ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L9:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <month_name@plt>
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L11:
  cmpl   $0xe,-0x4(%rbp)
  jle    L9
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D3
data_type: string

tool: D3: n=%d : %s


assistant: ```c
int main() {
  int i = 0;
  for (i = 0; i < 15; i++) {
    printf("n=%d : %s\n", i, month_name(i));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D6(%rip),%rbp
L1:
  mov    %ebx,%edi
  call   <month_name@plt>
  mov    %rax,%rdx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xf,%ebx
  jne    L1
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: string

tool: D6: n=%d : %s


assistant: ```c
int main() {
  int i = 0;
  for (i = 0; i < 15; i++) {
    printf("n=%d : %s\n", i, month_name(i));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %ebx,%edi
  call   <month_name@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %rax,%rdx
  call   <printf@plt>
  cmp    $0xf,%ebx
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
datatype: string

tool: D5: n=%d : %s


assistant: ```c
int main() {
  int i = 0;
  for (i = 0; i < 15; i++) {
    printf("n=%d : %s\n", i, month_name(i));
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %ebx,%edi
  call   <month_name@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  mov    %rax,%rdx
  call   <printf@plt>
  cmp    $0xf,%ebx
  jne    L9
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
datatype: string

tool: D2: n=%d : %s


assistant: ```c
int main() {
  int i = 0;
  for (i = 0; i < 15; i++) {
    printf("n=%d : %s\n", i, month_name(i));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<signatures_ok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int signatures_ok(uint8_t *store_hash) {
  (void)store_hash;
  return false;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<signatures_ok>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int signatures_ok(uint8_t *store_hash) {
  (void)store_hash;
  return false;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<signatures_ok>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int signatures_ok(uint8_t *store_hash) {
  (void)store_hash;
  return false;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<signatures_ok>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int signatures_ok(uint8_t *store_hash) {
  (void)store_hash;
  return false;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sha1_self_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: string

tool: D0: SHA-1 self-test not available



assistant: ```c
int sha1_self_test() {
  printf("SHA-1 self-test not available\n\n");
  return (1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sha1_self_test>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D6
datatype: string

tool: D6: SHA-1 self-test not available



assistant: ```c
int sha1_self_test() {
  printf("SHA-1 self-test not available\n\n");
  return (1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sha1_self_test>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
data_type: string

tool: D3: SHA-1 self-test not available



assistant: ```c
int sha1_self_test() {
  printf("SHA-1 self-test not available\n\n");
  return (1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sha1_self_test>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: SHA-1 self-test not available



assistant: ```c
int sha1_self_test() {
  printf("SHA-1 self-test not available\n\n");
  return (1);
}
```
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
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <fork@plt>
  cmp    $0xffffffff,%eax
  je     L2
  test   %eax,%eax
  jne    L24
  mov    $0xde,%esi
  lea    D96(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0xde,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0xde,%esi
  lea    D107(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0xde,%esi
  lea    D51(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0xde,%esi
  lea    D86(%rip),%rdi
  call   <printf@plt>
L13:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L2:
  lea    D16(%rip),%rdi
  call   <perror@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L24:
  mov    $0x3e7,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0x3e7,%esi
  lea    D58(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0x3e7,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0x3e7,%esi
  lea    D64(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <sleeprand@plt>
  mov    $0x3e7,%esi
  lea    D61(%rip),%rdi
  call   <printf@plt>
  jmp    L13
```

tool_calls: name: read_data
arguments: label: D96
type_guessed: string
name: read_data
arguments: label: D0
type_guessed: string
name: read_data
arguments: label: D107
type_guessed: string
name: read_data
arguments: label: D51
type_guessed: string
name: read_data
arguments: label: D86
type_guessed: string
name: read_data
arguments: label: D16
type_guessed: string
name: read_data
arguments: label: D93
type_guessed: string
name: read_data
arguments: label: D58
type_guessed: string
name: read_data
arguments: label: D8
type_guessed: string
name: read_data
arguments: label: D64
type_guessed: string
name: read_data
arguments: label: D61
type_guessed: string

tool: D96: 1. I am child: x = %d


tool: D0: 2. I am child: x = %d


tool: D107: 3. I am child: x = %d


tool: D51: 4. I am child: x = %d


tool: D86: 5. I am child: x = %d


tool: D16: Fork:

tool: D93: 1. I am parent: x = %d


tool: D58: 2. I am parent: x = %d


tool: D8: 3. I am parent: x = %d


tool: D64: 4. I am parent: x = %d


tool: D61: 5. I am parent: x = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int pid;
  int x = 999;
  if ((pid = fork()) == -1) {
    perror("Fork:");
    exit(-1);
  }
  if (pid == 0) {
    x = 222;
    printf("1. I am child: x = %d\n", x);
    sleeprand();
    printf("2. I am child: x = %d\n", x);
    sleeprand();
    printf("3. I am child: x = %d\n", x);
    sleeprand();
    printf("4. I am child: x = %d\n", x);
    sleeprand();
    printf("5. I am child: x = %d\n", x);
  } else {
    printf("1. I am parent: x = %d\n", x);
    sleeprand();
    printf("2. I am parent: x = %d\n", x);
    sleeprand();
    printf("3. I am parent: x = %d\n", x);
    sleeprand();
    printf("4. I am parent: x = %d\n", x);
    sleeprand();
    printf("5. I am parent: x = %d\n", x);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L16
  movl   $0x1,-0x8(%rbp)
  negl   -0x14(%rbp)
L16:
  movl   $0x0,-0xc(%rbp)
  jmp    L51
L7:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L51:
  cmpl   $0x0,-0x14(%rbp)
  jne    L7
  cmpl   $0x1,-0x8(%rbp)
  jne    L4
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  addl   $0x1,-0xc(%rbp)
L4:
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L44:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    -0xc(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0xc(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jl     L44
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void itoa(int n, char *str) {
  int i, minus = 0;
  if (n < 0) {
    minus = 1;
    n *= -1;
  }
  for (i = 0; n != 0; i++) {
    str[i] = n % 10 + '0';
    n /= 10;
  }
  if (1 == minus) {
    str[i] = '-';
    i++;
  }
  for (int j = 0; j < i / 2; j++) {
    char temp;
    temp = str[j];
    str[j] = str[i - j - 1];
    str[i - j - 1] = temp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoa>:
  endbr64
  mov    %rsi,%r8
  test   %edi,%edi
  js     L3
  je     L36
  mov    $0x0,%r9d
L49:
  mov    %r8,%rsi
  mov    $0x0,%ecx
L55:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,(%rsi)
  mov    %eax,%edi
  mov    %ecx,%edx
  add    $0x1,%ecx
  add    $0x1,%rsi
  test   %eax,%eax
  jne    L55
  cmp    $0x1,%r9d
  je     L13
L31:
  mov    %ecx,%edi
  shr    $0x1f,%edi
  add    %ecx,%edi
  sar    %edi
  cmp    $0x1,%ecx
  jle    L36
  movslq %ecx,%rcx
  lea    -0x1(%r8,%rcx,1),%rdx
  mov    $0x0,%eax
L35:
  movzbl (%r8,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,(%r8,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%edi
  jg     L35
L36:
  ret
L3:
  neg    %edi
  mov    $0x1,%r9d
  jmp    L49
L13:
  movslq %ecx,%rcx
  movb   $0x2d,(%r8,%rcx,1)
  lea    0x2(%rdx),%ecx
  jmp    L31
```

assistant: ```c
void itoa(int n, char *str) {
  int i, minus = 0;
  if (n < 0) {
    minus = 1;
    n *= -1;
  }
  for (i = 0; n != 0; i++) {
    str[i] = n % 10 + '0';
    n /= 10;
  }
  if (1 == minus) {
    str[i] = '-';
    i++;
  }
  for (int j = 0; j < i / 2; j++) {
    char temp;
    temp = str[j];
    str[j] = str[i - j - 1];
    str[i - j - 1] = temp;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<itoa>:
  endbr64
  mov    %rsi,%r8
  test   %edi,%edi
  js     L64
  mov    $0x0,%r10d
  je     L50
L79:
  mov    %r8,%rsi
  xor    %ecx,%ecx
  mov    $0xcccccccd,%r9d
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %edi,%eax
  add    $0x1,%rsi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,-0x1(%rsi)
  mov    %eax,%edi
  mov    %ecx,%eax
  add    $0x1,%ecx
  test   %edi,%edi
  jne    L4
  cmp    $0x1,%r10d
  je     L60
  mov    %ecx,%edi
  sar    %edi
  je     L67
L63:
  movslq %ecx,%rcx
  xor    %eax,%eax
  lea    -0x1(%r8,%rcx,1),%rdx
  nopl   (%rax)
L77:
  movzbl (%rdx),%esi
  movzbl (%r8,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L77
L67:
  ret
  nopl   0x0(%rax)
L64:
  neg    %edi
  mov    $0x1,%r10d
  jmp    L79
  nopw   0x0(%rax,%rax,1)
L60:
  movslq %ecx,%rcx
  movb   $0x2d,(%r8,%rcx,1)
  lea    0x2(%rax),%ecx
  mov    %ecx,%edi
  sar    %edi
  jmp    L63
L50:
  ret
```

assistant: ```c
void itoa(int n, char *str) {
  int i, minus = 0;
  if (n < 0) {
    minus = 1;
    n *= -1;
  }
  for (i = 0; n != 0; i++) {
    str[i] = n % 10 + '0';
    n /= 10;
  }
  if (1 == minus) {
    str[i] = '-';
    i++;
  }
  for (int j = 0; j < i / 2; j++) {
    char temp;
    temp = str[j];
    str[j] = str[i - j - 1];
    str[i - j - 1] = temp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<itoa>:
  endbr64
  mov    %rsi,%r8
  test   %edi,%edi
  js     L38
  mov    $0x0,%r10d
  je     L13
L22:
  mov    %r8,%rsi
  xor    %ecx,%ecx
  mov    $0xcccccccd,%r9d
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %edi,%eax
  add    $0x1,%rsi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,-0x1(%rsi)
  mov    %eax,%edi
  mov    %ecx,%eax
  add    $0x1,%ecx
  test   %edi,%edi
  jne    L11
  cmp    $0x1,%r10d
  je     L54
  mov    %ecx,%edi
  sar    %edi
  je     L27
L7:
  movslq %ecx,%rcx
  xor    %eax,%eax
  lea    -0x1(%r8,%rcx,1),%rdx
  nopl   (%rax)
L28:
  movzbl (%rdx),%esi
  movzbl (%r8,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L28
L27:
  ret
  nopl   0x0(%rax)
L38:
  neg    %edi
  mov    $0x1,%r10d
  jmp    L22
  nopw   0x0(%rax,%rax,1)
L54:
  movslq %ecx,%rcx
  movb   $0x2d,(%r8,%rcx,1)
  lea    0x2(%rax),%ecx
  mov    %ecx,%edi
  sar    %edi
  jmp    L7
L13:
  ret
```

assistant: ```c
void itoa(int n, char *str) {
  int i, minus = 0;
  if (n < 0) {
    minus = 1;
    n *= -1;
  }
  for (i = 0; n != 0; i++) {
    str[i] = n % 10 + '0';
    n /= 10;
  }
  if (1 == minus) {
    str[i] = '-';
    i++;
  }
  for (int j = 0; j < i / 2; j++) {
    char temp;
    temp = str[j];
    str[j] = str[i - j - 1];
    str[i - j - 1] = temp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  jmp    L17
L4:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
L17:
  mov    -0x20(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L4
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D46
type_guessed: string
name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D57
type_guessed: string
name: unpack_data
arguments: data_tag: D23
type_guessed: string
name: unpack_data
arguments: data_tag: D10
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string

tool: D46: while Loop Example 5


tool: D5: How many numbers would you like to enter? 

tool: D57: %d

tool: D23: Enter value %d: 

tool: D10: %lf

tool: D11: The sum of the values is %4.4lf.


assistant: ```c
int main() {
  int numValuesEnteredSoFar = 0;
  int numValuesToEnter;
  double numEntered, sum;
  printf("while Loop Example 5\n\n");
  printf("How many numbers would you like to enter? ");
  scanf("%d", &numValuesToEnter);
  sum = 0.0;
  while (numValuesEnteredSoFar < numValuesToEnter) {
    numValuesEnteredSoFar++;
    printf("Enter value %d: ", numValuesEnteredSoFar);
    scanf("%lf", &numEntered);
    sum = sum + numEntered;
  }
  printf("The sum of the values is %4.4lf.\n", sum);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <puts@plt>
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L36
  mov    D6(%rip),%rbp
  mov    $0x0,%ebx
  lea    D34(%rip),%r14
  lea    0x10(%rsp),%r13
  lea    D77(%rip),%r12
L9:
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movq   %rbp,%xmm1
  addsd  0x10(%rsp),%xmm1
  movq   %xmm1,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L9
L38:
  movq   %rbp,%xmm0
  lea    D32(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  mov    D6(%rip),%rbp
  jmp    L38
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_infered: string
name: unpack_data
arguments: data_label: D4
type_infered: string
name: unpack_data
arguments: data_label: D39
type_infered: string
name: unpack_data
arguments: data_label: D6
type_infered: f64
name: unpack_data
arguments: data_label: D34
type_infered: string
name: unpack_data
arguments: data_label: D77
type_infered: string
name: unpack_data
arguments: data_label: D32
type_infered: string

tool: D8: while Loop Example 5


tool: D4: How many numbers would you like to enter? 

tool: D39: %d

tool: D6: 0

tool: D34: Enter value %d: 

tool: D77: %lf

tool: D32: The sum of the values is %4.4lf.


assistant: ```c
int main() {
  int numValuesEnteredSoFar = 0;
  int numValuesToEnter;
  double numEntered, sum;
  printf("while Loop Example 5\n\n");
  printf("How many numbers would you like to enter? ");
  scanf("%d", &numValuesToEnter);
  sum = 0.0;
  while (numValuesEnteredSoFar < numValuesToEnter) {
    numValuesEnteredSoFar++;
    printf("Enter value %d: ", numValuesEnteredSoFar);
    scanf("%lf", &numEntered);
    sum = sum + numEntered;
  }
  printf("The sum of the values is %4.4lf.\n", sum);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D17(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D50(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1c(%rsp),%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x1c(%rsp),%eax
  pxor   %xmm0,%xmm0
  test   %eax,%eax
  jle    L26
  xor    %ebx,%ebx
  lea    D24(%rip),%r13
  lea    0x20(%rsp),%r12
  lea    D43(%rip),%rbp
  nopl   0x0(%rax)
L0:
  add    $0x1,%ebx
  mov    %r13,%rdi
  xor    %eax,%eax
  movsd  %xmm0,0x8(%rsp)
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  addsd  0x20(%rsp),%xmm0
  cmp    %ebx,0x1c(%rsp)
  jg     L0
L26:
  lea    D29(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D17
data_type: string
name: parse_data
arguments: tag: D50
data_type: string
name: parse_data
arguments: tag: D49
data_type: string
name: parse_data
arguments: tag: D24
data_type: string
name: parse_data
arguments: tag: D43
data_type: string
name: parse_data
arguments: tag: D29
data_type: string

tool: D17: while Loop Example 5


tool: D50: How many numbers would you like to enter? 

tool: D49: %d

tool: D24: Enter value %d: 

tool: D43: %lf

tool: D29: The sum of the values is %4.4lf.


assistant: ```c
int main() {
  int numValuesEnteredSoFar = 0;
  int numValuesToEnter;
  double numEntered, sum;
  printf("while Loop Example 5\n\n");
  printf("How many numbers would you like to enter? ");
  scanf("%d", &numValuesToEnter);
  sum = 0.0;
  while (numValuesEnteredSoFar < numValuesToEnter) {
    numValuesEnteredSoFar++;
    printf("Enter value %d: ", numValuesEnteredSoFar);
    scanf("%lf", &numEntered);
    sum = sum + numEntered;
  }
  printf("The sum of the values is %4.4lf.\n", sum);
  return 0;
}
```