ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<exact_square>:
  endbr64
  test   %edi,%edi
  js     L3
  mov    $0x1,%eax
  je     L12
  mov    $0x1,%edx
  nopl   0x0(%rax)
L6:
  cmp    %edx,%edi
  je     L34
  add    $0x1,%eax
  mov    %eax,%edx
  imul   %eax,%edx
  cmp    %edi,%edx
  jle    L6
L3:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  mov    $0x1,%eax
  ret
L12:
  ret
```

assistant: ```c
int exact_square(int num1) {
  int isit, mult = 0;
  while (1) {
    if (mult * mult > num1) {
      isit = 0;
      break;
    }
    if (mult * mult == num1) {
      isit = 1;
      break;
    } else
      mult++;
  }
  return isit;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<printline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x18(%rbp)
  jg     L16
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L35
L16:
  mov    $0x2e,%edi
  call   <putchar@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L37
L14:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L37:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L14
  mov    $0x20,%edi
  call   <putchar@plt>
L35:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_guessed: string
name: unpack_data
arguments: data_label: D19
type_guessed: string
name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D2: %2d

tool: D19: : 

tool: D3: ..

assistant: ```c
void printline(int line, int level) {
  printf("%2d", line);
  if (level <= 0)
    printf(": ");
  else {
    printf(".");
    for (int i = 0; i < level; i++)
      printf("..");
    printf(" ");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<printline>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L22
  mov    $0x2e,%edi
  call   <putchar@plt>
  mov    $0x0,%ebx
  lea    D2(%rip),%r12
L1:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L1
  mov    $0x20,%edi
  call   <putchar@plt>
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L3
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_infered: string
name: extract_symbolic_data
arguments: tag: D2
type_infered: string
name: extract_symbolic_data
arguments: tag: D17
type_infered: string

tool: D8: %2d

tool: D2: ..

tool: D17: : 

assistant: ```c
void printline(int line, int level) {
  printf("%2d", line);
  if (level <= 0)
    printf(": ");
  else {
    printf(".");
    for (int i = 0; i < level; i++)
      printf("..");
    printf(" ");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printline>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%esi
  lea    D12(%rip),%rdi
  push   %rbx
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L15
  mov    $0x2e,%edi
  xor    %ebx,%ebx
  lea    D9(%rip),%r12
  call   <putchar@plt>
  nopl   0x0(%rax)
L9:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L9
  pop    %rbx
  mov    $0x20,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L15:
  pop    %rbx
  lea    D27(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
datatype: string
name: disassemble_data
arguments: data_tag: D9
datatype: string
name: disassemble_data
arguments: data_tag: D27
datatype: string

tool: D12: %2d

tool: D9: ..

tool: D27: : 

assistant: ```c
void printline(int line, int level) {
  printf("%2d", line);
  if (level <= 0)
    printf(": ");
  else {
    printf(".");
    for (int i = 0; i < level; i++)
      printf("..");
    printf(" ");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printline>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %esi,%ebp
  mov    %edi,%esi
  lea    D9(%rip),%rdi
  push   %rbx
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L6
  mov    $0x2e,%edi
  xor    %ebx,%ebx
  lea    D0(%rip),%r12
  call   <putchar@plt>
  nopl   0x0(%rax)
L7:
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jne    L7
  pop    %rbx
  mov    $0x20,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  pop    %rbx
  lea    D18(%rip),%rdi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    1070 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type_infered: string
name: disassemble_data
arguments: label: D0
type_infered: string
name: disassemble_data
arguments: label: D18
type_infered: string

tool: D9: %2d

tool: D0: ..

tool: D18: : 

assistant: ```c
void printline(int line, int level) {
  printf("%2d", line);
  if (level <= 0)
    printf(": ");
  else {
    printf(".");
    for (int i = 0; i < level; i++)
      printf("..");
    printf(" ");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fn100>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,%ecx
  shl    $0x2,%ecx
  add    %edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  lea    0x5(%rdx),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
type: string

tool: D8: fn100

assistant: ```c
char *fn100() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn100");
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fn100>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%rbx
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string

tool: D9: fn100

assistant: ```c
char *fn100() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn100");
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fn100>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x4,%esi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  call   <calloc@plt>
  lea    D0(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type_infered: string

tool: D0: fn100

assistant: ```c
char *fn100() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn100");
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fn100>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x4,%esi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  lea    0x5(%rax),%edi
  call   <calloc@plt>
  lea    D6(%rip),%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D6: fn100

assistant: ```c
char *fn100() {
  char *result;
  result = calloc(5 + rand() % 5, 4);
  strcpy((char *)result, "fn100");
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fn2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L9:
  jmp    L9
```

assistant: ```c
void fn2() {
  for (;;)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn2>:
  endbr64
L8:
  jmp    L8
```

assistant: ```c
void fn2() {
  for (;;)
    ;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fn2>:
  endbr64
  nopl   0x0(%rax)
L5:
  jmp    L5
```

assistant: ```c
void fn2() {
  for (;;)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fn2>:
  endbr64
  nopl   0x0(%rax)
L9:
  jmp    L9
```

assistant: ```c
void fn2() {
  for (;;)
    ;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<random_add>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0xffffffffa57eb503,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  imul   $0x63,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L12
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <push@plt>
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L0
L12:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x4(%rbp),%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <append@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
L0:
  nop
  leave
  ret
```

assistant: ```c
void random_add(struct node **head, struct node **tail, int x) {
  int add;
  add = rand() % 99;
  if (*head == NULL) {
    *head = push(*head, x);
    *tail = *head;
  } else {
    *tail = append(*head, *tail, add);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<random_add>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  mov    $0x0,%eax
  call   <rand@plt>
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L12
  movslq %eax,%rcx
  imul   $0xffffffffa57eb503,%rcx,%rcx
  shr    $0x20,%rcx
  add    %eax,%ecx
  sar    $0x6,%ecx
  cltd
  sub    %edx,%ecx
  imul   $0x63,%ecx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    (%rbx),%rsi
  call   <append@plt>
L7:
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    %r12d,%esi
  call   <push@plt>
  mov    %rax,0x0(%rbp)
  jmp    L7
```

assistant: ```c
void random_add(struct node **head, struct node **tail, int x) {
  int add;
  add = rand() % 99;
  if (*head == NULL) {
    *head = push(*head, x);
    *tail = *head;
  } else {
    *tail = append(*head, *tail, add);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<random_add>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <rand@plt>
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L8
  movslq %eax,%rcx
  cltd
  mov    (%rbx),%rsi
  imul   $0xffffffffa57eb503,%rcx,%rcx
  shr    $0x20,%rcx
  add    %eax,%ecx
  sar    $0x6,%ecx
  sub    %edx,%ecx
  imul   $0x63,%ecx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  call   <append@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L8:
  mov    %r12d,%esi
  call   <push@plt>
  mov    %rax,0x0(%rbp)
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void random_add(struct node **head, struct node **tail, int x) {
  int add;
  add = rand() % 99;
  if (*head == NULL) {
    *head = push(*head, x);
    *tail = *head;
  } else {
    *tail = append(*head, *tail, add);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<random_add>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <rand@plt>
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L5
  movslq %eax,%rcx
  cltd
  mov    (%rbx),%rsi
  imul   $0xffffffffa57eb503,%rcx,%rcx
  shr    $0x20,%rcx
  add    %eax,%ecx
  sar    $0x6,%ecx
  sub    %edx,%ecx
  imul   $0x63,%ecx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  call   <append@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L5:
  mov    %r12d,%esi
  call   <push@plt>
  mov    %rax,0x0(%rbp)
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void random_add(struct node **head, struct node **tail, int x) {
  int add;
  add = rand() % 99;
  if (*head == NULL) {
    *head = push(*head, x);
    *tail = *head;
  } else {
    *tail = append(*head, *tail, add);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<monitor_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L1
L7:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <monitor_put@plt>
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void monitor_write(char *c) {
  int i;
  while (c[i]) {
    monitor_put(c[i++]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<monitor_write>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L12
  push   %rbx
  lea    0x1(%rdi),%rbx
L0:
  movsbl %al,%edi
  call   <monitor_put@plt>
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L0
  pop    %rbx
  ret
L12:
  ret
```

assistant: ```c
void monitor_write(char *c) {
  int i;
  while (c[i]) {
    monitor_put(c[i++]);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<monitor_write>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L19
  push   %rbx
  lea    0x1(%rdi),%rbx
L1:
  movsbl %al,%edi
  add    $0x1,%rbx
  call   <monitor_put@plt>
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L1
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L19:
  ret
```

assistant: ```c
void monitor_write(char *c) {
  int i;
  while (c[i]) {
    monitor_put(c[i++]);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<monitor_write>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L5
  push   %rbx
  lea    0x1(%rdi),%rbx
L4:
  movsbl %al,%edi
  add    $0x1,%rbx
  call   <monitor_put@plt>
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  jne    L4
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  ret
```

assistant: ```c
void monitor_write(char *c) {
  int i;
  while (c[i]) {
    monitor_put(c[i++]);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
L64:
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x28(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L86
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L86
  movsd  -0x20(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L86
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L86
  movsd  -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L86
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  je     L93
L86:
  addl   $0x1,-0x2c(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L51
L30:
  movsd  -0x20(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  D25(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x30(%rbp)
L51:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x30(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  comisd %xmm1,%xmm0
  ja     L30
  movsd  -0x18(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  mov    -0x28(%rbp),%rdx
  movsd  -0x10(%rbp),%xmm2
  mov    -0x2c(%rbp),%eax
  movapd %xmm2,%xmm3
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L64
L93:
  nop
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
type: string
name: read_data
arguments: data_tag: D25
type: f64
name: read_data
arguments: data_tag: D6
type: string

tool: D5: %lf%lf%lf

tool: D25: 100

tool: D6: Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf


assistant: ```c
int main() {
  double p, r, t;
  int i = 0, j = 0;
  double ans;
  while (1) {
    scanf("%lf%lf%lf", &p, &r, &t);
    if (p == 0 && r == 0 && t == 0)
      break;
    j++;
    ans = p;
    for (i = 0; i < t; i++) {
      ans += ((int)floor(ans * r / t)) / 100.0;
    }
    printf("Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf\n", j, p, r, t, ans);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  lea    D32(%rip),%r12
  jmp    L80
L51:
  add    $0x1,%ebp
  movsd  0x20(%rsp),%xmm2
  mov    $0x0,%ebx
  comisd D33(%rip),%xmm2
  jbe    L97
L95:
  movsd  0x8(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  divsd  %xmm2,%xmm0
  call   <floor@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D0(%rip),%xmm0
  movapd %xmm0,%xmm5
  addsd  0x8(%rsp),%xmm5
  movsd  %xmm5,0x8(%rsp)
  add    $0x1,%ebx
  movsd  0x20(%rsp),%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  comisd %xmm0,%xmm2
  ja     L95
L97:
  movsd  0x8(%rsp),%xmm3
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm0
  mov    %ebp,%esi
  lea    D29(%rip),%rdi
  call   <printf@plt>
L80:
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm6
  movsd  %xmm6,0x8(%rsp)
  pxor   %xmm7,%xmm7
  ucomisd %xmm7,%xmm6
  jp     L51
  jne    L51
  ucomisd 0x18(%rsp),%xmm7
  jp     L51
  jne    L51
  ucomisd 0x20(%rsp),%xmm7
  jp     L51
  jne    L51
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D32
type_guessed: string
name: unpack_data
arguments: label: D33
type_guessed: byte[16]
name: unpack_data
arguments: label: D0
type_guessed: f64
name: unpack_data
arguments: label: D29
type_guessed: string

tool: D32: %lf%lf%lf

tool: {"D33": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D0: 100

tool: D29: Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf


assistant: ```c
int main() {
  double p, r, t;
  int i = 0, j = 0;
  double ans;
  while (1) {
    scanf("%lf%lf%lf", &p, &r, &t);
    if (p == 0 && r == 0 && t == 0)
      break;
    j++;
    ans = p;
    for (i = 0; i < t; i++) {
      ans += ((int)floor(ans * r / t)) / 100.0;
    }
    printf("Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf\n", j, p, r, t, ans);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D26(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  lea    0x10(%rsp),%r12
  nopw   0x0(%rax,%rax,1)
L58:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm3
  pxor   %xmm4,%xmm4
  movsd  0x18(%rsp),%xmm1
  movsd  0x20(%rsp),%xmm2
  ucomisd %xmm4,%xmm3
  jp     L32
  jne    L32
  pxor   %xmm6,%xmm6
  ucomisd %xmm6,%xmm1
  jp     L32
  jne    L32
  ucomisd %xmm6,%xmm2
  jp     L32
  je     L92
L32:
  pxor   %xmm5,%xmm5
  add    $0x1,%ebx
  xor    %r15d,%r15d
  comisd %xmm5,%xmm2
  jbe    L14
  nopw   0x0(%rax,%rax,1)
L45:
  mulsd  %xmm3,%xmm1
  movsd  %xmm3,0x8(%rsp)
  add    $0x1,%r15d
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  call   <floor@plt>
  movsd  0x8(%rsp),%xmm3
  pxor   %xmm0,%xmm0
  movsd  0x20(%rsp),%xmm2
  cvtsi2sd %eax,%xmm0
  divsd  D21(%rip),%xmm0
  movsd  0x18(%rsp),%xmm1
  addsd  %xmm0,%xmm3
  pxor   %xmm0,%xmm0
  cvtsi2sd %r15d,%xmm0
  comisd %xmm0,%xmm2
  ja     L45
  movsd  0x10(%rsp),%xmm0
L31:
  mov    %ebx,%esi
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L58
  nopl   0x0(%rax)
L14:
  movapd %xmm3,%xmm0
  jmp    L31
L92:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D26
type_guessed: string
name: disassemble_data
arguments: label: D21
type_guessed: f64
name: disassemble_data
arguments: label: D20
type_guessed: string

tool: D26: %lf%lf%lf

tool: D21: 100

tool: D20: Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf


assistant: ```c
int main() {
  double p, r, t;
  int i = 0, j = 0;
  double ans;
  while (1) {
    scanf("%lf%lf%lf", &p, &r, &t);
    if (p == 0 && r == 0 && t == 0)
      break;
    j++;
    ans = p;
    for (i = 0; i < t; i++) {
      ans += ((int)floor(ans * r / t)) / 100.0;
    }
    printf("Case %d. $%.2lf at %.2lf%% APR compounded %.0lf times yields $%.2lf\n", j, p, r, t, ans);
  }
  return 0;
}
```