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
<seq2xy>:
  endbr64
  cmp    $0x30c5cf,%edx
  ja     L10
  mov    %edx,%ecx
  imul   $0x78fc217,%rcx,%rcx
  mov    %rcx,%rax
  shr    $0x2a,%rcx
  shr    $0x2a,%rax
  lea    (%rcx,%rcx,2),%ecx
  imul   $0x876c,%eax,%r8d
  mov    %edx,%eax
  shl    $0x3,%ecx
  mov    %ecx,(%rsi)
  sub    %r8d,%eax
  cmp    $0x1aff,%eax
  jbe    L2
  lea    -0x1b00(%rax),%edx
  mov    $0xaa00aa01,%r8d
  mov    %rdx,%rax
  imul   %r8,%rdx
  shr    $0x2a,%rdx
  lea    0x6(%rcx,%rdx,1),%ecx
  imul   $0x606,%edx,%edx
  mov    %ecx,(%rsi)
  sub    %edx,%eax
  mov    %eax,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %eax,%edx
  movl   $0x6,(%rdi)
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x28,%rdx
  add    %edx,(%rsi)
  lea    (%rdx,%rdx,8),%edx
  shl    $0x7,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,8),%ecx
  lea    (%rdx,%rdx,2),%edx
  add    %ecx,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rdx,8),%eax
  add    %eax,(%rdi)
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  movl   $0xffffffff,(%rdi)
  movl   $0xffffffff,(%rsi)
  ret
```

assistant: ```c
void seq2xy(int *x, int *y, unsigned seq) {
  unsigned rep;
  if (seq >= ((long)(((24 - 2 * 3) * ((2 * 2 + (24 * (65 - 1) + 2 * 3)) - 2 * 2)) + ((2 * 3) * ((24 - 2 * 3) * (65 - 1)))) * (93 - 1) + ((2 * 3) * ((24 - 2 * 3) * (65 - 1))))) {
    *x = -1;
    *y = -1;
    return;
  }
  rep = seq / (((24 - 2 * 3) * ((2 * 2 + (24 * (65 - 1) + 2 * 3)) - 2 * 2)) + ((2 * 3) * ((24 - 2 * 3) * (65 - 1))));
  seq = seq % (((24 - 2 * 3) * ((2 * 2 + (24 * (65 - 1) + 2 * 3)) - 2 * 2)) + ((2 * 3) * ((24 - 2 * 3) * (65 - 1))));
  *y = ((2 * 3) + (24 - 2 * 3)) * rep;
  if (seq >= ((2 * 3) * ((24 - 2 * 3) * (65 - 1)))) {
    *y += (2 * 3);
    seq -= ((2 * 3) * ((24 - 2 * 3) * (65 - 1)));
    *y += seq / ((2 * 2 + (24 * (65 - 1) + 2 * 3)) - 2 * 2);
    *x = seq % ((2 * 2 + (24 * (65 - 1) + 2 * 3)) - 2 * 2);
  } else {
    unsigned gap;
    *x = 2 * 3;
    *y += seq / ((24 - 2 * 3) * (65 - 1));
    seq %= ((24 - 2 * 3) * (65 - 1));
    gap = seq / (24 - 2 * 3);
    *x += gap * 24;
    seq %= (24 - 2 * 3);
    *x += seq;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fat_make_shortname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L48
  mov    $0x0,%eax
  jmp    L66
L48:
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movl   $0x0,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L17
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
L17:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L74
L72:
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L74:
  cmpl   $0x7,-0x14(%rbp)
  jle    L72
  mov    -0x28(%rbp),%rax
  mov    $0x2e,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L70
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
L70:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L79
L76:
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L79:
  cmpl   $0xa,-0x14(%rbp)
  jle    L76
  mov    -0x10(%rbp),%rax
L66:
  leave
  ret
```

assistant: ```c
char *fat_make_shortname(const char *longname) {
  if (!longname)
    return 0;
  char *shortname = calloc(11, 1);
  strncpy(shortname, longname, 8);
  int i = 0;
  char *dot = strchr(shortname, '.');
  if (dot) {
    dot[0] = '\0';
  }
  i = strlen(shortname);
  while (i < 8)
    shortname[i++] = ' ';
  dot = strrchr(longname, '.');
  if (dot)
    strncpy(&shortname[8], &dot[1], 3);
  i = strlen(shortname);
  while (i < 11)
    shortname[i++] = ' ';
  return shortname;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fat_make_shortname>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L45
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x2e,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L56
  movb   $0x0,(%rax)
L56:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jg     L14
  movslq %eax,%rcx
  lea    (%rbx,%rcx,1),%rdx
  lea    0x1(%rbx,%rcx,1),%rcx
  mov    $0x7,%esi
  sub    %eax,%esi
  lea    (%rcx,%rsi,1),%rax
L53:
  movb   $0x20,(%rdx)
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L53
L14:
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L21
  lea    0x1(%rax),%rsi
  lea    0x8(%rbx),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
L21:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0xa,%eax
  jg     L29
  movslq %eax,%rcx
  lea    (%rbx,%rcx,1),%rdx
  lea    0x1(%rbx,%rcx,1),%rcx
  mov    $0xa,%esi
  sub    %eax,%esi
  lea    (%rcx,%rsi,1),%rax
L43:
  movb   $0x20,(%rdx)
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L43
L29:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L45:
  mov    %rdi,%rbx
  jmp    L29
```

assistant: ```c
char *fat_make_shortname(const char *longname) {
  if (!longname)
    return 0;
  char *shortname = calloc(11, 1);
  strncpy(shortname, longname, 8);
  int i = 0;
  char *dot = strchr(shortname, '.');
  if (dot) {
    dot[0] = '\0';
  }
  i = strlen(shortname);
  while (i < 8)
    shortname[i++] = ' ';
  dot = strrchr(longname, '.');
  if (dot)
    strncpy(&shortname[8], &dot[1], 3);
  i = strlen(shortname);
  while (i < 11)
    shortname[i++] = ' ';
  return shortname;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fat_make_shortname>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L17
  mov    %rdi,%rbp
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rbp,%rsi
  mov    $0x8,%edx
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x2e,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L18
  movb   $0x0,(%rax)
L18:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jg     L24
  mov    $0x8,%edx
  mov    $0x20,%esi
  sub    %eax,%edx
  cltq
  lea    (%r12,%rax,1),%rdi
  call   <memset@plt>
L24:
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L14
  lea    0x1(%rax),%rsi
  lea    0x8(%r12),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
L14:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0xa,%eax
  jg     L44
  mov    $0xb,%edx
  mov    $0x20,%esi
  sub    %eax,%edx
  cltq
  lea    (%r12,%rax,1),%rdi
  call   <memset@plt>
L44:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L17:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *fat_make_shortname(const char *longname) {
  if (!longname)
    return 0;
  char *shortname = calloc(11, 1);
  strncpy(shortname, longname, 8);
  int i = 0;
  char *dot = strchr(shortname, '.');
  if (dot) {
    dot[0] = '\0';
  }
  i = strlen(shortname);
  while (i < 8)
    shortname[i++] = ' ';
  dot = strrchr(longname, '.');
  if (dot)
    strncpy(&shortname[8], &dot[1], 3);
  i = strlen(shortname);
  while (i < 11)
    shortname[i++] = ' ';
  return shortname;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fat_make_shortname>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L38
  mov    %rdi,%rbp
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rbp,%rsi
  mov    $0x8,%edx
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    $0x2e,%esi
  mov    %r12,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L22
  movb   $0x0,(%rax)
L22:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0x7,%eax
  jg     L11
  mov    $0x8,%edx
  mov    $0x20,%esi
  sub    %eax,%edx
  cltq
  lea    (%r12,%rax,1),%rdi
  call   <memset@plt>
L11:
  mov    $0x2e,%esi
  mov    %rbp,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L12
  lea    0x1(%rax),%rsi
  lea    0x8(%r12),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
L12:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    $0xa,%eax
  jg     L27
  mov    $0xb,%edx
  mov    $0x20,%esi
  sub    %eax,%edx
  cltq
  lea    (%r12,%rax,1),%rdi
  call   <memset@plt>
L27:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L38:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *fat_make_shortname(const char *longname) {
  if (!longname)
    return 0;
  char *shortname = calloc(11, 1);
  strncpy(shortname, longname, 8);
  int i = 0;
  char *dot = strchr(shortname, '.');
  if (dot) {
    dot[0] = '\0';
  }
  i = strlen(shortname);
  while (i < 8)
    shortname[i++] = ' ';
  dot = strrchr(longname, '.');
  if (dot)
    strncpy(&shortname[8], &dot[1], 3);
  i = strlen(shortname);
  while (i < 11)
    shortname[i++] = ' ';
  return shortname;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_vec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L4:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L37
L24:
  mov    -0x4(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L37:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L24
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L4
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type: string
name: disassemble_data
arguments: tag: D15
type: string

tool: D1:     

tool: D15: % 7.3f 

assistant: ```c
void print_vec(float cb[], int d, int e) {
  int i, j;
  for (j = 0; j < e; j++) {
    printf("    ");
    for (i = 0; i < d; i++)
      printf("% 7.3f ", cb[j * d + i]);
    printf("\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_vec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %edx,0x10(%rsp)
  test   %edx,%edx
  jle    L33
  mov    %esi,%r15d
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  lea    -0x1(%rsi),%eax
  mov    %eax,0x14(%rsp)
  lea    0x4(%rdi),%rax
  mov    %rax,0x18(%rsp)
  lea    D9(%rip),%r12
L2:
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r15d,%r15d
  jle    L36
  movslq %r14d,%rdx
  mov    0x8(%rsp),%rax
  lea    (%rax,%rdx,4),%rbx
  mov    0x14(%rsp),%eax
  add    %rdx,%rax
  mov    0x18(%rsp),%rcx
  lea    (%rcx,%rax,4),%rbp
L31:
  pxor   %xmm0,%xmm0
  cvtss2sd (%rbx),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L31
L36:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r13d
  add    %r15d,%r14d
  cmp    %r13d,0x10(%rsp)
  jne    L2
L33:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
type_guessed: string
name: unpack_data
arguments: tag: D18
type_guessed: string

tool: D9: % 7.3f 

tool: D18:     

assistant: ```c
void print_vec(float cb[], int d, int e) {
  int i, j;
  for (j = 0; j < e; j++) {
    printf("    ");
    for (i = 0; i < d; i++)
      printf("% 7.3f ", cb[j * d + i]);
    printf("\n");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_vec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %edx,0x4(%rsp)
  test   %edx,%edx
  jle    L32
  mov    %esi,%r12d
  xor    %r14d,%r14d
  lea    D2(%rip),%rbp
  xor    %r13d,%r13d
  nop
L11:
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L22
  mov    0x8(%rsp),%rdx
  movslq %r14d,%rcx
  lea    -0x1(%r12),%eax
  add    %rcx,%rax
  lea    (%rdx,%rcx,4),%r15
  lea    0x4(%rdx,%rax,4),%rbx
  nopl   0x0(%rax)
L30:
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  mov    $0x1,%eax
  add    $0x4,%r15
  cvtss2sd -0x4(%r15),%xmm0
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L30
L22:
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    %r12d,%r14d
  call   <putchar@plt>
  cmp    %r13d,0x4(%rsp)
  jne    L11
L32:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_label: D2
type: string
name: read_data
arguments: data_label: D1
type: string

tool: D2: % 7.3f 

tool: D1:     

assistant: ```c
void print_vec(float cb[], int d, int e) {
  int i, j;
  for (j = 0; j < e; j++) {
    printf("    ");
    for (i = 0; i < d; i++)
      printf("% 7.3f ", cb[j * d + i]);
    printf("\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_vec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %edx,0x4(%rsp)
  test   %edx,%edx
  jle    L1
  mov    %esi,%r12d
  xor    %r14d,%r14d
  lea    D8(%rip),%rbp
  xor    %r13d,%r13d
  nop
L21:
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L25
  mov    0x8(%rsp),%rdx
  movslq %r14d,%rcx
  lea    -0x1(%r12),%eax
  add    %rcx,%rax
  lea    (%rdx,%rcx,4),%r15
  lea    0x4(%rdx,%rax,4),%rbx
  nopl   0x0(%rax)
L10:
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  mov    $0x1,%eax
  add    $0x4,%r15
  cvtss2sd -0x4(%r15),%xmm0
  call   <printf@plt>
  cmp    %r15,%rbx
  jne    L10
L25:
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    %r12d,%r14d
  call   <putchar@plt>
  cmp    %r13d,0x4(%rsp)
  jne    L21
L1:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type: string
name: parse_data
arguments: data_label: D16
type: string

tool: D8: % 7.3f 

tool: D16:     

assistant: ```c
void print_vec(float cb[], int d, int e) {
  int i, j;
  for (j = 0; j < e; j++) {
    printf("    ");
    for (i = 0; i < d; i++)
      printf("% 7.3f ", cb[j * d + i]);
    printf("\n");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bi_tree_insert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L14
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <bi_tree_update_impl@plt>
  jmp    L19
L14:
  nop
L19:
  leave
  ret
```

assistant: ```c
void bi_tree_insert(struct bi_tree *bi, int val, int idx) {
  if (bi == NULL)
    return;
  bi_tree_update_impl(bi, val, idx + 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bi_tree_insert>:
  endbr64
  test   %rdi,%rdi
  je     L4
  sub    $0x8,%rsp
  add    $0x1,%edx
  call   <bi_tree_update_impl@plt>
  add    $0x8,%rsp
  ret
L4:
  ret
```

assistant: ```c
void bi_tree_insert(struct bi_tree *bi, int val, int idx) {
  if (bi == NULL)
    return;
  bi_tree_update_impl(bi, val, idx + 1);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bi_tree_insert>:
  endbr64
  test   %rdi,%rdi
  je     L7
  add    $0x1,%edx
  jmp    1050 <bi_tree_update_impl@plt>
  nopl   0x0(%rax)
L7:
  ret
```

assistant: ```c
void bi_tree_insert(struct bi_tree *bi, int val, int idx) {
  if (bi == NULL)
    return;
  bi_tree_update_impl(bi, val, idx + 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<bi_tree_insert>:
  endbr64
  test   %rdi,%rdi
  je     L6
  add    $0x1,%edx
  jmp    1050 <bi_tree_update_impl@plt>
  nopl   0x0(%rax)
L6:
  ret
```

assistant: ```c
void bi_tree_insert(struct bi_tree *bi, int val, int idx) {
  if (bi == NULL)
    return;
  bi_tree_update_impl(bi, val, idx + 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<__GC_KEEPALIVE__>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D39(%rip),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    D39(%rip),%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    D39(%rip),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    D39(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D7(%rip)
  nop
  leave
  ret
```

assistant: ```c
void __attribute__((used)) __GC_KEEPALIVE__() {
  static int times = 1;
  void *x = malloc(times);
  free(x);
  x = calloc(1, times);
  free(x);
  x = calloc(times, 1);
  free(x);
  times++;
}
```
user: What is the c source code of the assembly code below:

```asm
<__GC_KEEPALIVE__>:
  endbr64
  sub    $0x8,%rsp
  mov    D18(%rip),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    D18(%rip),%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x1,%esi
  mov    D18(%rip),%edi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,D18(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((used)) __GC_KEEPALIVE__() {
  static int times = 1;
  void *x = malloc(times);
  free(x);
  x = calloc(1, times);
  free(x);
  x = calloc(times, 1);
  free(x);
  times++;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<__GC_KEEPALIVE__>:
  endbr64
  sub    $0x8,%rsp
  mov    D35(%rip),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    D35(%rip),%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    D35(%rip),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,D35(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((used)) __GC_KEEPALIVE__() {
  static int times = 1;
  void *x = malloc(times);
  free(x);
  x = calloc(1, times);
  free(x);
  x = calloc(times, 1);
  free(x);
  times++;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<__GC_KEEPALIVE__>:
  endbr64
  sub    $0x8,%rsp
  mov    D25(%rip),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    D25(%rip),%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  mov    D25(%rip),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,D25(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((used)) __GC_KEEPALIVE__() {
  static int times = 1;
  void *x = malloc(times);
  free(x);
  x = calloc(1, times);
  free(x);
  x = calloc(times, 1);
  free(x);
  times++;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <InitGraphics@plt>
  movsd  D0(%rip),%xmm0
  mov    D0(%rip),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <MovePen@plt>
  movsd  D24(%rip),%xmm0
  movapd %xmm0,%xmm1
  mov    D22(%rip),%rax
  movq   %rax,%xmm0
  call   <DrawLine@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
datatype: f64
name: extract_symbolic_data
arguments: data_label: D24
datatype: f64
name: extract_symbolic_data
arguments: data_label: D22
datatype: f64

tool: D0: 0.5

tool: D24: 1

tool: D22: 0

assistant: ```c
main() {
  InitGraphics();
  MovePen(0.5, 0.5);
  DrawLine(0.0, 1.0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <InitGraphics@plt>
  movsd  D11(%rip),%xmm0
  movapd %xmm0,%xmm1
  call   <MovePen@plt>
  movsd  D6(%rip),%xmm1
  pxor   %xmm0,%xmm0
  call   <DrawLine@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D11
type_guessed: f64
name: disassemble_data
arguments: label: D6
type_guessed: f64

tool: D11: 0.5

tool: D6: 1

assistant: ```c
main() {
  InitGraphics();
  MovePen(0.5, 0.5);
  DrawLine(0.0, 1.0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <InitGraphics@plt>
  movsd  D9(%rip),%xmm0
  movapd %xmm0,%xmm1
  call   <MovePen@plt>
  movsd  D18(%rip),%xmm1
  pxor   %xmm0,%xmm0
  call   <DrawLine@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: f64
name: parse_data
arguments: data_tag: D18
data_type: f64

tool: D9: 0.5

tool: D18: 1

assistant: ```c
main() {
  InitGraphics();
  MovePen(0.5, 0.5);
  DrawLine(0.0, 1.0);
}
```
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
  xor    %eax,%eax
  call   <InitGraphics@plt>
  movsd  D18(%rip),%xmm0
  movapd %xmm0,%xmm1
  call   <MovePen@plt>
  movsd  D16(%rip),%xmm1
  pxor   %xmm0,%xmm0
  call   <DrawLine@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D18
type_guessed: f64
name: unpack_data
arguments: label: D16
type_guessed: f64

tool: D18: 0.5

tool: D16: 1

assistant: ```c
main() {
  InitGraphics();
  MovePen(0.5, 0.5);
  DrawLine(0.0, 1.0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<readInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type: string

tool: D0: %d

assistant: ```c
int readInt() {
  int n;
  scanf("%d", &n);
  return n;
}
```
user: What is the c source code of the assembly code below:

```asm
<readInt>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D1
data_type: string

tool: D1: %d

assistant: ```c
int readInt() {
  int n;
  scanf("%d", &n);
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readInt>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: %d

assistant: ```c
int readInt() {
  int n;
  scanf("%d", &n);
  return n;
}
```