ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<GoodSpot>:
  endbr64
  push   %r14
  mov    %esi,%r11d
  mov    %edx,%esi
  mov    %rdi,%r10
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x30(%rsp),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  neg    %eax
  lea    (%rax,%r8,1),%edi
  sub    %eax,%r8d
  cmp    %r8d,%edi
  jg     L19
  mov    0x38(%rsp),%eax
  mov    %edi,%r12d
  shr    $0x1f,%eax
  add    0x38(%rsp),%eax
  sar    %eax
  neg    %eax
  lea    (%rax,%r9,1),%ebx
  sub    %eax,%r9d
  mov    %ebx,%eax
  movslq %ebx,%r13
  shr    $0x1f,%eax
  cmp    %esi,%ebx
  setge  %bpl
  imul   %esi,%r12d
  or     %eax,%ebp
L90:
  cmp    %r9d,%ebx
  jg     L60
  test   %edi,%edi
  js     L65
  cmp    %edi,%r11d
  jle    L65
  test   %bpl,%bpl
  jne    L65
  movslq %r12d,%rdx
  mov    %ebx,%eax
  add    %r13,%rdx
  add    %r10,%rdx
  jmp    L94
L67:
  add    $0x1,%eax
  cmp    %r9d,%eax
  jg     L60
  add    $0x1,%rdx
  cmp    %eax,%esi
  jle    L65
L94:
  movzbl (%rdx),%r14d
  cmp    %ecx,%r14d
  je     L67
L65:
  xor    %eax,%eax
L68:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L60:
  add    $0x1,%edi
  add    %esi,%r12d
  cmp    %r8d,%edi
  jle    L90
L19:
  mov    $0x1,%eax
  jmp    L68
```

assistant: ```c
int GoodSpot(InImage, ROWS, COLS, Label, r, c, height, width)
unsigned char *InImage;
int ROWS, COLS, Label, r, c, width, height;
{
  int x, y;
  for (y = r - height / 2; y <= r + height / 2; y++)
    for (x = c - width / 2; x <= c + width / 2; x++)
      if (y < 0 || y >= ROWS || x < 0 || x >= COLS || InImage[y * COLS + x] != Label)
        return (0);
  return (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<plugin_api_info_locale_cb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <setlocale@plt>
  leave
  ret
```

assistant: ```c
const char *plugin_api_info_locale_cb(void *data, const char *info_name, const char *arguments) {
  (void)data;
  (void)info_name;
  (void)arguments;
  return setlocale(LC_MESSAGES, NULL);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<plugin_api_info_locale_cb>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <setlocale@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *plugin_api_info_locale_cb(void *data, const char *info_name, const char *arguments) {
  (void)data;
  (void)info_name;
  (void)arguments;
  return setlocale(LC_MESSAGES, NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<plugin_api_info_locale_cb>:
  endbr64
  mov    D3(%rip),%rax
  xor    %esi,%esi
  mov    (%rax),%edi
  jmp    1050 <setlocale@plt>
```

assistant: ```c
const char *plugin_api_info_locale_cb(void *data, const char *info_name, const char *arguments) {
  (void)data;
  (void)info_name;
  (void)arguments;
  return setlocale(LC_MESSAGES, NULL);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<plugin_api_info_locale_cb>:
  endbr64
  mov    D0(%rip),%rax
  xor    %esi,%esi
  mov    (%rax),%edi
  jmp    1050 <setlocale@plt>
```

assistant: ```c
const char *plugin_api_info_locale_cb(void *data, const char *info_name, const char *arguments) {
  (void)data;
  (void)info_name;
  (void)arguments;
  return setlocale(LC_MESSAGES, NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ultohex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x0,-0x8(%rbp)
  movb   $0x0,-0x15(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L21
L0:
  addl   $0x1,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x4,%rax
  mov    %rax,-0x8(%rbp)
L21:
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  addl   $0x2,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L36
L115:
  mov    -0x28(%rbp),%rax
  and    $0xf,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x9,-0xc(%rbp)
  jle    L112
  mov    -0xc(%rbp),%eax
  sub    $0xa,%eax
  cmp    $0x5,%eax
  ja     L110
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D36(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D36(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movb   $0x61,-0x15(%rbp)
  jmp    L23
  movb   $0x62,-0x15(%rbp)
  jmp    L23
  movb   $0x63,-0x15(%rbp)
  jmp    L23
  movb   $0x64,-0x15(%rbp)
  jmp    L23
  movb   $0x65,-0x15(%rbp)
  jmp    L23
  movb   $0x66,-0x15(%rbp)
  jmp    L23
L112:
  mov    -0xc(%rbp),%eax
  add    $0x30,%eax
  mov    %al,-0x15(%rbp)
  jmp    L23
L110:
  nop
L23:
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    D39(%rip),%rcx
  movzbl -0x15(%rbp),%edx
  mov    %dl,(%rax,%rcx,1)
  mov    -0x28(%rbp),%rax
  shr    $0x4,%rax
  mov    %rax,-0x28(%rbp)
L36:
  cmpq   $0x0,-0x28(%rbp)
  jne    L115
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    D39(%rip),%rdx
  movb   $0x78,(%rax,%rdx,1)
  mov    -0x10(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    D39(%rip),%rdx
  movb   $0x30,(%rax,%rdx,1)
  mov    -0x14(%rbp),%eax
  cltq
  lea    D39(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  lea    D39(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *ultohex(unsigned long num) {
  int len = 0, rem = 0, i;
  unsigned long num_cpy = 0;
  char c = '\0';
  for (num_cpy = num; num_cpy != 0; num_cpy /= 16)
    len++;
  len = len + 2;
  static char hex[20];
  i = len - 1;
  for (; num != 0; num /= 16) {
    rem = num % 16;
    if (rem >= 10) {
      switch (rem) {
      case 10:
        c = 'a';
        break;
      case 11:
        c = 'b';
        break;
      case 12:
        c = 'c';
        break;
      case 13:
        c = 'd';
        break;
      case 14:
        c = 'e';
        break;
      case 15:
        c = 'f';
        break;
      default:
        break;
      }
    } else
      c = '0' + rem;
    hex[i--] = c;
  }
  hex[i--] = 'x';
  hex[i--] = '0';
  hex[len] = '\0';
  return hex;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ultohex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L49
  mov    %rdi,%rcx
  mov    %rdi,%rsi
  mov    $0x0,%edx
L105:
  mov    %edx,%eax
  add    $0x1,%edx
  mov    %rsi,%rdi
  shr    $0x4,%rsi
  cmp    $0xf,%rdi
  ja     L105
  lea    0x3(%rax),%r9d
  add    $0x1,%eax
  cltq
  mov    $0x0,%esi
  lea    D0(%rip),%r8
  mov    $0x66,%r13d
  mov    $0x65,%r12d
  mov    $0x64,%ebp
  mov    $0x63,%ebx
  mov    $0x62,%r11d
  mov    $0x61,%r10d
  lea    D22(%rip),%rdi
  jmp    L35
  mov    %r10d,%esi
  jmp    L32
  mov    %r11d,%esi
  jmp    L32
  mov    %ebx,%esi
  jmp    L32
  mov    %ebp,%esi
  jmp    L32
  mov    %r12d,%esi
  jmp    L32
  mov    %r13d,%esi
  jmp    L32
L19:
  lea    0x30(%rdx),%esi
L32:
  mov    %eax,%r14d
  mov    %sil,(%rax,%rdi,1)
  mov    %rcx,%rdx
  shr    $0x4,%rdx
  sub    $0x1,%rax
  cmp    $0xf,%rcx
  jbe    L109
  mov    %rdx,%rcx
L35:
  mov    %ecx,%edx
  and    $0xf,%edx
  cmp    $0x9,%edx
  jle    L19
  sub    $0xa,%edx
  cmp    $0x5,%edx
  ja     L32
  mov    %edx,%edx
  movslq (%r8,%rdx,4),%rdx
  add    %r8,%rdx
  notrack jmp *%rdx
L49:
  mov    $0x2,%r9d
  mov    $0x1,%r14d
L109:
  lea    D25(%rip),%rax
  movslq %r14d,%rdx
  movb   $0x78,(%rax,%rdx,1)
  sub    $0x1,%r14d
  movslq %r14d,%r14
  movb   $0x30,(%rax,%r14,1)
  movslq %r9d,%r9
  movb   $0x0,(%rax,%r9,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ultohex(unsigned long num) {
  int len = 0, rem = 0, i;
  unsigned long num_cpy = 0;
  char c = '\0';
  for (num_cpy = num; num_cpy != 0; num_cpy /= 16)
    len++;
  len = len + 2;
  static char hex[20];
  i = len - 1;
  for (; num != 0; num /= 16) {
    rem = num % 16;
    if (rem >= 10) {
      switch (rem) {
      case 10:
        c = 'a';
        break;
      case 11:
        c = 'b';
        break;
      case 12:
        c = 'c';
        break;
      case 13:
        c = 'd';
        break;
      case 14:
        c = 'e';
        break;
      case 15:
        c = 'f';
        break;
      default:
        break;
      }
    } else
      c = '0' + rem;
    hex[i--] = c;
  }
  hex[i--] = 'x';
  hex[i--] = '0';
  hex[len] = '\0';
  return hex;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ultohex>:
  endbr64
  test   %rdi,%rdi
  je     L32
  mov    %rdi,%rdx
  xor    %eax,%eax
  xchg   %ax,%ax
L4:
  mov    %eax,%ecx
  add    $0x1,%eax
  shr    $0x4,%rdx
  jne    L4
  lea    0x3(%rcx),%esi
  cltq
  lea    D12(%rip),%r8
  mov    $0x61,%r9d
  nopl   (%rax)
L14:
  mov    %edi,%edx
  and    $0xf,%edx
  lea    0x30(%rdx),%ecx
  cmp    $0x9,%edx
  jle    L30
  lea    -0xb(%rdx),%ecx
  add    $0x57,%edx
  cmp    $0x5,%ecx
  mov    %edx,%ecx
  cmovae %r9d,%ecx
L30:
  mov    %cl,(%r8,%rax,1)
  movslq %eax,%rdx
  sub    $0x1,%rax
  shr    $0x4,%rdi
  jne    L14
  lea    -0x1(%rdx),%ecx
L1:
  lea    D2(%rip),%rax
  movb   $0x78,(%rax,%rdx,1)
  movslq %ecx,%rdx
  movslq %esi,%rcx
  movb   $0x30,(%rax,%rdx,1)
  movb   $0x0,(%rax,%rcx,1)
  ret
  nopw   0x0(%rax,%rax,1)
L32:
  xor    %ecx,%ecx
  mov    $0x2,%esi
  mov    $0x1,%edx
  jmp    L1
```

assistant: ```c
char *ultohex(unsigned long num) {
  int len = 0, rem = 0, i;
  unsigned long num_cpy = 0;
  char c = '\0';
  for (num_cpy = num; num_cpy != 0; num_cpy /= 16)
    len++;
  len = len + 2;
  static char hex[20];
  i = len - 1;
  for (; num != 0; num /= 16) {
    rem = num % 16;
    if (rem >= 10) {
      switch (rem) {
      case 10:
        c = 'a';
        break;
      case 11:
        c = 'b';
        break;
      case 12:
        c = 'c';
        break;
      case 13:
        c = 'd';
        break;
      case 14:
        c = 'e';
        break;
      case 15:
        c = 'f';
        break;
      default:
        break;
      }
    } else
      c = '0' + rem;
    hex[i--] = c;
  }
  hex[i--] = 'x';
  hex[i--] = '0';
  hex[len] = '\0';
  return hex;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ultohex>:
  endbr64
  test   %rdi,%rdi
  je     L6
  mov    %rdi,%rdx
  xor    %eax,%eax
  xchg   %ax,%ax
L46:
  mov    %eax,%ecx
  add    $0x1,%eax
  shr    $0x4,%rdx
  jne    L46
  lea    0x3(%rcx),%esi
  cltq
  lea    D4(%rip),%r8
  mov    $0x61,%r9d
  nopl   (%rax)
L3:
  mov    %edi,%edx
  and    $0xf,%edx
  lea    0x30(%rdx),%ecx
  cmp    $0x9,%edx
  jle    L34
  lea    -0xb(%rdx),%ecx
  add    $0x57,%edx
  cmp    $0x5,%ecx
  mov    %edx,%ecx
  cmovae %r9d,%ecx
L34:
  mov    %cl,(%r8,%rax,1)
  movslq %eax,%rdx
  sub    $0x1,%rax
  shr    $0x4,%rdi
  jne    L3
  lea    -0x1(%rdx),%ecx
L15:
  lea    D2(%rip),%rax
  movb   $0x78,(%rax,%rdx,1)
  movslq %ecx,%rdx
  movslq %esi,%rcx
  movb   $0x30,(%rax,%rdx,1)
  movb   $0x0,(%rax,%rcx,1)
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %ecx,%ecx
  mov    $0x2,%esi
  mov    $0x1,%edx
  jmp    L15
```

assistant: ```c
char *ultohex(unsigned long num) {
  int len = 0, rem = 0, i;
  unsigned long num_cpy = 0;
  char c = '\0';
  for (num_cpy = num; num_cpy != 0; num_cpy /= 16)
    len++;
  len = len + 2;
  static char hex[20];
  i = len - 1;
  for (; num != 0; num /= 16) {
    rem = num % 16;
    if (rem >= 10) {
      switch (rem) {
      case 10:
        c = 'a';
        break;
      case 11:
        c = 'b';
        break;
      case 12:
        c = 'c';
        break;
      case 13:
        c = 'd';
        break;
      case 14:
        c = 'e';
        break;
      case 15:
        c = 'f';
        break;
      default:
        break;
      }
    } else
      c = '0' + rem;
    hex[i--] = c;
  }
  hex[i--] = 'x';
  hex[i--] = '0';
  hex[len] = '\0';
  return hex;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<THREAD_C_main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x30d40,%edi
  call   <flow1@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
void *THREAD_C_main(void *arg) {
  flow1(200000);
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<THREAD_C_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x30d40,%edi
  call   <flow1@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *THREAD_C_main(void *arg) {
  flow1(200000);
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<THREAD_C_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x30d40,%edi
  call   <flow1@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *THREAD_C_main(void *arg) {
  flow1(200000);
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<THREAD_C_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x30d40,%edi
  call   <flow1@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *THREAD_C_main(void *arg) {
  flow1(200000);
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0xffffffff,-0x2c(%rbp)
  lea    -0x34(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x34(%rbp),%eax
  test   %eax,%eax
  jle    L30
  mov    -0x34(%rbp),%eax
  cmp    $0x64,%eax
  jle    L49
L30:
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L8
L20:
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  mov    -0x28(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  mov    $0x2,%edi
  call   <solve@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L49:
  mov    -0x30(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x30(%rbp)
  mov    -0x34(%rbp),%edx
  cmp    %edx,%eax
  jl     L20
  movl   $0x0,-0x2c(%rbp)
L8:
  mov    -0x2c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
data_type: string
name: parse_data
arguments: data_label: D29
data_type: string
name: parse_data
arguments: data_label: D32
data_type: string
name: parse_data
arguments: data_label: D30
data_type: string

tool: D4: %d

tool: D29: Illegal number of tests (1 <= T <= 100)

tool: D32: %lf %lf %lf

tool: D30: Case #%d: %.7f


assistant: ```c
int main(void) {
  int num_tests, test = 0, ret = -1;
  double round, farm_cost, farm_rate, cap;
  scanf("%d", &num_tests);
  if (num_tests < 1 || num_tests > 100) {
    printf("Illegal number of tests (1 <= T <= 100)\n");
    goto out;
  }
  while (test++ < num_tests) {
    scanf("%lf %lf %lf", &farm_cost, &farm_rate, &cap);
    round = solve(2, farm_cost, farm_rate, cap);
    printf("Case #%d: %.7f\n", test, round);
  }
  ret = 0;
out:
  return ret;
}
```
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    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  cmp    $0x63,%eax
  ja     L27
  mov    $0x1,%ebx
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  lea    D27(%rip),%r12
  lea    D23(%rip),%rbp
L31:
  lea    0x10(%rsp),%rsi
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  mov    $0x2,%edi
  call   <solve@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0xc(%rsp)
  jg     L31
  mov    $0x0,%eax
L8:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  lea    D31(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%eax
  jmp    L8
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type: string
name: extract_symbolic_data
arguments: data_tag: D27
type: string
name: extract_symbolic_data
arguments: data_tag: D23
type: string
name: extract_symbolic_data
arguments: data_tag: D31
type: string

tool: D2: %d

tool: D27: %lf %lf %lf

tool: D23: Case #%d: %.7f


tool: D31: Illegal number of tests (1 <= T <= 100)

assistant: ```c
int main(void) {
  int num_tests, test = 0, ret = -1;
  double round, farm_cost, farm_rate, cap;
  scanf("%d", &num_tests);
  if (num_tests < 1 || num_tests > 100) {
    printf("Illegal number of tests (1 <= T <= 100)\n");
    goto out;
  }
  while (test++ < num_tests) {
    scanf("%lf %lf %lf", &farm_cost, &farm_rate, &cap);
    round = solve(2, farm_cost, farm_rate, cap);
    printf("Case #%d: %.7f\n", test, round);
  }
  ret = 0;
out:
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D21(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  cmp    $0x63,%eax
  ja     L21
  mov    $0x1,%ebx
  lea    0x20(%rsp),%r15
  lea    0x18(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D12(%rip),%r12
  lea    D36(%rip),%rbp
  nopl   (%rax)
L10:
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  mov    $0x2,%edi
  movsd  0x10(%rsp),%xmm0
  call   <solve@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0xc(%rsp)
  jg     L10
  xor    %eax,%eax
L11:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L11
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D21
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D12
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D36
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D37
type_infered: string

tool: D21: %d

tool: D12: %lf %lf %lf

tool: D36: Case #%d: %.7f


tool: D37: Illegal number of tests (1 <= T <= 100)

assistant: ```c
int main(void) {
  int num_tests, test = 0, ret = -1;
  double round, farm_cost, farm_rate, cap;
  scanf("%d", &num_tests);
  if (num_tests < 1 || num_tests > 100) {
    printf("Illegal number of tests (1 <= T <= 100)\n");
    goto out;
  }
  while (test++ < num_tests) {
    scanf("%lf %lf %lf", &farm_cost, &farm_rate, &cap);
    round = solve(2, farm_cost, farm_rate, cap);
    printf("Case #%d: %.7f\n", test, round);
  }
  ret = 0;
out:
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D36(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  sub    $0x1,%eax
  cmp    $0x63,%eax
  ja     L10
  mov    $0x1,%ebx
  lea    0x20(%rsp),%r15
  lea    0x18(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    D34(%rip),%r12
  lea    D23(%rip),%rbp
  nopl   (%rax)
L8:
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movsd  0x20(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm1
  mov    $0x2,%edi
  movsd  0x10(%rsp),%xmm0
  call   <solve@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,0xc(%rsp)
  jg     L8
  xor    %eax,%eax
L13:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L13
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D36
type_infered: string
name: read_data
arguments: tag: D34
type_infered: string
name: read_data
arguments: tag: D23
type_infered: string
name: read_data
arguments: tag: D6
type_infered: string

tool: D36: %d

tool: D34: %lf %lf %lf

tool: D23: Case #%d: %.7f


tool: D6: Illegal number of tests (1 <= T <= 100)

assistant: ```c
int main(void) {
  int num_tests, test = 0, ret = -1;
  double round, farm_cost, farm_rate, cap;
  scanf("%d", &num_tests);
  if (num_tests < 1 || num_tests > 100) {
    printf("Illegal number of tests (1 <= T <= 100)\n");
    goto out;
  }
  while (test++ < num_tests) {
    scanf("%lf %lf %lf", &farm_cost, &farm_rate, &cap);
    round = solve(2, farm_cost, farm_rate, cap);
    printf("Case #%d: %.7f\n", test, round);
  }
  ret = 0;
out:
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  mov    $0x0,%eax
  call   <abort@plt>
L1:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *alloc(size_t size) {
  void *buffer = malloc(size);
  if (!buffer)
    abort();
  return buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<alloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L9
L4:
  mov    %rbx,%rax
  pop    %rbx
  ret
L9:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L4
```

assistant: ```c
void *alloc(size_t size) {
  void *buffer = malloc(size);
  if (!buffer)
    abort();
  return buffer;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<alloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
L16:
  mov    %r12,%rax
  pop    %r12
  ret
L2:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L16
```

assistant: ```c
void *alloc(size_t size) {
  void *buffer = malloc(size);
  if (!buffer)
    abort();
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<alloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L16
L2:
  mov    %r12,%rax
  pop    %r12
  ret
L16:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L2
```

assistant: ```c
void *alloc(size_t size) {
  void *buffer = malloc(size);
  if (!buffer)
    abort();
  return buffer;
}
```