ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L22
L26:
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  mov    -0x10(%rbp),%edx
  mov    %eax,%esi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jge    L17
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L17:
  mov    -0x10(%rbp),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <pow@plt>
  cltd
  idivl  -0x10(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  -0xc(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L22:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L26
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type: string
name: unpack_data
arguments: label: D40
type: string
name: unpack_data
arguments: label: D43
type: string
name: unpack_data
arguments: label: D39
type: string
name: unpack_data
arguments: label: D11
type: string

tool: D7: Enter the length of the series: 

tool: D40: %d

tool: D43: %d/%d 

tool: D39: + 

tool: D11: = %.f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    printf("%d/%d ", i * i, i);
    if (i < n)
      printf("+ ");
    sum += pow(i, 2) / i;
  }
  printf("= %.f\n", sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L39
  pxor   %xmm2,%xmm2
  movss  %xmm2,0xc(%rsp)
  mov    $0x1,%ebx
  lea    D3(%rip),%rbp
  lea    D12(%rip),%r12
  jmp    L29
L18:
  mov    $0x2,%esi
  mov    %ebx,%edi
  call   <pow@plt>
  cltd
  idiv   %ebx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jl     L51
L29:
  mov    %ebx,%esi
  imul   %ebx,%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jle    L18
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L18
L39:
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
L51:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  lea    D1(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
data_type: string
name: read_data
arguments: tag: D2
data_type: string
name: read_data
arguments: tag: D3
data_type: string
name: read_data
arguments: tag: D12
data_type: string
name: read_data
arguments: tag: D1
data_type: string

tool: D7: Enter the length of the series: 

tool: D2: %d

tool: D3: %d/%d 

tool: D12: + 

tool: D1: = %.f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    printf("%d/%d ", i * i, i);
    if (i < n)
      printf("+ ");
    sum += pow(i, 2) / i;
  }
  printf("= %.f\n", sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D8(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L55
  pxor   %xmm2,%xmm2
  mov    $0x1,%ebx
  lea    D15(%rip),%rbp
  movss  %xmm2,0xc(%rsp)
  lea    D7(%rip),%r12
  jmp    L41
L29:
  mov    %ebx,%edi
  mov    $0x2,%esi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cltd
  idiv   %ebx
  add    $0x1,%ebx
  cvtsi2ss %eax,%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  cmp    %ebx,0x14(%rsp)
  jl     L59
L41:
  mov    %ebx,%esi
  xor    %eax,%eax
  mov    %ebx,%edx
  mov    %rbp,%rdi
  imul   %ebx,%esi
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jle    L29
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L29
L55:
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
  nopl   0x0(%rax)
L59:
  pxor   %xmm0,%xmm0
  lea    D39(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd 0xc(%rsp),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D8
data_type: string
name: read_data
arguments: label: D18
data_type: string
name: read_data
arguments: label: D15
data_type: string
name: read_data
arguments: label: D7
data_type: string
name: read_data
arguments: label: D39
data_type: string

tool: D8: Enter the length of the series: 

tool: D18: %d

tool: D15: %d/%d 

tool: D7: + 

tool: D39: = %.f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    printf("%d/%d ", i * i, i);
    if (i < n)
      printf("+ ");
    sum += pow(i, 2) / i;
  }
  printf("= %.f\n", sum);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D12(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D47(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L18
  pxor   %xmm3,%xmm3
  mov    $0x1,%ebx
  lea    D13(%rip),%rbp
  movss  %xmm3,0xc(%rsp)
  lea    D16(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%esi
  xor    %eax,%eax
  mov    %ebx,%edx
  mov    %rbp,%rdi
  imul   %ebx,%esi
  call   <printf@plt>
  cmp    %ebx,0x14(%rsp)
  jg     L29
  mov    %ebx,%edi
  mov    $0x2,%esi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cltd
  idiv   %ebx
  add    $0x1,%ebx
  cvtsi2ss %eax,%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  cmp    %ebx,0x14(%rsp)
  jge    L25
L13:
  pxor   %xmm0,%xmm0
  lea    D30(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd 0xc(%rsp),%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L29:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%edi
  mov    $0x2,%esi
  call   <pow@plt>
  pxor   %xmm0,%xmm0
  cltd
  idiv   %ebx
  add    $0x1,%ebx
  cvtsi2ss %eax,%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  cmp    0x14(%rsp),%ebx
  jle    L25
  jmp    L13
L18:
  pxor   %xmm4,%xmm4
  movss  %xmm4,0xc(%rsp)
  jmp    L13
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D47
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D13
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D16
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D30
type_guessed: string

tool: D12: Enter the length of the series: 

tool: D47: %d

tool: D13: %d/%d 

tool: D16: + 

tool: D30: = %.f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    printf("%d/%d ", i * i, i);
    if (i < n)
      printf("+ ");
    sum += pow(i, 2) / i;
  }
  printf("= %.f\n", sum);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getNibble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L45
  cmpb   $0x39,-0x4(%rbp)
  jg     L45
  movsbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L13
L45:
  cmpb   $0x40,-0x4(%rbp)
  jle    L35
  cmpb   $0x46,-0x4(%rbp)
  jg     L35
  movsbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L13
L35:
  cmpb   $0x60,-0x4(%rbp)
  jle    L27
  cmpb   $0x66,-0x4(%rbp)
  jg     L27
  movsbl -0x4(%rbp),%eax
  sub    $0x57,%eax
  jmp    L13
L27:
  mov    $0xffffffff,%eax
L13:
  pop    %rbp
  ret
```

assistant: ```c
int getNibble(char x) {
  if ((x >= '0') && (x <= '9')) {
    return (x - '0');
  }
  if ((x >= 'A') && (x <= 'F')) {
    return ((x - 'A') + 10);
  }
  if ((x >= 'a') && (x <= 'f')) {
    return ((x - 'a') + 10);
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L17
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L8
  lea    -0x61(%rdi),%edx
  movsbl %dil,%eax
  sub    $0x57,%eax
  cmp    $0x6,%dl
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
L17:
  movsbl %dil,%edi
  lea    -0x30(%rdi),%eax
  ret
L8:
  movsbl %dil,%edi
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
int getNibble(char x) {
  if ((x >= '0') && (x <= '9')) {
    return (x - '0');
  }
  if ((x >= 'A') && (x <= 'F')) {
    return ((x - 'A') + 10);
  }
  if ((x >= 'a') && (x <= 'f')) {
    return ((x - 'a') + 10);
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L12
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L4
  lea    -0x61(%rdi),%edx
  lea    -0x57(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  movsbl %al,%eax
  ret
```

assistant: ```c
int getNibble(char x) {
  if ((x >= '0') && (x <= '9')) {
    return (x - '0');
  }
  if ((x >= 'A') && (x <= 'F')) {
    return ((x - 'A') + 10);
  }
  if ((x >= 'a') && (x <= 'f')) {
    return ((x - 'a') + 10);
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getNibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L9
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L15
  lea    -0x61(%rdi),%edx
  lea    -0x57(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0xffffffff,%edx
  cmovae %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  movsbl %al,%eax
  ret
```

assistant: ```c
int getNibble(char x) {
  if ((x >= '0') && (x <= '9')) {
    return (x - '0');
  }
  if ((x >= 'A') && (x <= 'F')) {
    return ((x - 'A') + 10);
  }
  if ((x >= 'a') && (x <= 'f')) {
    return ((x - 'a') + 10);
  }
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<printMonth>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L50
L8:
  movl   $0x0,-0x4(%rbp)
  jmp    L52
L13:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  jle    L20
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  sub    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L43
L20:
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L43:
  addl   $0x1,-0x4(%rbp)
L52:
  cmpl   $0x6,-0x4(%rbp)
  jle    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L50:
  cmpl   $0x5,-0x8(%rbp)
  jle    L8
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D16
type_guessed: string

tool: D7: %3i

tool: D16:    

assistant: ```c
void printMonth(int month[6][7]) {
  int i, k;
  for (i = 0; i < 6; i++) {
    for (k = 0; k < 7; k++) {
      if (month[i][k] > 0)
        printf("%3i", month[i][k]);
      else
        printf("   ");
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<printMonth>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    0x1c(%rdi),%rbp
  lea    0xc4(%rdi),%r14
  lea    D18(%rip),%r13
  lea    D14(%rip),%r12
  jmp    L57
L1:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L58:
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  je     L46
L3:
  mov    (%rbx),%esi
  test   %esi,%esi
  jg     L1
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L58
L46:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1c,%rbp
  cmp    %r14,%rbp
  je     L56
L57:
  lea    -0x1c(%rbp),%rbx
  jmp    L3
L56:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D18
type: string
name: unpack_data
arguments: data_label: D14
type: string

tool: D18:    

tool: D14: %3i

assistant: ```c
void printMonth(int month[6][7]) {
  int i, k;
  for (i = 0; i < 6; i++) {
    for (k = 0; k < 7; k++) {
      if (month[i][k] > 0)
        printf("%3i", month[i][k]);
      else
        printf("   ");
    }
    printf("\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<printMonth>:
  endbr64
  push   %r14
  lea    0xc4(%rdi),%r14
  push   %r13
  lea    D2(%rip),%r13
  push   %r12
  lea    D17(%rip),%r12
  push   %rbp
  lea    0x1c(%rdi),%rbp
  push   %rbx
  nopl   (%rax)
L35:
  lea    -0x1c(%rbp),%rbx
  jmp    L22
  xchg   %ax,%ax
L36:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  je     L9
L22:
  mov    (%rbx),%esi
  test   %esi,%esi
  jg     L36
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L22
L9:
  mov    $0xa,%edi
  lea    0x1c(%rbx),%rbp
  call   <putchar@plt>
  cmp    %r14,%rbp
  jne    L35
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: string
name: unpack_data
arguments: tag: D17
data_type: string

tool: D2:    

tool: D17: %3i

assistant: ```c
void printMonth(int month[6][7]) {
  int i, k;
  for (i = 0; i < 6; i++) {
    for (k = 0; k < 7; k++) {
      if (month[i][k] > 0)
        printf("%3i", month[i][k]);
      else
        printf("   ");
    }
    printf("\n");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<printMonth>:
  endbr64
  push   %r14
  lea    0xc4(%rdi),%r14
  push   %r13
  lea    D8(%rip),%r13
  push   %r12
  lea    D6(%rip),%r12
  push   %rbp
  lea    0x1c(%rdi),%rbp
  push   %rbx
  nopl   (%rax)
L47:
  lea    -0x1c(%rbp),%rbx
  jmp    L35
  xchg   %ax,%ax
L25:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  je     L24
L35:
  mov    (%rbx),%esi
  test   %esi,%esi
  jg     L25
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L35
L24:
  mov    $0xa,%edi
  lea    0x1c(%rbx),%rbp
  call   <putchar@plt>
  cmp    %r14,%rbp
  jne    L47
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type_infered: string
name: unpack_data
arguments: data_tag: D6
type_infered: string

tool: D8:    

tool: D6: %3i

assistant: ```c
void printMonth(int month[6][7]) {
  int i, k;
  for (i = 0; i < 6; i++) {
    for (k = 0; k < 7; k++) {
      if (month[i][k] > 0)
        printf("%3i", month[i][k]);
      else
        printf("   ");
    }
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<xdbg_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L35
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L26
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3e,%al
  je     L26
  mov    -0x20(%rbp),%rax
  mov    (%rax),%esi
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  and    $0x4,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  xor    %eax,%esi
  mov    %esi,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L35:
  cmpl   $0x10,-0x4(%rbp)
  jle    L3
L26:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int xdbg_hash(const char *keyword, int *hash) {
  int i;
  *hash = 0;
  for (i = 1; i <= 16; i++) {
    if (keyword[i] == 0 || keyword[i] == '>')
      break;
    1 ? 0 : ("%c", keyword[i]);
    *hash ^= (keyword[i] & 15) << ((i & 1) << 2);
  }
  1 ? 0 : (" => hash: 0x%02x len: %d\n", *hash, i - 1);
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xdbg_hash>:
  endbr64
  movl   $0x0,(%rsi)
  add    $0x1,%rdi
  mov    $0x1,%edx
L28:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L15
  cmp    $0x3e,%al
  je     L15
  and    $0xf,%eax
  lea    0x0(,%rdx,4),%ecx
  and    $0x4,%ecx
  shl    %cl,%eax
  xor    %eax,(%rsi)
  add    $0x1,%edx
  add    $0x1,%rdi
  cmp    $0x11,%edx
  jne    L28
L15:
  mov    %edx,%eax
  ret
```

assistant: ```c
int xdbg_hash(const char *keyword, int *hash) {
  int i;
  *hash = 0;
  for (i = 1; i <= 16; i++) {
    if (keyword[i] == 0 || keyword[i] == '>')
      break;
    1 ? 0 : ("%c", keyword[i]);
    *hash ^= (keyword[i] & 15) << ((i & 1) << 2);
  }
  1 ? 0 : (" => hash: 0x%02x len: %d\n", *hash, i - 1);
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<xdbg_hash>:
  endbr64
  movl   $0x0,(%rsi)
  mov    $0x1,%r8d
L11:
  movzbl (%rdi,%r8,1),%edx
  mov    %r8d,%eax
  test   %dl,%dl
  je     L18
  cmp    $0x3e,%dl
  je     L18
  lea    0x0(,%r8,4),%ecx
  and    $0xf,%edx
  add    $0x1,%r8
  and    $0x4,%ecx
  shl    %cl,%edx
  xor    %edx,(%rsi)
  cmp    $0x11,%r8
  jne    L11
  mov    $0x11,%eax
L18:
  ret
```

assistant: ```c
int xdbg_hash(const char *keyword, int *hash) {
  int i;
  *hash = 0;
  for (i = 1; i <= 16; i++) {
    if (keyword[i] == 0 || keyword[i] == '>')
      break;
    1 ? 0 : ("%c", keyword[i]);
    *hash ^= (keyword[i] & 15) << ((i & 1) << 2);
  }
  1 ? 0 : (" => hash: 0x%02x len: %d\n", *hash, i - 1);
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RNApdist_cmdline_parser_required>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
int RNApdist_cmdline_parser_required(struct RNApdist_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RNApdist_cmdline_parser_required>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int RNApdist_cmdline_parser_required(struct RNApdist_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<RNApdist_cmdline_parser_required>:
  endbr64
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int RNApdist_cmdline_parser_required(struct RNApdist_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RNApdist_cmdline_parser_required>:
  endbr64
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
int RNApdist_cmdline_parser_required(struct RNApdist_args_info *args_info, const char *prog_name) {
  (void)(args_info);
  (void)(prog_name);
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<player_response>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L27
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <base_n_bit@plt>
  jmp    L20
L27:
  cmpl   $0x1,-0x4(%rbp)
  jne    L23
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  call   <floor@plt>
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    %ecx,%esi
  mov    $0x3,%edi
  call   <base_n_bit@plt>
  jmp    L20
L23:
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
L20:
  leave
  ret
```

assistant: ```c
int player_response(int player, int strategy, int input) {
  if (player == 0) {
    return base_n_bit(2, strategy, input);
  } else if (player == 1) {
    return base_n_bit(3, floor(strategy / pow(2, 2)), input);
  } else {
    exit(EXIT_FAILURE);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<player_response>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebx
  mov    %edx,%ebp
  test   %edi,%edi
  je     L9
  cmp    $0x1,%edi
  je     L38
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  mov    $0x2,%edi
  call   <base_n_bit@plt>
  jmp    L34
L38:
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%edi
  call   <floor@plt>
  mov    %eax,%esi
  mov    %ebp,%edx
  mov    $0x3,%edi
  call   <base_n_bit@plt>
  jmp    L34
```

assistant: ```c
int player_response(int player, int strategy, int input) {
  if (player == 0) {
    return base_n_bit(2, strategy, input);
  } else if (player == 1) {
    return base_n_bit(3, floor(strategy / pow(2, 2)), input);
  } else {
    exit(EXIT_FAILURE);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<player_response>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L18
  cmp    $0x1,%edi
  je     L1
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    10a0 <exit@plt>
  xchg   %ax,%ax
L18:
  add    $0x8,%rsp
  mov    $0x2,%edi
  pop    %rbp
  pop    %r12
  jmp    1080 <base_n_bit@plt>
  nopl   0x0(%rax)
L1:
  mov    $0x2,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%r8d
  mov    %ebp,%eax
  cltd
  idiv   %r8d
  mov    %eax,%edi
  call   <floor@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    $0x3,%edi
  pop    %rbp
  mov    %eax,%esi
  pop    %r12
  jmp    1080 <base_n_bit@plt>
```

assistant: ```c
int player_response(int player, int strategy, int input) {
  if (player == 0) {
    return base_n_bit(2, strategy, input);
  } else if (player == 1) {
    return base_n_bit(3, floor(strategy / pow(2, 2)), input);
  } else {
    exit(EXIT_FAILURE);
  }
}
```