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

```asm
<minimum>:
  endbr64
  movd   (%rdi),%xmm3
  movdqu 0x4(%rdi),%xmm1
  movdqu 0x4(%rdi),%xmm4
  movdqu 0x14(%rdi),%xmm5
  pshufd $0x0,%xmm3,%xmm0
  mov    0x24(%rdi),%edx
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm4,%xmm1
  por    %xmm0,%xmm1
  movdqu 0x14(%rdi),%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm1
  pandn  %xmm5,%xmm0
  por    %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x8,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm2,%xmm1
  por    %xmm0,%xmm1
  movdqa %xmm1,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm1
  pandn  %xmm2,%xmm0
  por    %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%eax
  cmovg  %edx,%eax
  ret
```

assistant: ```c
int minimum(int values[10]) {
  int minValue, i;
  minValue = values[0];
  for (i = 1; i < 10; ++i)
    if (values[i] < minValue)
      minValue = values[i];
  return minValue;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TranspositionGetColumns>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
L5:
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  test   %eax,%eax
  jle    L5
  mov    -0xc(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L5
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

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

tool: D7: %d

assistant: ```c
int TranspositionGetColumns(int letters, int *key_columns) {
  int tmp = 0;
  do {
    scanf("%d", &tmp);
  } while (tmp < 1 || tmp >= letters);
  *key_columns = tmp;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<TranspositionGetColumns>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    D7(%rip),%rbp
L6:
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jge    L6
  test   %eax,%eax
  jle    L6
  mov    %eax,(%r12)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
datatype: string

tool: D7: %d

assistant: ```c
int TranspositionGetColumns(int letters, int *key_columns) {
  int tmp = 0;
  do {
    scanf("%d", &tmp);
  } while (tmp < 1 || tmp >= letters);
  *key_columns = tmp;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<TranspositionGetColumns>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%r12
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jge    L18
  test   %eax,%eax
  jle    L18
  mov    %eax,0x0(%r13)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
datatype: string

tool: D8: %d

assistant: ```c
int TranspositionGetColumns(int letters, int *key_columns) {
  int tmp = 0;
  do {
    scanf("%d", &tmp);
  } while (tmp < 1 || tmp >= letters);
  *key_columns = tmp;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TranspositionGetColumns>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%r12
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebx,%eax
  jge    L4
  test   %eax,%eax
  jle    L4
  mov    %eax,0x0(%r13)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_guessed: string

tool: D2: %d

assistant: ```c
int TranspositionGetColumns(int letters, int *key_columns) {
  int tmp = 0;
  do {
    scanf("%d", &tmp);
  } while (tmp < 1 || tmp >= letters);
  *key_columns = tmp;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_oled_dat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x420,%rsp
  mov    %edi,-0x414(%rbp)
  mov    %rsi,-0x420(%rbp)
  mov    %edx,-0x418(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0xff,-0x418(%rbp)
  jle    L10
  mov    $0xffffffff,%eax
  jmp    L11
L10:
  lea    -0x410(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x80,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x40,-0x410(%rbp)
  lea    -0x410(%rbp),%rax
  add    $0x4,%rax
  mov    -0x418(%rbp),%edx
  mov    -0x420(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x418(%rbp),%eax
  lea    0x1(%rax),%edx
  lea    -0x410(%rbp),%rcx
  mov    -0x414(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
L11:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int _oled_dat(const int fd, const uint8_t *dat, const uint8_t len) {
  if (len > 255)
    return -1;
  uint8_t buf[256] = {0x40};
  memcpy(buf + 1, dat, len);
  return write(fd, buf, 1 + len);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_oled_dat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  cmp    $0xff,%edx
  jg     L15
  mov    %edi,%ebp
  mov    %edx,%ebx
  mov    %rsp,%rdi
  mov    $0x80,%ecx
  rep stos %rax,%es:(%rdi)
  movl   $0x40,(%rsp)
  mov    %rsp,%r12
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  lea    0x1(%rbx),%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
L26:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L25
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  mov    $0xffffffff,%eax
  jmp    L26
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _oled_dat(const int fd, const uint8_t *dat, const uint8_t len) {
  if (len > 255)
    return -1;
  uint8_t buf[256] = {0x40};
  memcpy(buf + 1, dat, len);
  return write(fd, buf, 1 + len);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<_oled_dat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  cmp    $0xff,%edx
  jg     L25
  mov    %edi,%ebp
  mov    $0x80,%ecx
  mov    %rsp,%rdi
  mov    %edx,%ebx
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%r12
  lea    0x4(%rsp),%rdi
  movl   $0x40,(%rsp)
  call   <memcpy@plt>
  lea    0x1(%rbx),%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
L17:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L25:
  mov    $0xffffffff,%eax
  jmp    L17
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _oled_dat(const int fd, const uint8_t *dat, const uint8_t len) {
  if (len > 255)
    return -1;
  uint8_t buf[256] = {0x40};
  memcpy(buf + 1, dat, len);
  return write(fd, buf, 1 + len);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_oled_dat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  cmp    $0xff,%edx
  jg     L27
  mov    %edi,%ebp
  mov    $0x80,%ecx
  mov    %rsp,%rdi
  mov    %edx,%ebx
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%r12
  lea    0x4(%rsp),%rdi
  movl   $0x40,(%rsp)
  call   <memcpy@plt>
  lea    0x1(%rbx),%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <write@plt>
L5:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L27:
  mov    $0xffffffff,%eax
  jmp    L5
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int _oled_dat(const int fd, const uint8_t *dat, const uint8_t len) {
  if (len > 255)
    return -1;
  uint8_t buf[256] = {0x40};
  memcpy(buf + 1, dat, len);
  return write(fd, buf, 1 + len);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<qsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x28(%rbp)
L5:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <newgap@plt>
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L12
L38:
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x48(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    -0x50(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  test   %eax,%eax
  jle    L46
  mov    -0x48(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memswap@plt>
  movl   $0x1,-0x2c(%rbp)
L46:
  addq   $0x1,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  add    %rax,-0x18(%rbp)
L12:
  mov    -0x40(%rbp),%rax
  sub    -0x28(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jb     L38
  cmpq   $0x1,-0x28(%rbp)
  ja     L5
  cmpl   $0x0,-0x2c(%rbp)
  jne    L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  size_t gap = nmemb;
  size_t i, j;
  char *p1, *p2;
  int swapped;
  do {
    gap = newgap(gap);
    swapped = 0;
    for (i = 0, p1 = base; i < nmemb - gap; i++, p1 += size) {
      j = i + gap;
      if (compar(p1, p2 = (char *)base + j * size) > 0) {
        memswap(p1, p2, size);
        swapped = 1;
      }
    }
  } while (gap > 1 || swapped);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rdx,%r13
  mov    %rcx,%r15
  mov    %rsi,0x8(%rsp)
  jmp    L45
L35:
  add    $0x1,%r12
  add    %r13,%rbp
  add    %r13,%rbx
  cmp    %r14,%r12
  je     L39
L42:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   *%r15
  test   %eax,%eax
  jle    L35
  mov    %r13,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memswap@plt>
  movl   $0x1,0x4(%rsp)
  jmp    L35
L41:
  movl   $0x0,0x4(%rsp)
L39:
  cmpq   $0x1,0x8(%rsp)
  ja     L45
  cmpl   $0x0,0x4(%rsp)
  je     L71
L45:
  mov    0x8(%rsp),%rdi
  call   <newgap@plt>
  mov    %rax,0x8(%rsp)
  mov    0x10(%rsp),%r14
  sub    %rax,%r14
  je     L41
  imul   %r13,%rax
  mov    %rax,%rbx
  mov    0x18(%rsp),%rbp
  add    %rbp,%rbx
  movl   $0x0,0x4(%rsp)
  mov    $0x0,%r12d
  jmp    L42
L71:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  size_t gap = nmemb;
  size_t i, j;
  char *p1, *p2;
  int swapped;
  do {
    gap = newgap(gap);
    swapped = 0;
    for (i = 0, p1 = base; i < nmemb - gap; i++, p1 += size) {
      j = i + gap;
      if (compar(p1, p2 = (char *)base + j * size) > 0) {
        memswap(p1, p2, size);
        swapped = 1;
      }
    }
  } while (gap > 1 || swapped);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rsi,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L15:
  mov    0x8(%rsp),%rdi
  call   <newgap@plt>
  mov    0x10(%rsp),%r12
  mov    %rax,0x8(%rsp)
  sub    %rax,%r12
  je     L23
  imul   %rbp,%rax
  mov    0x18(%rsp),%r15
  movl   $0x0,0x4(%rsp)
  xor    %ebx,%ebx
  lea    (%r15,%rax,1),%r14
L29:
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   *%r13
  test   %eax,%eax
  jle    L38
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <memswap@plt>
  movl   $0x1,0x4(%rsp)
L38:
  add    $0x1,%rbx
  add    %rbp,%r15
  add    %rbp,%r14
  cmp    %r12,%rbx
  jne    L29
  movzbl 0x4(%rsp),%r15d
  and    $0x1,%r15d
L9:
  cmpq   $0x1,0x8(%rsp)
  ja     L15
  test   %r15b,%r15b
  jne    L15
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  xor    %r15d,%r15d
  jmp    L9
```

assistant: ```c
void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  size_t gap = nmemb;
  size_t i, j;
  char *p1, *p2;
  int swapped;
  do {
    gap = newgap(gap);
    swapped = 0;
    for (i = 0, p1 = base; i < nmemb - gap; i++, p1 += size) {
      j = i + gap;
      if (compar(p1, p2 = (char *)base + j * size) > 0) {
        memswap(p1, p2, size);
        swapped = 1;
      }
    }
  } while (gap > 1 || swapped);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %rsi,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L16:
  mov    0x8(%rsp),%rdi
  call   <newgap@plt>
  mov    0x10(%rsp),%r12
  mov    %rax,0x8(%rsp)
  sub    %rax,%r12
  je     L18
  imul   %rbp,%rax
  mov    0x18(%rsp),%r15
  movl   $0x0,0x4(%rsp)
  xor    %ebx,%ebx
  lea    (%r15,%rax,1),%r14
L44:
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   *%r13
  test   %eax,%eax
  jle    L53
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <memswap@plt>
  movl   $0x1,0x4(%rsp)
L53:
  add    $0x1,%rbx
  add    %rbp,%r15
  add    %rbp,%r14
  cmp    %r12,%rbx
  jne    L44
  movzbl 0x4(%rsp),%r15d
  and    $0x1,%r15d
L3:
  cmpq   $0x1,0x8(%rsp)
  ja     L16
  test   %r15b,%r15b
  jne    L16
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  xor    %r15d,%r15d
  jmp    L3
```

assistant: ```c
void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *)) {
  size_t gap = nmemb;
  size_t i, j;
  char *p1, *p2;
  int swapped;
  do {
    gap = newgap(gap);
    swapped = 0;
    for (i = 0, p1 = base; i < nmemb - gap; i++, p1 += size) {
      j = i + gap;
      if (compar(p1, p2 = (char *)base + j * size) > 0) {
        memswap(p1, p2, size);
        swapped = 1;
      }
    }
  } while (gap > 1 || swapped);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D24
datatype: string
name: retrieve_data
arguments: label: D29
datatype: string
name: retrieve_data
arguments: label: D5
datatype: string

tool: D24: Enter a date (mm/dd/yy): 

tool: D29: %d/%d/%d

tool: D5: You entered the date %.2d%.2d%.4d


assistant: ```c
int main(void) {
  int month, day, year;
  printf("Enter a date (mm/dd/yy): ");
  scanf("%d/%d/%d", &month, &day, &year);
  printf("You entered the date %.2d%.2d%.4d\n", month, day, year);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D23
datatype: string
name: unpack_data
arguments: data_label: D0
datatype: string
name: unpack_data
arguments: data_label: D12
datatype: string

tool: D23: Enter a date (mm/dd/yy): 

tool: D0: %d/%d/%d

tool: D12: You entered the date %.2d%.2d%.4d


assistant: ```c
int main(void) {
  int month, day, year;
  printf("Enter a date (mm/dd/yy): ");
  scanf("%d/%d/%d", &month, &day, &year);
  printf("You entered the date %.2d%.2d%.4d\n", month, day, year);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D27(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  xor    %eax,%eax
  mov    0xc(%rsp),%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
type: string
name: extract_symbolic_data
arguments: label: D25
type: string
name: extract_symbolic_data
arguments: label: D21
type: string

tool: D27: Enter a date (mm/dd/yy): 

tool: D25: %d/%d/%d

tool: D21: You entered the date %.2d%.2d%.4d


assistant: ```c
int main(void) {
  int month, day, year;
  printf("Enter a date (mm/dd/yy): ");
  scanf("%d/%d/%d", &month, &day, &year);
  printf("You entered the date %.2d%.2d%.4d\n", month, day, year);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D6(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  xor    %eax,%eax
  mov    0xc(%rsp),%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
type_guessed: string
name: parse_data
arguments: label: D10
type_guessed: string
name: parse_data
arguments: label: D12
type_guessed: string

tool: D6: Enter a date (mm/dd/yy): 

tool: D10: %d/%d/%d

tool: D12: You entered the date %.2d%.2d%.4d


assistant: ```c
int main(void) {
  int month, day, year;
  printf("Enter a date (mm/dd/yy): ");
  scanf("%d/%d/%d", &month, &day, &year);
  printf("You entered the date %.2d%.2d%.4d\n", month, day, year);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cftfx41>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    %rcx,-0x18(%rbp)
  cmpl   $0x80,-0x4(%rbp)
  jne    L0
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf161@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x80(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  sub    $0xffffffffffffff80,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf162@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x100,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf161@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x180,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf161@plt>
  jmp    L13
L0:
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf081@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x40,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf082@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  sub    $0xffffffffffffff80,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf081@plt>
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x20(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0xc0,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <cftf081@plt>
L13:
  nop
  leave
  ret
```

assistant: ```c
void cftfx41(int n, float *a, int nw, float *w) {
  void cftf161(float *a, float *w);
  void cftf162(float *a, float *w);
  void cftf081(float *a, float *w);
  void cftf082(float *a, float *w);
  if (n == 128) {
    cftf161(a, &w[nw - 8]);
    cftf162(&a[32], &w[nw - 32]);
    cftf161(&a[64], &w[nw - 8]);
    cftf161(&a[96], &w[nw - 8]);
  } else {
    cftf081(a, &w[nw - 8]);
    cftf082(&a[16], &w[nw - 8]);
    cftf081(&a[32], &w[nw - 8]);
    cftf081(&a[48], &w[nw - 8]);
  }
}
```