ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<buildNvals>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  lea    0x0(,%rbp,8),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L15
  xor    %r14d,%r14d
  mov    $0xcccccccd,%r13d
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %r14d,%ebx
  mov    $0xa,%edi
  imul   %r13,%rbx
  shr    $0x23,%rbx
  mov    %ebx,%esi
  call   <pow@plt>
  lea    (%rbx,%rbx,4),%eax
  mov    %r14d,%edx
  add    %eax,%eax
  movapd %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  sub    %eax,%edx
  cvtsi2sd %edx,%xmm0
  addsd  D9(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%r12,%r14,8)
  add    $0x1,%r14
  cmp    %r14,%rbp
  jne    L2
L15:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
datatype: f64

tool: D9: 0.1

assistant: ```c
double *buildNvals(int n) {
  int i;
  double *vals = malloc(sizeof(*vals) * n);
  for (i = 0; i < n; i++) {
    vals[i] = (0.1 + (double)(i % 10)) * pow(10, (i / 10));
  }
  return vals;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L2
L12:
  addq   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L12
L11:
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L23
L44:
  subq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  imul   -0x14(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x14(%rbp)
L23:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jne    L44
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
const char *get_n(const char *str, unsigned *nret) {
  const char *curr = str;
  while (*curr && isdigit(*curr))
    ++curr;
  const char *tmp = curr;
  unsigned n = 0, i = 1;
  while (curr != str) {
    n += (*--curr - '0') * i;
    i *= 10;
  }
  *nret = n;
  return tmp;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<get_n>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  movzbl (%rdi),%edi
  mov    %rbp,%rbx
  mov    $0x0,%esi
  test   %dil,%dil
  je     L46
L59:
  movsbl %dil,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L15
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L59
L15:
  cmp    %rbx,%rbp
  je     L5
  mov    %rbx,%rcx
  mov    $0x1,%eax
  mov    $0x0,%esi
L42:
  sub    $0x1,%rcx
  movsbl (%rcx),%edx
  sub    $0x30,%edx
  imul   %eax,%edx
  add    %edx,%esi
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  cmp    %rcx,%rbp
  jne    L42
L46:
  mov    %esi,(%r12)
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x0,%esi
  jmp    L46
```

assistant: ```c
const char *get_n(const char *str, unsigned *nret) {
  const char *curr = str;
  while (*curr && isdigit(*curr))
    ++curr;
  const char *tmp = curr;
  unsigned n = 0, i = 1;
  while (curr != str) {
    n += (*--curr - '0') * i;
    i *= 10;
  }
  *nret = n;
  return tmp;
}
```
user: What is the c source code of the assembly code below:

```asm
<get_n>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  mov    %rbx,%r12
  test   %dil,%dil
  jne    L52
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L36:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L35
L52:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L36
L35:
  cmp    %r12,%rbx
  je     L15
  mov    %r12,%rcx
  mov    $0x1,%eax
  xor    %esi,%esi
  nopl   0x0(%rax,%rax,1)
L20:
  movsbl -0x1(%rcx),%edx
  sub    $0x1,%rcx
  sub    $0x30,%edx
  imul   %eax,%edx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  add    %edx,%esi
  cmp    %rcx,%rbx
  jne    L20
  mov    %r12,%rax
  pop    %rbx
  mov    %esi,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L15:
  xor    %esi,%esi
  mov    %r12,%rax
  pop    %rbx
  mov    %esi,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *get_n(const char *str, unsigned *nret) {
  const char *curr = str;
  while (*curr && isdigit(*curr))
    ++curr;
  const char *tmp = curr;
  unsigned n = 0, i = 1;
  while (curr != str) {
    n += (*--curr - '0') * i;
    i *= 10;
  }
  *nret = n;
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_n>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  mov    %rbx,%r12
  test   %dil,%dil
  jne    L3
  jmp    L55
  nopl   0x0(%rax,%rax,1)
L13:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L20
L3:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L13
L20:
  cmp    %r12,%rbx
  je     L55
  mov    %r12,%rcx
  mov    $0x1,%eax
  xor    %esi,%esi
  nopl   0x0(%rax,%rax,1)
L18:
  movsbl -0x1(%rcx),%edx
  sub    $0x1,%rcx
  sub    $0x30,%edx
  imul   %eax,%edx
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  add    %edx,%esi
  cmp    %rcx,%rbx
  jne    L18
  mov    %r12,%rax
  pop    %rbx
  mov    %esi,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L55:
  xor    %esi,%esi
  mov    %r12,%rax
  pop    %rbx
  mov    %esi,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
const char *get_n(const char *str, unsigned *nret) {
  const char *curr = str;
  while (*curr && isdigit(*curr))
    ++curr;
  const char *tmp = curr;
  unsigned n = 0, i = 1;
  while (curr != str) {
    n += (*--curr - '0') * i;
    i *= 10;
  }
  *nret = n;
  return tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<genSuppliedDate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0x431bde82d7b634db,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x12,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %al,-0x1e(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0x346dc5d63886594b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0xb,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,%rcx
  movabs $0xa3d70a3d70a3d70b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rax
  sar    $0x6,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %dl,-0x1d(%rbp)
  mov    -0x18(%rbp),%rcx
  movabs $0x346dc5d63886594b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0xb,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  imul   $0x2710,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  movzbl %al,%eax
  add    $0x30,%eax
  mov    %eax,-0x1c(%rbp)
  movzbl -0x1d(%rbp),%edx
  mov    $0x25,%ecx
  mov    %ecx,%eax
  mul    %dl
  shr    $0x8,%ax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  shr    %al
  add    %ecx,%eax
  shr    $0x2,%al
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  sub    %edx,%eax
  mov    %al,-0x1d(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %al,-0xb(%rbp)
  movzbl -0x1e(%rbp),%eax
  mov    %al,-0xa(%rbp)
  movzbl -0x1d(%rbp),%eax
  mov    %al,-0x9(%rbp)
  lea    -0xb(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int genSuppliedDate(char *argv[], unsigned char *dateData) {
  int year;
  unsigned char month;
  unsigned char day;
  long int a = strtol(argv[1], NULL, 10);
  month = a / 1000000;
  day = a / 10000 % 100;
  year = a % 10000;
  year = (year & 0xff) + 0x30;
  day = (day / 7) * 7;
  unsigned char combined[] = {year, month, day};
  memcpy(dateData, &combined, 4);
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<genSuppliedDate>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rdi),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %rax,%rsi
  movabs $0x346dc5d63886594b,%rdx
  imul   %rdx
  sar    $0xb,%rdx
  mov    %rsi,%rdi
  sar    $0x3f,%rdi
  mov    %rdx,%rcx
  sub    %rdi,%rcx
  imul   $0x2710,%rcx,%rax
  mov    %rsi,%r8
  sub    %rax,%r8
  movzbl %r8b,%r8d
  movabs $0xa3d70a3d70a3d70b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  add    %rcx,%rdx
  sar    $0x6,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    $0x25,%edx
  mov    %edx,%eax
  mul    %cl
  shr    $0x8,%ax
  sub    %eax,%ecx
  shr    %cl
  add    %eax,%ecx
  shr    $0x2,%cl
  add    $0x30,%r8d
  mov    %r8b,0x5(%rsp)
  movabs $0x431bde82d7b634db,%rdx
  mov    %rsi,%rax
  imul   %rdx
  sar    $0x12,%rdx
  sub    %rdi,%rdx
  mov    %dl,0x6(%rsp)
  lea    0x0(,%rcx,8),%eax
  sub    %ecx,%eax
  mov    %al,0x7(%rsp)
  lea    0x5(%rsp),%rsi
  mov    $0x4,%edx
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x1,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int genSuppliedDate(char *argv[], unsigned char *dateData) {
  int year;
  unsigned char month;
  unsigned char day;
  long int a = strtol(argv[1], NULL, 10);
  month = a / 1000000;
  day = a / 10000 % 100;
  year = a % 10000;
  year = (year & 0xff) + 0x30;
  day = (day / 7) * 7;
  unsigned char combined[] = {year, month, day};
  memcpy(dateData, &combined, 4);
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<genSuppliedDate>:
  endbr64
  push   %rbp
  mov    $0xa,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  sub    $0x10,%rsp
  mov    0x8(%rdi),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strtol@plt>
  movabs $0x346dc5d63886594b,%rdx
  mov    %rax,%rsi
  imul   %rdx
  mov    %rsi,%rdi
  mov    %rsi,%r8
  sar    $0x3f,%rdi
  sar    $0xb,%rdx
  mov    %rdx,%rcx
  movabs $0xa3d70a3d70a3d70b,%rdx
  sub    %rdi,%rcx
  imul   $0x2710,%rcx,%rax
  sub    %rax,%r8
  mov    %rcx,%rax
  imul   %rdx
  mov    %rcx,%rax
  movzbl %r8b,%r8d
  sar    $0x3f,%rax
  add    $0x30,%r8d
  mov    %r8b,0x5(%rsp)
  add    %rcx,%rdx
  sar    $0x6,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  movabs $0x431bde82d7b634db,%rdx
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    $0x25,%eax
  mul    %cl
  shr    $0x8,%ax
  sub    %eax,%ecx
  shr    %cl
  add    %eax,%ecx
  mov    %rsi,%rax
  lea    0x5(%rsp),%rsi
  imul   %rdx
  shr    $0x2,%cl
  lea    0x0(,%rcx,8),%eax
  sub    %ecx,%eax
  sar    $0x12,%rdx
  mov    %al,0x7(%rsp)
  sub    %rdi,%rdx
  mov    %rbp,%rdi
  mov    %dl,0x6(%rsp)
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  mov    $0x1,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int genSuppliedDate(char *argv[], unsigned char *dateData) {
  int year;
  unsigned char month;
  unsigned char day;
  long int a = strtol(argv[1], NULL, 10);
  month = a / 1000000;
  day = a / 10000 % 100;
  year = a % 10000;
  year = (year & 0xff) + 0x30;
  day = (day / 7) * 7;
  unsigned char combined[] = {year, month, day};
  memcpy(dateData, &combined, 4);
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<genSuppliedDate>:
  endbr64
  push   %rbp
  mov    $0xa,%edx
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    0x8(%rdi),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <strtol@plt>
  movabs $0x346dc5d63886594b,%rdx
  mov    %rax,%rsi
  imul   %rdx
  mov    %rsi,%rdi
  mov    %rsi,%r8
  sar    $0x3f,%rdi
  sar    $0xb,%rdx
  mov    %rdx,%rcx
  movabs $0xa3d70a3d70a3d70b,%rdx
  sub    %rdi,%rcx
  imul   $0x2710,%rcx,%rax
  sub    %rax,%r8
  mov    %rcx,%rax
  imul   %rdx
  mov    %rcx,%rax
  movzbl %r8b,%r8d
  sar    $0x3f,%rax
  add    $0x30,%r8d
  mov    %r8b,%bl
  add    %rcx,%rdx
  sar    $0x6,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  movabs $0x431bde82d7b634db,%rdx
  lea    (%rax,%rax,4),%rax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    $0x25,%eax
  mul    %cl
  shr    $0x8,%ax
  sub    %eax,%ecx
  shr    %cl
  add    %eax,%ecx
  mov    %rsi,%rax
  lea    0x4(%rsp),%rsi
  imul   %rdx
  shr    $0x2,%cl
  lea    0x0(,%rcx,8),%eax
  sub    %ecx,%eax
  sar    $0x12,%rdx
  mov    %al,0x6(%rsp)
  sub    %rdi,%rdx
  mov    %rbp,%rdi
  mov    %dl,%bh
  mov    $0x4,%edx
  mov    %bx,0x4(%rsp)
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int genSuppliedDate(char *argv[], unsigned char *dateData) {
  int year;
  unsigned char month;
  unsigned char day;
  long int a = strtol(argv[1], NULL, 10);
  month = a / 1000000;
  day = a / 10000 % 100;
  year = a % 10000;
  year = (year & 0xff) + 0x30;
  day = (day / 7) * 7;
  unsigned char combined[] = {year, month, day};
  memcpy(dateData, &combined, 4);
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D8(%rip),%r9
  lea    D32(%rip),%r8
  lea    D21(%rip),%rax
  mov    %rax,%rcx
  lea    D36(%rip),%rax
  mov    %rax,%rdx
  lea    D29(%rip),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L13:
  mov    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %edx,%edi
  mov    %eax,%ecx
  sar    %cl,%edi
  mov    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %edx,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    -0x4(%rbp),%eax
  mov    $0xffffffff,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    -0x4(%rbp),%eax
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    $0xffffffff,%ecx
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L7:
  cmpl   $0x41,-0x4(%rbp)
  jle    L13
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type: string
name: unpack_data
arguments: tag: D32
type: string
name: unpack_data
arguments: tag: D21
type: string
name: unpack_data
arguments: tag: D36
type: string
name: unpack_data
arguments: tag: D29
type: string
name: unpack_data
arguments: tag: D25
type: string
name: unpack_data
arguments: tag: D3
type: string

tool: D8: 1 >> e

tool: D32: 1 << e

tool: D21: ~0 >> e

tool: D36: ~0 << e

tool: D29: e

tool: D25: | %.2s | %.32s | %.32s | %.32s | %.32s | 


tool: D3: | %2d | %32u | %32u | %32u | %32u |


assistant: ```c
int main(void) {
  int e;
  printf("| %.2s | %.32s | %.32s | %.32s | %.32s | \n", "e", "~0 << e", "~0 >> e", "1 << e", "1 >> e");
  for (e = 0; e < 66; ++e) {
    printf("| %2d | %32u | %32u | %32u | %32u |\n", e, ~0 << e, ~0 >> e, 1 << e, 1 >> e);
  }
  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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D4(%rip),%r9
  lea    D5(%rip),%r8
  lea    D53(%rip),%rcx
  lea    D52(%rip),%rdx
  lea    D42(%rip),%rsi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  mov    $0x1,%r12d
  mov    $0xffffffff,%ebp
  lea    D22(%rip),%r13
L3:
  mov    %ebp,%edx
  mov    %ebx,%ecx
  shl    %cl,%edx
  mov    %r12d,%r9d
  sar    %cl,%r9d
  mov    %r12d,%r8d
  shl    %cl,%r8d
  mov    %ebp,%ecx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x42,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
type_infered: string
name: extract_symbolic_data
arguments: tag: D5
type_infered: string
name: extract_symbolic_data
arguments: tag: D53
type_infered: string
name: extract_symbolic_data
arguments: tag: D52
type_infered: string
name: extract_symbolic_data
arguments: tag: D42
type_infered: string
name: extract_symbolic_data
arguments: tag: D21
type_infered: string
name: extract_symbolic_data
arguments: tag: D22
type_infered: string

tool: D4: 1 >> e

tool: D5: 1 << e

tool: D53: ~0 >> e

tool: D52: ~0 << e

tool: D42: e

tool: D21: | %.2s | %.32s | %.32s | %.32s | %.32s | 


tool: D22: | %2d | %32u | %32u | %32u | %32u |


assistant: ```c
int main(void) {
  int e;
  printf("| %.2s | %.32s | %.32s | %.32s | %.32s | \n", "e", "~0 << e", "~0 >> e", "1 << e", "1 >> e");
  for (e = 0; e < 66; ++e) {
    printf("| %2d | %32u | %32u | %32u | %32u |\n", e, ~0 << e, ~0 >> e, 1 << e, 1 >> e);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D32(%rip),%r9
  xor    %eax,%eax
  lea    D47(%rip),%r8
  push   %rbp
  lea    D23(%rip),%rcx
  lea    D15(%rip),%rdx
  xor    %r12d,%r12d
  push   %rbx
  lea    D35(%rip),%rsi
  lea    D36(%rip),%rdi
  mov    $0xffffffff,%ebx
  lea    D37(%rip),%rbp
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L6:
  mov    $0x1,%r8d
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r8d,%r9d
  shl    %cl,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  sar    %cl,%r9d
  shl    %cl,%r8d
  add    $0x1,%r12d
  mov    $0xffffffff,%ecx
  call   <printf@plt>
  cmp    $0x42,%r12d
  jne    L6
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D32
type_guessed: string
name: parse_data
arguments: label: D47
type_guessed: string
name: parse_data
arguments: label: D23
type_guessed: string
name: parse_data
arguments: label: D15
type_guessed: string
name: parse_data
arguments: label: D35
type_guessed: string
name: parse_data
arguments: label: D36
type_guessed: string
name: parse_data
arguments: label: D37
type_guessed: string

tool: D32: 1 >> e

tool: D47: 1 << e

tool: D23: ~0 >> e

tool: D15: ~0 << e

tool: D35: e

tool: D36: | %.2s | %.32s | %.32s | %.32s | %.32s | 


tool: D37: | %2d | %32u | %32u | %32u | %32u |


assistant: ```c
int main(void) {
  int e;
  printf("| %.2s | %.32s | %.32s | %.32s | %.32s | \n", "e", "~0 << e", "~0 >> e", "1 << e", "1 >> e");
  for (e = 0; e < 66; ++e) {
    printf("| %2d | %32u | %32u | %32u | %32u |\n", e, ~0 << e, ~0 >> e, 1 << e, 1 >> e);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D5(%rip),%r9
  xor    %eax,%eax
  lea    D10(%rip),%r8
  push   %rbp
  lea    D11(%rip),%rcx
  lea    D39(%rip),%rdx
  xor    %r12d,%r12d
  push   %rbx
  lea    D31(%rip),%rsi
  lea    D45(%rip),%rdi
  mov    $0xffffffff,%ebx
  lea    D3(%rip),%rbp
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%r8d
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %r8d,%r9d
  shl    %cl,%edx
  mov    %rbp,%rdi
  xor    %eax,%eax
  sar    %cl,%r9d
  shl    %cl,%r8d
  add    $0x1,%r12d
  mov    $0xffffffff,%ecx
  call   <printf@plt>
  cmp    $0x42,%r12d
  jne    L9
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
datatype: string
name: retrieve_data
arguments: data_tag: D10
datatype: string
name: retrieve_data
arguments: data_tag: D11
datatype: string
name: retrieve_data
arguments: data_tag: D39
datatype: string
name: retrieve_data
arguments: data_tag: D31
datatype: string
name: retrieve_data
arguments: data_tag: D45
datatype: string
name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D5: 1 >> e

tool: D10: 1 << e

tool: D11: ~0 >> e

tool: D39: ~0 << e

tool: D31: e

tool: D45: | %.2s | %.32s | %.32s | %.32s | %.32s | 


tool: D3: | %2d | %32u | %32u | %32u | %32u |


assistant: ```c
int main(void) {
  int e;
  printf("| %.2s | %.32s | %.32s | %.32s | %.32s | \n", "e", "~0 << e", "~0 >> e", "1 << e", "1 >> e");
  for (e = 0; e < 66; ++e) {
    printf("| %2d | %32u | %32u | %32u | %32u |\n", e, ~0 << e, ~0 >> e, 1 << e, 1 >> e);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<pmod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <fmod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L3
  movsd  -0x8(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L3:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double pmod(x, y)
double x, y;
{
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<pmod>:
  endbr64
  push   %rbx
  movq   %xmm1,%rbx
  call   <fmod@plt>
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L4
L19:
  pop    %rbx
  ret
L4:
  movq   %rbx,%xmm2
  addsd  %xmm2,%xmm0
  jmp    L19
```

assistant: ```c
double pmod(x, y)
double x, y;
{
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L0
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L0:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(x, y)
double x, y;
{
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L8
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L8:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(x, y)
double x, y;
{
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```