ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<matchRgbColor>:
  endbr64
  push   %r12
  lea    D29(%rip),%rsi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <matchString@plt>
  test   %eax,%eax
  jne    L77
L42:
  xor    %eax,%eax
L59:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L77:
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L42
  lea    D22(%rip),%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <matchString@plt>
  test   %eax,%eax
  je     L42
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L42
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <matchString@plt>
  test   %eax,%eax
  je     L42
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L42
  lea    D17(%rip),%rsi
  mov    %rbp,%rdi
  call   <matchString@plt>
  test   %eax,%eax
  je     L42
  mov    (%rsp),%rax
  mov    %rax,(%rbx)
  mov    $0x1,%eax
  jmp    L59
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D29
type_guessed: string
name: retrieve_data
arguments: label: D22
type_guessed: string
name: retrieve_data
arguments: label: D17
type_guessed: string

tool: D29: rgb(

tool: D22: ,

tool: D17: )

assistant: ```c
int matchRgbColor(const char **strptr) {
  const char *str = *strptr;
  if (!matchString(&str, "rgb("))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ","))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ","))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ")"))
    return 0;
  *strptr = str;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<matchRgbColor>:
  endbr64
  push   %r12
  lea    D27(%rip),%rsi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    (%rdi),%rax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  call   <matchString@plt>
  test   %eax,%eax
  jne    L18
L39:
  xor    %eax,%eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L82
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L18:
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L39
  lea    D22(%rip),%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <matchString@plt>
  test   %eax,%eax
  je     L39
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L39
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <matchString@plt>
  test   %eax,%eax
  je     L39
  mov    %rbp,%rdi
  call   <matchDigitPlus@plt>
  test   %eax,%eax
  je     L39
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  call   <matchString@plt>
  test   %eax,%eax
  je     L39
  mov    (%rsp),%rax
  mov    %rax,(%rbx)
  mov    $0x1,%eax
  jmp    L20
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D27
type_guessed: string
name: retrieve_data
arguments: tag: D22
type_guessed: string
name: retrieve_data
arguments: tag: D3
type_guessed: string

tool: D27: rgb(

tool: D22: ,

tool: D3: )

assistant: ```c
int matchRgbColor(const char **strptr) {
  const char *str = *strptr;
  if (!matchString(&str, "rgb("))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ","))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ","))
    return 0;
  if (!matchDigitPlus(&str))
    return 0;
  if (!matchString(&str, ")"))
    return 0;
  *strptr = str;
  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    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  cmpl   $0x6,-0x34(%rbp)
  je     L8
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%eax
  jmp    L5
L8:
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x28,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  jmp    L37
L34:
  mov    -0x24(%rbp),%ecx
  mov    -0x28(%rbp),%edx
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  mov    %ecx,%esi
  mov    %edx,%edi
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <sho_wf@plt>
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  -0x20(%rbp),%xmm0
  addsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
L37:
  movsd  -0x8(%rbp),%xmm0
  comisd -0x20(%rbp),%xmm0
  jae    L34
  mov    $0x0,%eax
L5:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_infered: string
name: retrieve_data
arguments: tag: D6
type_infered: string

tool: D9: Usage: ./sho_print mw n l rstep rmax

tool: D6: %.5lf	%.8le


assistant: ```c
int main(int argc, char *argv[]) {
  int n, l;
  double mw, r, rstep, rmax;
  if (argc != 6) {
    printf("Usage: ./sho_print mw n l rstep rmax\n");
    return 1;
  }
  mw = atof(argv[1]);
  n = atoi(argv[2]);
  l = atoi(argv[3]);
  rstep = atof(argv[4]);
  rmax = atof(argv[5]);
  for (r = 0.; r <= rmax; r += rstep)
    printf("%.5lf\t%.8le\n", r, sho_wf(r, mw, n, l));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  cmp    $0x6,%edi
  je     L42
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L4:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  call   <atof@plt>
  movq   %xmm0,%r13
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%r12d
  mov    0x18(%rbx),%rdi
  call   <atoi@plt>
  mov    %eax,%ebp
  mov    0x20(%rbx),%rdi
  call   <atof@plt>
  movsd  %xmm0,0x8(%rsp)
  mov    0x28(%rbx),%rdi
  call   <atof@plt>
  movq   %xmm0,%rbx
  pxor   %xmm5,%xmm5
  movsd  %xmm5,(%rsp)
  lea    D19(%rip),%r14
  comisd %xmm5,%xmm0
  jb     L34
L27:
  mov    %ebp,%esi
  mov    %r12d,%edi
  movq   %r13,%xmm1
  movsd  (%rsp),%xmm0
  call   <sho_wf@plt>
  mov    %eax,%esi
  movsd  (%rsp),%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  (%rsp),%xmm2
  addsd  0x8(%rsp),%xmm2
  movsd  %xmm2,(%rsp)
  movq   %rbx,%xmm4
  comisd %xmm2,%xmm4
  jae    L27
  mov    $0x0,%eax
  jmp    L4
L34:
  mov    $0x0,%eax
  jmp    L4
```

tool_calls: name: unpack_data
arguments: data_tag: D14
data_type: string
name: unpack_data
arguments: data_tag: D19
data_type: string

tool: D14: Usage: ./sho_print mw n l rstep rmax

tool: D19: %.5lf	%.8le


assistant: ```c
int main(int argc, char *argv[]) {
  int n, l;
  double mw, r, rstep, rmax;
  if (argc != 6) {
    printf("Usage: ./sho_print mw n l rstep rmax\n");
    return 1;
  }
  mw = atof(argv[1]);
  n = atoi(argv[2]);
  l = atoi(argv[3]);
  rstep = atof(argv[4]);
  rmax = atof(argv[5]);
  for (r = 0.; r <= rmax; r += rstep)
    printf("%.5lf\t%.8le\n", r, sho_wf(r, mw, n, l));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  cmp    $0x6,%edi
  je     L1
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L33:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L1:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%r12
  call   <atof@plt>
  mov    0x10(%r12),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <atoi@plt>
  mov    0x18(%r12),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  mov    0x20(%r12),%rdi
  mov    %eax,%ebx
  call   <atof@plt>
  mov    0x28(%r12),%rdi
  lea    D18(%rip),%r12
  movsd  %xmm0,0x18(%rsp)
  call   <atof@plt>
  movsd  0x8(%rsp),%xmm1
  movq   %xmm0,%r14
  pxor   %xmm0,%xmm0
  movq   %r14,%xmm4
  comisd %xmm0,%xmm4
  jb     L10
  nopl   0x0(%rax)
L0:
  mov    %ebx,%esi
  mov    %ebp,%edi
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm0,0x8(%rsp)
  call   <sho_wf@plt>
  movsd  0x8(%rsp),%xmm0
  mov    %r12,%rdi
  mov    %eax,%esi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  movq   %r14,%xmm3
  addsd  0x18(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm1
  comisd %xmm0,%xmm3
  jae    L0
L10:
  xor    %eax,%eax
  jmp    L33
```

tool_calls: name: unpack_data
arguments: label: D12
type_infered: string
name: unpack_data
arguments: label: D18
type_infered: string

tool: D12: Usage: ./sho_print mw n l rstep rmax

tool: D18: %.5lf	%.8le


assistant: ```c
int main(int argc, char *argv[]) {
  int n, l;
  double mw, r, rstep, rmax;
  if (argc != 6) {
    printf("Usage: ./sho_print mw n l rstep rmax\n");
    return 1;
  }
  mw = atof(argv[1]);
  n = atoi(argv[2]);
  l = atoi(argv[3]);
  rstep = atof(argv[4]);
  rmax = atof(argv[5]);
  for (r = 0.; r <= rmax; r += rstep)
    printf("%.5lf\t%.8le\n", r, sho_wf(r, mw, n, l));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  cmp    $0x6,%edi
  je     L1
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%eax
L38:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L1:
  mov    0x8(%rsi),%rdi
  mov    %rsi,%r12
  call   <atof@plt>
  mov    0x10(%r12),%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <atoi@plt>
  mov    0x18(%r12),%rdi
  mov    %eax,%ebp
  call   <atoi@plt>
  mov    0x20(%r12),%rdi
  mov    %eax,%ebx
  call   <atof@plt>
  mov    0x28(%r12),%rdi
  lea    D3(%rip),%r12
  movsd  %xmm0,0x18(%rsp)
  call   <atof@plt>
  movsd  0x8(%rsp),%xmm1
  movq   %xmm0,%r14
  pxor   %xmm0,%xmm0
  movq   %r14,%xmm4
  comisd %xmm0,%xmm4
  jb     L30
  nopl   0x0(%rax)
L24:
  mov    %ebx,%esi
  mov    %ebp,%edi
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm0,0x8(%rsp)
  call   <sho_wf@plt>
  movsd  0x8(%rsp),%xmm0
  mov    %r12,%rdi
  mov    %eax,%esi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm0
  movq   %r14,%xmm3
  addsd  0x18(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm1
  comisd %xmm0,%xmm3
  jae    L24
L30:
  xor    %eax,%eax
  jmp    L38
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string
name: parse_data
arguments: label: D3
datatype: string

tool: D2: Usage: ./sho_print mw n l rstep rmax

tool: D3: %.5lf	%.8le


assistant: ```c
int main(int argc, char *argv[]) {
  int n, l;
  double mw, r, rstep, rmax;
  if (argc != 6) {
    printf("Usage: ./sho_print mw n l rstep rmax\n");
    return 1;
  }
  mw = atof(argv[1]);
  n = atoi(argv[2]);
  l = atoi(argv[3]);
  rstep = atof(argv[4]);
  rmax = atof(argv[5]);
  for (r = 0.; r <= rmax; r += rstep)
    printf("%.5lf\t%.8le\n", r, sho_wf(r, mw, n, l));
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<arikkei_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %edx,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  cmpl   $0x0,-0x50(%rbp)
  jns    L10
  mov    -0x38(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  negl   -0x50(%rbp)
L10:
  movl   $0x0,-0x34(%rbp)
L12:
  mov    -0x50(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%edx
  addl   $0x1,-0x34(%rbp)
  mov    $0x20,%eax
  sub    -0x34(%rbp),%eax
  cltq
  mov    %dl,-0x30(%rbp,%rax,1)
  mov    -0x50(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x50(%rbp)
  cmpl   $0x0,-0x50(%rbp)
  jg     L12
  mov    $0x20,%eax
  sub    -0x34(%rbp),%eax
  lea    -0x30(%rbp),%rdx
  cltq
  lea    (%rdx,%rax,1),%rsi
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x34(%rbp),%eax
  add    %eax,-0x38(%rbp)
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
unsigned int arikkei_itoa(unsigned char *buf, unsigned int len, int val) {
  char c[32];
  int p, i;
  p = 0;
  if (val < 0) {
    buf[p++] = '-';
    val = -val;
  }
  i = 0;
  do {
    c[32 - (++i)] = '0' + (val % 10);
    val /= 10;
  } while (val > 0);
  memcpy(buf + p, &c[32 - i], i);
  p += i;
  buf[p] = 0;
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<arikkei_itoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  test   %edx,%edx
  js     L37
L28:
  lea    0x1f(%rsp),%rdi
  mov    $0x0,%ebx
  mov    $0x20,%r9d
L14:
  add    $0x1,%ebx
  mov    %r9d,%r8d
  sub    %ebx,%r8d
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,(%rdi)
  mov    %edx,%ecx
  mov    %eax,%edx
  sub    $0x1,%rdi
  cmp    $0x9,%ecx
  jg     L14
  movslq %r8d,%r8
  lea    (%rsp,%r8,1),%rsi
  movslq %ebp,%rdi
  add    %r12,%rdi
  mov    %ebx,%edx
  call   <memcpy@plt>
  lea    0x0(%rbp,%rbx,1),%eax
  movslq %eax,%rdx
  movb   $0x0,(%r12,%rdx,1)
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  movb   $0x2d,(%rdi)
  neg    %edx
  mov    $0x1,%ebp
  jmp    L28
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int arikkei_itoa(unsigned char *buf, unsigned int len, int val) {
  char c[32];
  int p, i;
  p = 0;
  if (val < 0) {
    buf[p++] = '-';
    val = -val;
  }
  i = 0;
  do {
    c[32 - (++i)] = '0' + (val % 10);
    val /= 10;
  } while (val > 0);
  memcpy(buf + p, &c[32 - i], i);
  p += i;
  buf[p] = 0;
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<arikkei_itoa>:
  endbr64
  push   %r12
  mov    %rdi,%r9
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jns    L7
  movb   $0x2d,(%rdi)
  neg    %edx
  lea    0x1(%rdi),%r9
  mov    $0x1,%r12d
L7:
  lea    0x1f(%rsp),%rdi
  xor    %ebx,%ebx
  mov    $0x20,%r11d
  mov    $0xcccccccd,%r10d
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %edx,%eax
  mov    %edx,%ecx
  add    $0x1,%ebx
  mov    %r11d,%r8d
  imul   %r10,%rax
  sub    %ebx,%r8d
  sub    $0x1,%rdi
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,0x1(%rdi)
  mov    %edx,%ecx
  mov    %eax,%edx
  cmp    $0x9,%ecx
  jg     L15
  movslq %r8d,%r8
  mov    %ebx,%edx
  mov    %r9,%rdi
  lea    (%rsp,%r8,1),%rsi
  call   <memcpy@plt>
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rdx
  movb   $0x0,0x0(%rbp,%rdx,1)
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int arikkei_itoa(unsigned char *buf, unsigned int len, int val) {
  char c[32];
  int p, i;
  p = 0;
  if (val < 0) {
    buf[p++] = '-';
    val = -val;
  }
  i = 0;
  do {
    c[32 - (++i)] = '0' + (val % 10);
    val /= 10;
  } while (val > 0);
  memcpy(buf + p, &c[32 - i], i);
  p += i;
  buf[p] = 0;
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<arikkei_itoa>:
  endbr64
  push   %r12
  mov    %rdi,%r9
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jns    L12
  movb   $0x2d,(%rdi)
  neg    %edx
  lea    0x1(%rdi),%r9
  mov    $0x1,%r12d
L12:
  lea    0x1f(%rsp),%rdi
  xor    %ebx,%ebx
  mov    $0x20,%r11d
  mov    $0xcccccccd,%r10d
  nopl   0x0(%rax,%rax,1)
L23:
  mov    %edx,%eax
  mov    %edx,%ecx
  add    $0x1,%ebx
  mov    %r11d,%r8d
  imul   %r10,%rax
  sub    %ebx,%r8d
  sub    $0x1,%rdi
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,0x1(%rdi)
  mov    %edx,%ecx
  mov    %eax,%edx
  cmp    $0x9,%ecx
  jg     L23
  movslq %r8d,%r8
  mov    %ebx,%edx
  mov    %r9,%rdi
  lea    (%rsp,%r8,1),%rsi
  call   <memcpy@plt>
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rdx
  movb   $0x0,0x0(%rbp,%rdx,1)
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int arikkei_itoa(unsigned char *buf, unsigned int len, int val) {
  char c[32];
  int p, i;
  p = 0;
  if (val < 0) {
    buf[p++] = '-';
    val = -val;
  }
  i = 0;
  do {
    c[32 - (++i)] = '0' + (val % 10);
    val /= 10;
  } while (val > 0);
  memcpy(buf + p, &c[32 - i], i);
  p += i;
  buf[p] = 0;
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <open_max@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_guessed: string

tool: D7: open_max :%d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("open_max :%d\n", open_max());
  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
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <open_max@plt>
  mov    %eax,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
datatype: string

tool: D7: open_max :%d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("open_max :%d\n", open_max());
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <open_max@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
data_type: string

tool: D5: open_max :%d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("open_max :%d\n", open_max());
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <open_max@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
data_type: string

tool: D5: open_max :%d


assistant: ```c
int main(int argc, char *argv[]) {
  printf("open_max :%d\n", open_max());
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
void getcwd() {
  _PROCEDURE_LINKAGE_TABLE_();
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
void getcwd() {
  _PROCEDURE_LINKAGE_TABLE_();
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getcwd>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <_PROCEDURE_LINKAGE_TABLE_@plt>
```

assistant: ```c
void getcwd() {
  _PROCEDURE_LINKAGE_TABLE_();
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getcwd>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <_PROCEDURE_LINKAGE_TABLE_@plt>
```

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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x61000(%rsp),%r11
L8:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L8
  sub    $0xbd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x61bd0(%rbp),%rax
  mov    $0x186a0,%esi
  mov    %rax,%rdi
  call   <preparePrimeSieve@plt>
L33:
  lea    -0x140(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  movzbl -0x140(%rbp),%eax
  cmp    $0x30,%al
  jne    L40
  movzbl -0x13f(%rbp),%eax
  test   %al,%al
  je     L3
L40:
  lea    -0x61bd0(%rbp),%rdx
  lea    -0x140(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <findMaxPrimeSubstr@plt>
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L33
L3:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: string

tool: D2: %d


assistant: ```c
int main(void) {
  int primeSieve[100000 + 1];
  char strNum[300 + 1];
  preparePrimeSieve(primeSieve, 100000);
  while (1) {
    gets(strNum);
    if (strNum[0] == '0' && strNum[1] == '\0')
      break;
    printf("%d\n", findMaxPrimeSubstr(strNum, primeSieve));
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  lea    -0x61000(%rsp),%r11
L34:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L34
  sub    $0xbd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x61bc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x186a0,%esi
  call   <preparePrimeSieve@plt>
  lea    D3(%rip),%rbx
  jmp    L48
L18:
  mov    %rsp,%rsi
  lea    0x61a90(%rsp),%rdi
  call   <findMaxPrimeSubstr@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
L48:
  lea    0x61a90(%rsp),%rdi
  call   <gets@plt>
  cmpb   $0x30,0x61a90(%rsp)
  jne    L18
  cmpb   $0x0,0x61a91(%rsp)
  jne    L18
  mov    0x61bc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  mov    $0x0,%eax
  add    $0x61bd0,%rsp
  pop    %rbx
  ret
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
datatype: string

tool: D3: %d


assistant: ```c
int main(void) {
  int primeSieve[100000 + 1];
  char strNum[300 + 1];
  preparePrimeSieve(primeSieve, 100000);
  while (1) {
    gets(strNum);
    if (strNum[0] == '0' && strNum[1] == '\0')
      break;
    printf("%d\n", findMaxPrimeSubstr(strNum, primeSieve));
  }
  return 0;
}
```