov    0x0(%r13),%eax
  sub    %ebp,%eax
  mov    %eax,0x0(%r13)
  cltq
  movb   $0x0,(%r12,%rax,1)
  test   %r15,%r15
  je     L29
  subl   $0x1,(%r15)
  jmp    L29
L32:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D14
data_type: string
name: parse_data
arguments: tag: D20
data_type: string
name: parse_data
arguments: tag: D21
data_type: string

tool: D14: 


tool: D20: Bulk

tool: D21: Error

assistant: ```c
void process_buffer(struct iptrie *result, char *buffer, int buflen, int *offset, struct amp_asn_info *info, int *outstanding) {
  char *line;
  char *lineptr = NULL;
  int linelen;
  while (strchr(buffer, '\n') != NULL) {
    line = strtok_r(buffer, "\n", &lineptr);
    linelen = strlen(line) + 1;
    if (strncmp(line, "Bulk", 4) == 0 || strncmp(line, "Error", 5) == 0) {
      memmove(buffer, buffer + linelen, buflen - linelen);
      *offset = *offset - linelen;
      buffer[*offset] = '\0';
      continue;
    }
    add_parsed_line(result, line, info);
    memmove(buffer, buffer + linelen, buflen - linelen);
    *offset = *offset - linelen;
    buffer[*offset] = '\0';
    if (outstanding) {
      (*outstanding)--;
    }
  }
}
```