ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x2710,%edi
  call   <sleep@plt>
  mov    $0x2710,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
void *func(void *arg) {
  sleep(10000);
  sleep(10000);
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<func>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2710,%edi
  call   <sleep@plt>
  mov    $0x2710,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *func(void *arg) {
  sleep(10000);
  sleep(10000);
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<func>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2710,%edi
  call   <sleep@plt>
  mov    $0x2710,%edi
  call   <sleep@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *func(void *arg) {
  sleep(10000);
  sleep(10000);
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<expr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <term@plt>
  mov    $0x0,%eax
  call   <exprp@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void expr() {
  term();
  exprp();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<expr>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <term@plt>
  mov    $0x0,%eax
  call   <exprp@plt>
  add    $0x8,%rsp
  ret
```

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

```asm
<expr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <term@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <exprp@plt>
```

assistant: ```c
void expr() {
  term();
  exprp();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<expr>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <term@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <exprp@plt>
```

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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x38(%rbp)
  lea    D34(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D3(%rip),%rax
  mov    %rax,-0x28(%rbp)
  lea    D32(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x50(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,(%rdx)
  addl   $0x1,-0x44(%rbp)
  mov    -0x50(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L36
L46:
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L36:
  mov    -0x38(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L46
  movl   $0x5,-0x34(%rbp)
  movl   $0x0,-0x38(%rbp)
  jmp    L2
L1:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,8),%rdx
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x38(%rbp)
L2:
  mov    -0x38(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D34
type_infered: string
name: retrieve_data
arguments: data_label: D3
type_infered: string
name: retrieve_data
arguments: data_label: D32
type_infered: string
name: retrieve_data
arguments: data_label: D7
type_infered: string
name: retrieve_data
arguments: data_label: D59
type_infered: string
name: retrieve_data
arguments: data_label: D20
type_infered: string

tool: D34: California

tool: D3: Oregon

tool: D32: Washington

tool: D7: Texas

tool: D59: arg %d: %s


tool: D20: state %d: %s


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  char *states[] = {"California", "Oregon", "Washington", "Texas"};
  argv[2] = states[1];
  argc = argc + 1;
  states[4] = argv[1];
  for (i = 0; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  int num_states = 5;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s\n", i, states[i]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D35(%rip),%rax
  mov    %rax,(%rsp)
  lea    D55(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D8(%rip),%rcx
  mov    %rcx,0x10(%rsp)
  lea    D3(%rip),%rcx
  mov    %rcx,0x18(%rsp)
  mov    %rax,0x10(%rsi)
  test   %edi,%edi
  js     L21
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    $0x0,%ebx
  lea    D45(%rip),%r13
L10:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jge    L10
L21:
  mov    $0x0,%ebx
  lea    D26(%rip),%rbp
L31:
  mov    (%rsp,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x5,%rbx
  jne    L31
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D35
datatype: string
name: retrieve_data
arguments: tag: D55
datatype: string
name: retrieve_data
arguments: tag: D8
datatype: string
name: retrieve_data
arguments: tag: D3
datatype: string
name: retrieve_data
arguments: tag: D45
datatype: string
name: retrieve_data
arguments: tag: D26
datatype: string

tool: D35: California

tool: D55: Oregon

tool: D8: Washington

tool: D3: Texas

tool: D45: arg %d: %s


tool: D26: state %d: %s


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  char *states[] = {"California", "Oregon", "Washington", "Texas"};
  argv[2] = states[1];
  argc = argc + 1;
  states[4] = argv[1];
  for (i = 0; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  int num_states = 5;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s\n", i, states[i]);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  lea    D46(%rip),%rcx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D26(%rip),%rax
  mov    %rcx,0x10(%rsp)
  lea    D56(%rip),%rcx
  mov    %rcx,0x18(%rsp)
  mov    %rax,0x8(%rsp)
  mov    %rax,0x10(%rsi)
  test   %edi,%edi
  js     L40
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D12(%rip),%r13
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jge    L19
L40:
  xor    %ebx,%ebx
  lea    D58(%rip),%rdx
  lea    D51(%rip),%rbp
L5:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  je     L44
  mov    (%rsp,%rbx,8),%rdx
  jmp    L5
L44:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D46
type_infered: string
name: disassemble_data
arguments: data_label: D26
type_infered: string
name: disassemble_data
arguments: data_label: D56
type_infered: string
name: disassemble_data
arguments: data_label: D12
type_infered: string
name: disassemble_data
arguments: data_label: D58
type_infered: string
name: disassemble_data
arguments: data_label: D51
type_infered: string

tool: D46: Washington

tool: D26: Oregon

tool: D56: Texas

tool: D12: arg %d: %s


tool: D58: California

tool: D51: state %d: %s


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  char *states[] = {"California", "Oregon", "Washington", "Texas"};
  argv[2] = states[1];
  argc = argc + 1;
  states[4] = argv[1];
  for (i = 0; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  int num_states = 5;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s\n", i, states[i]);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D56(%rip),%rcx
  lea    D34(%rip),%rdx
  push   %r12
  movq   %rcx,%xmm0
  lea    D1(%rip),%rcx
  movq   %rdx,%xmm2
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  lea    D26(%rip),%rax
  movq   %rax,%xmm1
  mov    %rax,0x10(%rsi)
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  movq   %rcx,%xmm0
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x10(%rsp)
  test   %edi,%edi
  js     L6
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D24(%rip),%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jge    L1
L6:
  xor    %ebx,%ebx
  mov    %rsp,%r12
  lea    D41(%rip),%rbp
L3:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  jne    L3
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D56
type: string
name: read_data
arguments: data_label: D34
type: string
name: read_data
arguments: data_label: D1
type: string
name: read_data
arguments: data_label: D26
type: string
name: read_data
arguments: data_label: D24
type: string
name: read_data
arguments: data_label: D41
type: string

tool: D56: California

tool: D34: Texas

tool: D1: Washington

tool: D26: Oregon

tool: D24: arg %d: %s


tool: D41: state %d: %s


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  char *states[] = {"California", "Oregon", "Washington", "Texas"};
  argv[2] = states[1];
  argc = argc + 1;
  states[4] = argv[1];
  for (i = 0; i < argc; i++) {
    printf("arg %d: %s\n", i, argv[i]);
  }
  int num_states = 5;
  for (i = 0; i < num_states; i++) {
    printf("state %d: %s\n", i, states[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x32,%edi
  call   <insert_end@plt>
  mov    $0x46,%edi
  call   <insert_end@plt>
  mov    $0x5a,%edi
  call   <insert_end@plt>
  mov    $0x6e,%edi
  call   <insert_end@plt>
  mov    $0x0,%eax
  call   <show_list@plt>
  mov    $0x1,%edi
  call   <delete_node@plt>
  mov    $0x0,%eax
  call   <show_list@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  insert_end(50);
  insert_end(70);
  insert_end(90);
  insert_end(110);
  show_list();
  delete_node(1);
  show_list();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%edi
  call   <insert_end@plt>
  mov    $0x46,%edi
  call   <insert_end@plt>
  mov    $0x5a,%edi
  call   <insert_end@plt>
  mov    $0x6e,%edi
  call   <insert_end@plt>
  mov    $0x0,%eax
  call   <show_list@plt>
  mov    $0x1,%edi
  call   <delete_node@plt>
  mov    $0x0,%eax
  call   <show_list@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  insert_end(50);
  insert_end(70);
  insert_end(90);
  insert_end(110);
  show_list();
  delete_node(1);
  show_list();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%edi
  call   <insert_end@plt>
  mov    $0x46,%edi
  call   <insert_end@plt>
  mov    $0x5a,%edi
  call   <insert_end@plt>
  mov    $0x6e,%edi
  call   <insert_end@plt>
  xor    %eax,%eax
  call   <show_list@plt>
  mov    $0x1,%edi
  call   <delete_node@plt>
  xor    %eax,%eax
  call   <show_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  insert_end(50);
  insert_end(70);
  insert_end(90);
  insert_end(110);
  show_list();
  delete_node(1);
  show_list();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x32,%edi
  call   <insert_end@plt>
  mov    $0x46,%edi
  call   <insert_end@plt>
  mov    $0x5a,%edi
  call   <insert_end@plt>
  mov    $0x6e,%edi
  call   <insert_end@plt>
  xor    %eax,%eax
  call   <show_list@plt>
  mov    $0x1,%edi
  call   <delete_node@plt>
  xor    %eax,%eax
  call   <show_list@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  insert_end(50);
  insert_end(70);
  insert_end(90);
  insert_end(110);
  show_list();
  delete_node(1);
  show_list();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D63(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x28(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  0x28(%rsp),%xmm5
  movsd  0x20(%rsp),%xmm4
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm2
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm0
  call   <Section1@plt>
  lea    D41(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x30(%rsp),%rdi
  call   <Section2@plt>
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <Section3@plt>
  mov    %rax,%rsi
  mov    %rax,0x30(%rsp)
  lea    D79(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D30
type: string
name: disassemble_data
arguments: data_tag: D63
type: string
name: disassemble_data
arguments: data_tag: D10
type: string
name: disassemble_data
arguments: data_tag: D41
type: string
name: disassemble_data
arguments: data_tag: D26
type: string
name: disassemble_data
arguments: data_tag: D54
type: string
name: disassemble_data
arguments: data_tag: D29
type: string
name: disassemble_data
arguments: data_tag: D79
type: string

tool: D30: 
1.- * * * Manhattan - Taxicab * * *
Insert first point coordinates (x,y,z) separated by commas: 

tool: D63: %lf,%lf,%lf

tool: D10: Insert second point coordinates (x,y,z) separated by commas: 

tool: D41: The Manhattan distance is: %g.



tool: D26: 2.- * * * Cents to Euros * * *
Insert the ammount of cents: 

tool: D54: %ld

tool: D29: 3.- * * * Euros to Cents * * *

tool: D79: Equivalent in cents: %ld



assistant: ```c
int main() {
  double x1, y1, z1, x2, y2, z2;
  long cents;
  printf("\n1.- * * * Manhattan - Taxicab * * *\nInsert first point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x1, &y1, &z1);
  printf("Insert second point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x2, &y2, &z2);
  printf("The Manhattan distance is: %g.\n\n", Section1(x1, y1, z1, x2, y2, z2));
  printf("2.- * * * Cents to Euros * * *\nInsert the ammount of cents: ");
  scanf("%ld", &cents);
  Section2(cents);
  printf("3.- * * * Euros to Cents * * *\n");
  cents = Section3();
  printf("Equivalent in cents: %ld\n\n", cents);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rdi
  lea    D43(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rcx
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D52(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x28(%rsp),%rcx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  call   <scanf@plt>
  movsd  0x28(%rsp),%xmm5
  movsd  (%rsp),%xmm0
  movsd  0x20(%rsp),%xmm4
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm2
  movsd  0x8(%rsp),%xmm1
  call   <Section1@plt>
  lea    D75(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x30(%rsp),%rdi
  call   <Section2@plt>
  lea    D9(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <Section3@plt>
  lea    D20(%rip),%rdi
  mov    %rax,0x30(%rsp)
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type_infered: string
name: disassemble_data
arguments: tag: D43
type_infered: string
name: disassemble_data
arguments: tag: D52
type_infered: string
name: disassemble_data
arguments: tag: D75
type_infered: string
name: disassemble_data
arguments: tag: D64
type_infered: string
name: disassemble_data
arguments: tag: D26
type_infered: string
name: disassemble_data
arguments: tag: D9
type_infered: string
name: disassemble_data
arguments: tag: D20
type_infered: string

tool: D5: 
1.- * * * Manhattan - Taxicab * * *
Insert first point coordinates (x,y,z) separated by commas: 

tool: D43: %lf,%lf,%lf

tool: D52: Insert second point coordinates (x,y,z) separated by commas: 

tool: D75: The Manhattan distance is: %g.



tool: D64: 2.- * * * Cents to Euros * * *
Insert the ammount of cents: 

tool: D26: %ld

tool: D9: 3.- * * * Euros to Cents * * *

tool: D20: Equivalent in cents: %ld



assistant: ```c
int main() {
  double x1, y1, z1, x2, y2, z2;
  long cents;
  printf("\n1.- * * * Manhattan - Taxicab * * *\nInsert first point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x1, &y1, &z1);
  printf("Insert second point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x2, &y2, &z2);
  printf("The Manhattan distance is: %g.\n\n", Section1(x1, y1, z1, x2, y2, z2));
  printf("2.- * * * Cents to Euros * * *\nInsert the ammount of cents: ");
  scanf("%ld", &cents);
  Section2(cents);
  printf("3.- * * * Euros to Cents * * *\n");
  cents = Section3();
  printf("Equivalent in cents: %ld\n\n", cents);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D42(%rip),%rdi
  lea    D29(%rip),%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rcx
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  lea    0x8(%rsp),%rdx
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x28(%rsp),%rcx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    0x20(%rsp),%rdx
  lea    0x18(%rsp),%rsi
  call   <scanf@plt>
  movsd  0x28(%rsp),%xmm5
  movsd  (%rsp),%xmm0
  movsd  0x20(%rsp),%xmm4
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm2
  movsd  0x8(%rsp),%xmm1
  call   <Section1@plt>
  lea    D21(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D72(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x30(%rsp),%rsi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x30(%rsp),%rdi
  call   <Section2@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <Section3@plt>
  lea    D73(%rip),%rdi
  mov    %rax,0x30(%rsp)
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D42
type_infered: string
name: read_data
arguments: data_label: D29
type_infered: string
name: read_data
arguments: data_label: D23
type_infered: string
name: read_data
arguments: data_label: D21
type_infered: string
name: read_data
arguments: data_label: D72
type_infered: string
name: read_data
arguments: data_label: D39
type_infered: string
name: read_data
arguments: data_label: D13
type_infered: string
name: read_data
arguments: data_label: D73
type_infered: string

tool: D42: 
1.- * * * Manhattan - Taxicab * * *
Insert first point coordinates (x,y,z) separated by commas: 

tool: D29: %lf,%lf,%lf

tool: D23: Insert second point coordinates (x,y,z) separated by commas: 

tool: D21: The Manhattan distance is: %g.



tool: D72: 2.- * * * Cents to Euros * * *
Insert the ammount of cents: 

tool: D39: %ld

tool: D13: 3.- * * * Euros to Cents * * *

tool: D73: Equivalent in cents: %ld



assistant: ```c
int main() {
  double x1, y1, z1, x2, y2, z2;
  long cents;
  printf("\n1.- * * * Manhattan - Taxicab * * *\nInsert first point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x1, &y1, &z1);
  printf("Insert second point coordinates (x,y,z) separated by commas: ");
  scanf("%lf,%lf,%lf", &x2, &y2, &z2);
  printf("The Manhattan distance is: %g.\n\n", Section1(x1, y1, z1, x2, y2, z2));
  printf("2.- * * * Cents to Euros * * *\nInsert the ammount of cents: ");
  scanf("%ld", &cents);
  Section2(cents);
  printf("3.- * * * Euros to Cents * * *\n");
  cents = Section3();
  printf("Equivalent in cents: %ld\n\n", cents);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_qnqstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L14
L5:
  addq   $0x1,-0x18(%rbp)
L14:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L5
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L5
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L67
  mov    -0x18(%rbp),%rax
  jmp    L45
L67:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  cmpb   $0x22,-0x1(%rbp)
  je     L70
  cmpb   $0x27,-0x1(%rbp)
  jne    L16
L70:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <get_qstring@plt>
  jmp    L45
L16:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <get_string@plt>
L45:
  leave
  ret
```

assistant: ```c
const char *get_qnqstring(const char *pnt, char **label) {
  char quote_char;
  while (*pnt == ' ' || *pnt == '\t')
    pnt++;
  if (*pnt == '\0')
    return pnt;
  quote_char = *pnt;
  if (quote_char == '"' || quote_char == '\'')
    return get_qstring(pnt, label);
  else
    return get_string(pnt, label);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_qnqstring>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L70
  cmp    $0x9,%dl
  jne    L50
L70:
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  je     L70
  cmp    $0x9,%dl
  je     L70
L50:
  mov    %rdi,%rax
  test   %dl,%dl
  je     L63
  sub    $0x8,%rsp
  cmp    $0x22,%dl
  je     L13
  cmp    $0x27,%dl
  je     L13
  call   <get_string@plt>
  jmp    L6
L13:
  call   <get_qstring@plt>
L6:
  add    $0x8,%rsp
  ret
L63:
  ret
```

assistant: ```c
const char *get_qnqstring(const char *pnt, char **label) {
  char quote_char;
  while (*pnt == ' ' || *pnt == '\t')
    pnt++;
  if (*pnt == '\0')
    return pnt;
  quote_char = *pnt;
  if (quote_char == '"' || quote_char == '\'')
    return get_qstring(pnt, label);
  else
    return get_string(pnt, label);
}
```