ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<MY_strtok_r>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  test   %rdi,%rdi
  je     L94
L13:
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L91
L67:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L58
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L67
L91:
  mov    $0x0,%ebx
  jmp    L122
L94:
  test   %rdx,%rdx
  je     L175
  mov    (%rdx),%rbx
  test   %rbx,%rbx
  jne    L13
  jmp    L122
L58:
  cmpb   $0x0,(%rbx)
  je     L104
  lea    0x1(%rbx),%r13
  movzbl 0x1(%rbx),%esi
  test   %sil,%sil
  je     L70
L20:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L79
  add    $0x1,%r13
  movzbl 0x0(%r13),%esi
  test   %sil,%sil
  jne    L20
L70:
  mov    %r13,%rax
  jmp    L154
L79:
  cmpb   $0x0,0x0(%r13)
  je     L23
  lea    0x1(%r13),%rax
  movb   $0x0,0x0(%r13)
L154:
  test   %r12,%r12
  je     L122
  mov    %rax,(%r12)
L122:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  mov    %r13,%rax
  jmp    L154
L175:
  mov    %rdx,%rbx
  jmp    L122
L104:
  mov    %rax,%rbx
  jmp    L122
```

assistant: ```c
char *MY_strtok_r(char *str, const char *delim, char **saveptr) {
  char *pbegin;
  char *pend = NULL;
  if (str) {
    pbegin = str;
  } else if (saveptr && *saveptr) {
    pbegin = *saveptr;
  } else {
    return NULL;
  }
  for (; *pbegin && strchr(delim, *pbegin) != NULL; pbegin++)
    ;
  if (!*pbegin) {
    return NULL;
  }
  for (pend = pbegin + 1; *pend && strchr(delim, *pend) == NULL; pend++)
    ;
  if (*pend) {
    *pend++ = '\0';
  }
  if (saveptr) {
    *saveptr = pend;
  }
  return pbegin;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<MY_strtok_r>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  jne    L101
  jmp    L104
  nopl   0x0(%rax)
L1:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L85
  add    $0x1,%r12
L101:
  movsbl (%r12),%esi
  test   %sil,%sil
  jne    L1
L44:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L85:
  cmpb   $0x0,(%r12)
  je     L44
  movsbl 0x1(%r12),%esi
  lea    0x1(%r12),%r13
  test   %sil,%sil
  jne    L65
  jmp    L121
  nopl   0x0(%rax)
L100:
  movsbl 0x1(%r13),%esi
  add    $0x1,%r13
  test   %sil,%sil
  je     L121
L65:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L100
  cmpb   $0x0,0x0(%r13)
  je     L121
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
L121:
  test   %rbp,%rbp
  je     L117
  mov    %r13,0x0(%rbp)
L117:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L104:
  test   %rdx,%rdx
  je     L44
  mov    (%rdx),%r12
  test   %r12,%r12
  jne    L101
  jmp    L44
```

assistant: ```c
char *MY_strtok_r(char *str, const char *delim, char **saveptr) {
  char *pbegin;
  char *pend = NULL;
  if (str) {
    pbegin = str;
  } else if (saveptr && *saveptr) {
    pbegin = *saveptr;
  } else {
    return NULL;
  }
  for (; *pbegin && strchr(delim, *pbegin) != NULL; pbegin++)
    ;
  if (!*pbegin) {
    return NULL;
  }
  for (pend = pbegin + 1; *pend && strchr(delim, *pend) == NULL; pend++)
    ;
  if (*pend) {
    *pend++ = '\0';
  }
  if (saveptr) {
    *saveptr = pend;
  }
  return pbegin;
}
```
user: What is the c source code of the assembly code below?

```asm
<MY_strtok_r>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  jne    L29
  jmp    L52
  nopl   0x0(%rax)
L106:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L104
  add    $0x1,%r12
L29:
  movsbl (%r12),%esi
  test   %sil,%sil
  jne    L106
L27:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L104:
  cmpb   $0x0,(%r12)
  je     L27
  movsbl 0x1(%r12),%esi
  lea    0x1(%r12),%r13
  test   %sil,%sil
  jne    L66
  jmp    L61
  nopl   0x0(%rax)
L19:
  movsbl 0x1(%r13),%esi
  add    $0x1,%r13
  test   %sil,%sil
  je     L61
L66:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L19
  cmpb   $0x0,0x0(%r13)
  je     L61
  movb   $0x0,0x0(%r13)
  add    $0x1,%r13
L61:
  test   %rbp,%rbp
  je     L91
  mov    %r13,0x0(%rbp)
L91:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L52:
  test   %rdx,%rdx
  je     L27
  mov    (%rdx),%r12
  test   %r12,%r12
  jne    L29
  jmp    L27
```

assistant: ```c
char *MY_strtok_r(char *str, const char *delim, char **saveptr) {
  char *pbegin;
  char *pend = NULL;
  if (str) {
    pbegin = str;
  } else if (saveptr && *saveptr) {
    pbegin = *saveptr;
  } else {
    return NULL;
  }
  for (; *pbegin && strchr(delim, *pbegin) != NULL; pbegin++)
    ;
  if (!*pbegin) {
    return NULL;
  }
  for (pend = pbegin + 1; *pend && strchr(delim, *pend) == NULL; pend++)
    ;
  if (*pend) {
    *pend++ = '\0';
  }
  if (saveptr) {
    *saveptr = pend;
  }
  return pbegin;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<move_part>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9d,-0x2c(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  setae  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  setae  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x0,-0x2c(%rbp)
  je     L0
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L0:
  mov    -0x28(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,4),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  nop
  leave
  ret
```

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

tool: D9: Moving size %d from %d to %d.


assistant: ```c
void move_part(int *A, size_t N, size_t target, size_t source, size_t size, int show_steps) {
  assert(source + size <= N);
  assert(target + size <= N);
  if (show_steps) {
    printf("Moving size %d from %d to %d.\n", size, source, target);
  }
  memmove(A + target, A + source, size * sizeof(int));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<move_part>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %rdx,%rbp
  mov    %rcx,%r12
  mov    %r8,%rbx
  mov    %r9d,%r15d
  lea    (%rcx,%r8,1),%rax
  cmp    %rsi,%rax
  setbe  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  lea    (%rbx,%rbp,1),%rax
  cmp    %r14,%rax
  setbe  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  test   %r15d,%r15d
  jne    L4
L6:
  lea    0x0(,%rbx,4),%rdx
  lea    0x0(%r13,%r12,4),%rsi
  lea    0x0(%r13,%rbp,4),%rdi
  call   <memmove@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %rbx,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L6
```

tool_calls: name: read_data
arguments: data_tag: D7
datatype: string

tool: D7: Moving size %d from %d to %d.


assistant: ```c
void move_part(int *A, size_t N, size_t target, size_t source, size_t size, int show_steps) {
  assert(source + size <= N);
  assert(target + size <= N);
  if (show_steps) {
    printf("Moving size %d from %d to %d.\n", size, source, target);
  }
  memmove(A + target, A + source, size * sizeof(int));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<move_part>:
  endbr64
  push   %r15
  lea    (%rcx,%r8,1),%rax
  mov    %r9d,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x8,%rsp
  cmp    %rsi,%rax
  setbe  %dil
  call   <assert@plt>
  lea    (%rbx,%rbp,1),%rax
  xor    %edi,%edi
  cmp    %r14,%rax
  setbe  %dil
  call   <assert@plt>
  test   %r15d,%r15d
  jne    L18
L9:
  add    $0x8,%rsp
  lea    0x0(%r13,%r12,4),%rsi
  lea    0x0(%r13,%rbp,4),%rdi
  lea    0x0(,%rbx,4),%rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <memmove@plt>
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  lea    D5(%rip),%rdi
  mov    %rbx,%rsi
  call   <printf@plt>
  jmp    L9
```

tool_calls: name: read_data
arguments: label: D5
type: string

tool: D5: Moving size %d from %d to %d.


assistant: ```c
void move_part(int *A, size_t N, size_t target, size_t source, size_t size, int show_steps) {
  assert(source + size <= N);
  assert(target + size <= N);
  if (show_steps) {
    printf("Moving size %d from %d to %d.\n", size, source, target);
  }
  memmove(A + target, A + source, size * sizeof(int));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<move_part>:
  endbr64
  push   %r15
  lea    (%rcx,%r8,1),%rax
  mov    %r9d,%r15d
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  xor    %edi,%edi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x8,%rsp
  cmp    %rsi,%rax
  setbe  %dil
  call   <assert@plt>
  lea    (%rbx,%rbp,1),%rax
  xor    %edi,%edi
  cmp    %r14,%rax
  setbe  %dil
  call   <assert@plt>
  test   %r15d,%r15d
  jne    L17
L11:
  add    $0x8,%rsp
  lea    0x0(%r13,%r12,4),%rsi
  lea    0x0(%r13,%rbp,4),%rdi
  lea    0x0(,%rbx,4),%rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <memmove@plt>
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  lea    D9(%rip),%rdi
  mov    %rbx,%rsi
  call   <printf@plt>
  jmp    L11
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string

tool: D9: Moving size %d from %d to %d.


assistant: ```c
void move_part(int *A, size_t N, size_t target, size_t source, size_t size, int show_steps) {
  assert(source + size <= N);
  assert(target + size <= N);
  if (show_steps) {
    printf("Moving size %d from %d to %d.\n", size, source, target);
  }
  memmove(A + target, A + source, size * sizeof(int));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<modequ>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rcx
  lea    -0x18(%rbp),%rdx
  mov    -0x2c(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <egcd@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x28(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L4
  mov    -0x2c(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  imul   %edx,%eax
  mov    -0x14(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jns    L46
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
L46:
  movl   $0x0,-0x10(%rbp)
  jmp    L0
L16:
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  je     L9
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L76
L9:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L76:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  addl   $0x1,-0x10(%rbp)
L0:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L16
  jmp    L70
L4:
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L70:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D24
datatype: string
name: parse_data
arguments: data_tag: D25
datatype: string
name: parse_data
arguments: data_tag: D21
datatype: string

tool: D24: %d 

tool: D25: %d


tool: D21: No Answer.

assistant: ```c
void modequ(int a, int b, int n) {
  int d, x, y, i;
  d = egcd(a, n, &x, &y);
  if (b % d == 0) {
    y = n / d;
    x = (x * (b / d)) % y;
    if (x < 0)
      x += y;
    for (i = 0; i < d; i++) {
      if (i != d - 1)
        printf("%d ", x);
      else
        printf("%d\n", x);
      x += y;
    }
  } else
    printf("No Answer.\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<modequ>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %esi,%r12d
  mov    %edx,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rdx
  mov    %r13d,%esi
  call   <egcd@plt>
  mov    %eax,%ebp
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  test   %edx,%edx
  jne    L49
  mov    %edx,%ebx
  mov    %r13d,%eax
  cltd
  idiv   %ebp
  mov    %eax,%r13d
  mov    %eax,0x4(%rsp)
  mov    %r12d,%eax
  cltd
  idiv   %ebp
  imul   (%rsp),%eax
  cltd
  idiv   %r13d
  lea    0x0(%r13,%rdx,1),%ecx
  test   %edx,%edx
  cmovs  %ecx,%edx
  mov    %edx,(%rsp)
  test   %ebp,%ebp
  jle    L47
  lea    -0x1(%rbp),%r12d
  lea    D7(%rip),%r14
  lea    D1(%rip),%r13
  jmp    L30
L55:
  mov    (%rsp),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L59:
  mov    0x4(%rsp),%eax
  add    %eax,(%rsp)
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  je     L47
L30:
  cmp    %ebx,%r12d
  je     L55
  mov    (%rsp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L59
L49:
  lea    D14(%rip),%rdi
  call   <puts@plt>
L47:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D7
datatype: string
name: unpack_data
arguments: label: D1
datatype: string
name: unpack_data
arguments: label: D14
datatype: string

tool: D7: %d


tool: D1: %d 

tool: D14: No Answer.

assistant: ```c
void modequ(int a, int b, int n) {
  int d, x, y, i;
  d = egcd(a, n, &x, &y);
  if (b % d == 0) {
    y = n / d;
    x = (x * (b / d)) % y;
    if (x < 0)
      x += y;
    for (i = 0; i < d; i++) {
      if (i != d - 1)
        printf("%d ", x);
      else
        printf("%d\n", x);
      x += y;
    }
  } else
    printf("No Answer.\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<modequ>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  mov    %r12d,%esi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  lea    0x4(%rsp),%rcx
  call   <egcd@plt>
  mov    %eax,%ebx
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L25
  mov    %eax,%ebp
  mov    %r12d,%eax
  lea    D14(%rip),%r13
  mov    %edx,%r14d
  cltd
  idiv   %ebx
  mov    %eax,%r12d
  mov    %eax,0x4(%rsp)
  mov    (%rsp),%eax
  imul   %ebp,%eax
  lea    -0x1(%rbx),%ebp
  cltd
  idiv   %r12d
  lea    (%r12,%rdx,1),%ecx
  test   %edx,%edx
  lea    D17(%rip),%r12
  cmovns %edx,%ecx
  mov    %ecx,(%rsp)
  mov    %ecx,%esi
  test   %ebx,%ebx
  jg     L49
  jmp    L64
  nopl   (%rax)
L38:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L59:
  mov    0x4(%rsp),%esi
  add    $0x1,%r14d
  add    (%rsp),%esi
  mov    %esi,(%rsp)
  cmp    %r14d,%ebx
  je     L64
L49:
  cmp    %r14d,%ebp
  jne    L38
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L25:
  lea    D12(%rip),%rdi
  call   <puts@plt>
L64:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D14
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string
name: parse_data
arguments: data_label: D12
type_infered: string

tool: D14: %d


tool: D17: %d 

tool: D12: No Answer.

assistant: ```c
void modequ(int a, int b, int n) {
  int d, x, y, i;
  d = egcd(a, n, &x, &y);
  if (b % d == 0) {
    y = n / d;
    x = (x * (b / d)) % y;
    if (x < 0)
      x += y;
    for (i = 0; i < d; i++) {
      if (i != d - 1)
        printf("%d ", x);
      else
        printf("%d\n", x);
      x += y;
    }
  } else
    printf("No Answer.\n");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<modequ>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  mov    %r12d,%esi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  lea    0x4(%rsp),%rcx
  call   <egcd@plt>
  mov    %eax,%ebx
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  jne    L96
  mov    %eax,%ebp
  mov    %r12d,%eax
  mov    %edx,%r14d
  cltd
  idiv   %ebx
  mov    %eax,%r12d
  mov    %eax,0x4(%rsp)
  mov    (%rsp),%eax
  imul   %ebp,%eax
  cltd
  idiv   %r12d
  mov    %edx,%esi
  test   %edx,%edx
  js     L73
L13:
  mov    %esi,(%rsp)
  lea    -0x1(%rbx),%ebp
  lea    D27(%rip),%r13
  lea    D21(%rip),%r12
  test   %ebx,%ebx
  jle    L41
  xchg   %ax,%ax
L91:
  cmp    %r14d,%ebp
  je     L17
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r14d
  call   <printf@plt>
  mov    (%rsp),%esi
  add    0x4(%rsp),%esi
  mov    %esi,(%rsp)
  cmp    %ebx,%r14d
  jne    L91
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L17:
  mov    %r13,%rdi
  xor    %eax,%eax
  lea    0x1(%rbp),%r14d
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  mov    %esi,(%rsp)
  cmp    %r14d,%ebx
  jne    L91
  jmp    L41
  nop
L73:
  add    %r12d,%esi
  jmp    L13
  nopl   (%rax)
L96:
  lea    D5(%rip),%rdi
  call   <puts@plt>
  jmp    L41
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D27
type: string
name: unpack_data
arguments: tag: D21
type: string
name: unpack_data
arguments: tag: D5
type: string

tool: D27: %d


tool: D21: %d 

tool: D5: No Answer.

assistant: ```c
void modequ(int a, int b, int n) {
  int d, x, y, i;
  d = egcd(a, n, &x, &y);
  if (b % d == 0) {
    y = n / d;
    x = (x * (b / d)) % y;
    if (x < 0)
      x += y;
    for (i = 0; i < d; i++) {
      if (i != d - 1)
        printf("%d ", x);
      else
        printf("%d\n", x);
      x += y;
    }
  } else
    printf("No Answer.\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stdlib_strided_dmeankbn2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L22
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  jmp    L3
L22:
  cmpl   $0x1,-0x4(%rbp)
  je     L0
  cmpl   $0x0,-0x8(%rbp)
  jne    L9
L0:
  mov    -0x10(%rbp),%rax
  movsd  (%rax),%xmm0
  jmp    L3
L9:
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <stdlib_strided_dsumkbn2@plt>
  movq   %xmm0,%rax
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x4(%rbp),%xmm1
  movq   %rax,%xmm0
  divsd  %xmm1,%xmm0
L3:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double stdlib_strided_dmeankbn2(const int64_t N, const double *X, const int64_t stride) {
  if (N <= 0) {
    return 0.0 / 0.0;
  }
  if (N == 1 || stride == 0) {
    return X[0];
  }
  return stdlib_strided_dsumkbn2(N, X, stride) / (double)N;
}
```
user: What is the c source code of the assembly code below:

```asm
<stdlib_strided_dmeankbn2>:
  endbr64
  test   %edi,%edi
  jle    L33
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%edi
  je     L37
  test   %edx,%edx
  je     L37
  call   <stdlib_strided_dsumkbn2@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  divsd  %xmm1,%xmm0
  jmp    L29
L33:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
L37:
  movsd  (%rsi),%xmm0
L29:
  pop    %rbx
  ret
```

assistant: ```c
double stdlib_strided_dmeankbn2(const int64_t N, const double *X, const int64_t stride) {
  if (N <= 0) {
    return 0.0 / 0.0;
  }
  if (N == 1 || stride == 0) {
    return X[0];
  }
  return stdlib_strided_dsumkbn2(N, X, stride) / (double)N;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<stdlib_strided_dmeankbn2>:
  endbr64
  test   %edi,%edi
  jle    L2
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%edi
  je     L23
  test   %edx,%edx
  je     L23
  call   <stdlib_strided_dsumkbn2@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  pop    %rbx
  divsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  movsd  (%rsi),%xmm0
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
```

assistant: ```c
double stdlib_strided_dmeankbn2(const int64_t N, const double *X, const int64_t stride) {
  if (N <= 0) {
    return 0.0 / 0.0;
  }
  if (N == 1 || stride == 0) {
    return X[0];
  }
  return stdlib_strided_dsumkbn2(N, X, stride) / (double)N;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stdlib_strided_dmeankbn2>:
  endbr64
  test   %edi,%edi
  jle    L16
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%edi
  je     L12
  test   %edx,%edx
  je     L12
  call   <stdlib_strided_dsumkbn2@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  pop    %rbx
  divsd  %xmm1,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  movsd  (%rsi),%xmm0
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L16:
  pxor   %xmm0,%xmm0
  divsd  %xmm0,%xmm0
  ret
```

assistant: ```c
double stdlib_strided_dmeankbn2(const int64_t N, const double *X, const int64_t stride) {
  if (N <= 0) {
    return 0.0 / 0.0;
  }
  if (N == 1 || stride == 0) {
    return X[0];
  }
  return stdlib_strided_dsumkbn2(N, X, stride) / (double)N;
}
```