ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<swap_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x24(%rbp)
  jmp    L23
L42:
  subl   $0x1,-0x24(%rbp)
L23:
  cmpl   $0x0,-0x24(%rbp)
  jle    L69
  mov    -0x24(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L42
L69:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%rax
  mov    $0x20,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L52
  mov    -0x18(%rbp),%rax
  sub    -0x38(%rbp),%rax
  mov    %eax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <alloca@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x38(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L22
L10:
  addq   $0x1,-0x18(%rbp)
L22:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L10
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  mov    -0x20(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x38(%rbp),%rax
  add    %rax,%rcx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  jmp    L71
L52:
  nop
L71:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void swap_words(unsigned char *buf) {
  int len = strlen(buf);
  while (len > 0 && isspace(buf[len - 1]))
    --len;
  buf[len] = 0;
  unsigned char *pos = strchr(buf, ' ');
  if (!pos)
    return;
  int len2 = (int)(pos - buf);
  unsigned char *buf1 = alloca(len + 1);
  memcpy(buf1, buf, len2);
  buf1[len2] = 0;
  while (isspace(*pos))
    ++pos;
  int len3 = strlen(pos);
  memmove(buf, pos, len3);
  buf[len3] = ' ';
  memcpy(buf + len3 + 1, buf1, len2 + 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L13
  cltq
  lea    -0x1(%r13,%rax,1),%r12
L48:
  movzbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L13
  sub    $0x1,%r12
  sub    $0x1,%ebx
  jne    L48
L13:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    $0x20,%esi
  mov    %r13,%rdi
  call   <strchr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L46
  mov    %rax,%r14
  sub    %r13,%r14
  lea    0x1(%rbx),%edi
  call   <alloca@plt>
  mov    %rax,%rbx
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %r14d,%rax
  movb   $0x0,(%rbx,%rax,1)
  jmp    L51
L34:
  add    $0x1,%r12
L51:
  movzbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L34
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %eax,%edx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <memmove@plt>
  movslq %r15d,%r15
  movb   $0x20,0x0(%r13,%r15,1)
  lea    0x1(%r14),%edx
  lea    0x1(%r13,%r15,1),%rdi
  mov    %rbx,%rsi
  call   <memcpy@plt>
L46:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void swap_words(unsigned char *buf) {
  int len = strlen(buf);
  while (len > 0 && isspace(buf[len - 1]))
    --len;
  buf[len] = 0;
  unsigned char *pos = strchr(buf, ' ');
  if (!pos)
    return;
  int len2 = (int)(pos - buf);
  unsigned char *buf1 = alloca(len + 1);
  memcpy(buf1, buf, len2);
  buf1[len2] = 0;
  while (isspace(*pos))
    ++pos;
  int len3 = strlen(pos);
  memmove(buf, pos, len3);
  buf[len3] = ' ';
  memcpy(buf + len3 + 1, buf1, len2 + 1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<swap_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  movslq %eax,%rbx
  mov    %rbx,%r12
  test   %ebx,%ebx
  jg     L26
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L88:
  lea    -0x1(%rbx),%r12d
  sub    $0x1,%rbx
  je     L59
L26:
  movzbl -0x1(%r13,%rbx,1),%edi
  mov    %ebx,%r12d
  call   <isspace@plt>
  test   %rax,%rax
  jne    L88
L52:
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    $0x20,%esi
  mov    %r13,%rdi
  call   <strchr@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L19
  lea    0x1(%r12),%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  sub    %r13,%rbx
  mov    %r13,%rsi
  mov    %ebx,%edx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  jmp    L72
  cs nopw 0x0(%rax,%rax,1)
L84:
  add    $0x1,%r14
L72:
  movzbl (%r14),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L84
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  movslq %eax,%r15
  mov    %r15d,%edx
  call   <memmove@plt>
  lea    0x1(%rbx),%edx
  lea    0x1(%r13,%r15,1),%rdi
  mov    %r12,%rsi
  movb   $0x20,0x0(%r13,%r15,1)
  call   <memcpy@plt>
L19:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L59:
  movslq %r12d,%rbx
  jmp    L52
L60:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void swap_words(unsigned char *buf) {
  int len = strlen(buf);
  while (len > 0 && isspace(buf[len - 1]))
    --len;
  buf[len] = 0;
  unsigned char *pos = strchr(buf, ' ');
  if (!pos)
    return;
  int len2 = (int)(pos - buf);
  unsigned char *buf1 = alloca(len + 1);
  memcpy(buf1, buf, len2);
  buf1[len2] = 0;
  while (isspace(*pos))
    ++pos;
  int len3 = strlen(pos);
  memmove(buf, pos, len3);
  buf[len3] = ' ';
  memcpy(buf + len3 + 1, buf1, len2 + 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<swap_words>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  movslq %eax,%rbx
  mov    %rbx,%r12
  test   %ebx,%ebx
  jg     L54
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L71:
  lea    -0x1(%rbx),%r12d
  sub    $0x1,%rbx
  je     L69
L54:
  movzbl -0x1(%r13,%rbx,1),%edi
  mov    %ebx,%r12d
  call   <isspace@plt>
  test   %rax,%rax
  jne    L71
L2:
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    $0x20,%esi
  mov    %r13,%rdi
  call   <strchr@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L11
  lea    0x1(%r12),%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  sub    %r13,%rbx
  mov    %r13,%rsi
  mov    %ebx,%edx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  movslq %ebx,%rax
  movb   $0x0,(%r12,%rax,1)
  jmp    L59
  cs nopw 0x0(%rax,%rax,1)
L34:
  add    $0x1,%r14
L59:
  movzbl (%r14),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L34
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  movslq %eax,%r15
  mov    %r15d,%edx
  call   <memmove@plt>
  lea    0x1(%rbx),%edx
  lea    0x1(%r13,%r15,1),%rdi
  mov    %r12,%rsi
  movb   $0x20,0x0(%r13,%r15,1)
  call   <memcpy@plt>
L11:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nop
L69:
  movslq %r12d,%rbx
  jmp    L2
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void swap_words(unsigned char *buf) {
  int len = strlen(buf);
  while (len > 0 && isspace(buf[len - 1]))
    --len;
  buf[len] = 0;
  unsigned char *pos = strchr(buf, ' ');
  if (!pos)
    return;
  int len2 = (int)(pos - buf);
  unsigned char *buf1 = alloca(len + 1);
  memcpy(buf1, buf, len2);
  buf1[len2] = 0;
  while (isspace(*pos))
    ++pos;
  int len3 = strlen(pos);
  memmove(buf, pos, len3);
  buf[len3] = ' ';
  memcpy(buf + len3 + 1, buf1, len2 + 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movsd  D45(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D7(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x64,-0x14(%rbp)
  movb   $0x61,-0x1a(%rbp)
  movb   $0x7a,-0x19(%rbp)
  movsd  -0x10(%rbp),%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L4
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  movsbl %al,%eax
  jmp    L14
L4:
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  movsbl %al,%eax
L14:
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%edx
  movsbl %dl,%edx
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%ecx
  movsbl %cl,%esi
  mov    %eax,%ecx
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  movsbl %al,%ecx
  mov    -0x14(%rbp),%eax
  movsbl %al,%edx
  mov    -0x18(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x19(%rbp),%edx
  movzbl -0x1a(%rbp),%eax
  cmp    %al,%dl
  cmovge %edx,%eax
  movsbl %al,%ecx
  movsbl -0x19(%rbp),%edx
  movsbl -0x1a(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D45
type_infered: f64
name: extract_symbolic_data
arguments: data_tag: D7
type_infered: f64
name: extract_symbolic_data
arguments: data_tag: D49
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D16
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string

tool: D45: 2

tool: D7: 100

tool: D49: max(%.2f,%.2f) = %.2f


tool: D16: max(%d,%d) = %d


tool: D1: max(%c,%c) = %c


assistant: ```c
int main(int argc, char *argv[]) {
  double dSmall = 2, dBig = 100;
  int iSmall = 2, iBig = 100;
  char cSmall = 'a', cBig = 'z';
  printf("max(%.2f,%.2f) = %.2f\n", dSmall, dBig, ((dSmall) > (dBig) ? (dSmall) : (dBig)));
  printf("max(%d,%d) = %d\n", iSmall, iBig, ((iSmall) > (iBig) ? (iSmall) : (iBig)));
  printf("max(%c,%c) = %c\n", cSmall, cBig, ((cSmall) > (cBig) ? (cSmall) : (cBig)));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D21(%rip),%rdi
  call   <printf@plt>
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x7a,%ecx
  mov    $0x7a,%edx
  mov    $0x61,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D21
data_type: string
name: unpack_data
arguments: data_label: D3
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string

tool: D21: max(%.2f,%.2f) = %.2f


tool: D3: max(%d,%d) = %d


tool: D13: max(%c,%c) = %c


assistant: ```c
int main(int argc, char *argv[]) {
  double dSmall = 2, dBig = 100;
  int iSmall = 2, iBig = 100;
  char cSmall = 'a', cBig = 'z';
  printf("max(%.2f,%.2f) = %.2f\n", dSmall, dBig, ((dSmall) > (dBig) ? (dSmall) : (dBig)));
  printf("max(%d,%d) = %d\n", iSmall, iBig, ((iSmall) > (iBig) ? (iSmall) : (iBig)));
  printf("max(%c,%c) = %c\n", cSmall, cBig, ((cSmall) > (cBig) ? (cSmall) : (cBig)));
  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
  sub    $0x8,%rsp
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    $0x7a,%ecx
  mov    $0x7a,%edx
  mov    $0x61,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D26
type: string
name: parse_data
arguments: label: D10
type: string
name: parse_data
arguments: label: D15
type: string

tool: D26: max(%.2f,%.2f) = %.2f


tool: D10: max(%d,%d) = %d


tool: D15: max(%c,%c) = %c


assistant: ```c
int main(int argc, char *argv[]) {
  double dSmall = 2, dBig = 100;
  int iSmall = 2, iBig = 100;
  char cSmall = 'a', cBig = 'z';
  printf("max(%.2f,%.2f) = %.2f\n", dSmall, dBig, ((dSmall) > (dBig) ? (dSmall) : (dBig)));
  printf("max(%d,%d) = %d\n", iSmall, iBig, ((iSmall) > (iBig) ? (iSmall) : (iBig)));
  printf("max(%c,%c) = %c\n", cSmall, cBig, ((cSmall) > (cBig) ? (cSmall) : (cBig)));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x64,%ecx
  mov    $0x64,%edx
  mov    $0x2,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x7a,%ecx
  mov    $0x7a,%edx
  mov    $0x61,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D18
type_infered: string
name: retrieve_data
arguments: label: D16
type_infered: string
name: retrieve_data
arguments: label: D1
type_infered: string

tool: D18: max(%.2f,%.2f) = %.2f


tool: D16: max(%d,%d) = %d


tool: D1: max(%c,%c) = %c


assistant: ```c
int main(int argc, char *argv[]) {
  double dSmall = 2, dBig = 100;
  int iSmall = 2, iBig = 100;
  char cSmall = 'a', cBig = 'z';
  printf("max(%.2f,%.2f) = %.2f\n", dSmall, dBig, ((dSmall) > (dBig) ? (dSmall) : (dBig)));
  printf("max(%d,%d) = %d\n", iSmall, iBig, ((iSmall) > (iBig) ? (iSmall) : (iBig)));
  printf("max(%c,%c) = %c\n", cSmall, cBig, ((cSmall) > (cBig) ? (cSmall) : (cBig)));
  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
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L15
L5:
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L15:
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L8
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  test   %eax,%eax
  jne    L5
L8:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
data_type: string
name: retrieve_data
arguments: data_tag: D15
data_type: string

tool: D0: %d


tool: D15: %d%d

assistant: ```c
int main(int argc, char const *argv[]) {
  int a, b;
  while (scanf("%d%d", &a, &b) && (a + b) != 0) {
    printf("%d\n", a + b);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  lea    D8(%rip),%rbx
  lea    D4(%rip),%r13
  jmp    L25
L32:
  mov    %r13,%rdi
  call   <printf@plt>
L25:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L21
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  jne    L32
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
type_guessed: string
name: retrieve_data
arguments: label: D4
type_guessed: string

tool: D8: %d%d

tool: D4: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int a, b;
  while (scanf("%d%d", &a, &b) && (a + b) != 0) {
    printf("%d\n", a + b);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L19
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  je     L21
  mov    %r13,%rdi
  call   <printf@plt>
L19:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L26
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D15
datatype: string
name: read_data
arguments: data_label: D1
datatype: string

tool: D15: %d


tool: D1: %d%d

assistant: ```c
int main(int argc, char const *argv[]) {
  int a, b;
  while (scanf("%d%d", &a, &b) && (a + b) != 0) {
    printf("%d\n", a + b);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D0(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L36
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  je     L8
  mov    %r13,%rdi
  call   <printf@plt>
L36:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L7
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type: string
name: disassemble_data
arguments: data_label: D0
type: string

tool: D10: %d


tool: D0: %d%d

assistant: ```c
int main(int argc, char const *argv[]) {
  int a, b;
  while (scanf("%d%d", &a, &b) && (a + b) != 0) {
    printf("%d\n", a + b);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoll@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpq   $0x5,-0x8(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  nop
  leave
  ret
```

assistant: ```c
void test1(char *array, char *endptr) {
  long long r = strtoll(array, &endptr, 3);
  assert(array + 2 == endptr);
  assert(r == 5);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  mov    $0x3,%edx
  call   <strtoll@plt>
  mov    %rax,%rbp
  add    $0x2,%rbx
  cmp    %rbx,0x8(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x5,%rbp
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void test1(char *array, char *endptr) {
  long long r = strtoll(array, &endptr, 3);
  assert(array + 2 == endptr);
  assert(r == 5);
}
```
user: What is the c source code of the assembly code below:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    $0x3,%edx
  push   %rbx
  mov    %rdi,%rbx
  add    $0x2,%rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  call   <strtoll@plt>
  xor    %edi,%edi
  cmp    %rbx,0x8(%rsp)
  mov    %rax,%rbp
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmp    $0x5,%rbp
  sete   %dil
  call   <assert@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void test1(char *array, char *endptr) {
  long long r = strtoll(array, &endptr, 3);
  assert(array + 2 == endptr);
  assert(r == 5);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    $0x3,%edx
  push   %rbx
  mov    %rdi,%rbx
  add    $0x2,%rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rsi
  call   <strtoll@plt>
  xor    %edi,%edi
  cmp    %rbx,0x8(%rsp)
  mov    %rax,%rbp
  sete   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmp    $0x5,%rbp
  sete   %dil
  call   <assert@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void test1(char *array, char *endptr) {
  long long r = strtoll(array, &endptr, 3);
  assert(array + 2 == endptr);
  assert(r == 5);
}
```
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
  mov    $0x0,%edi
  call   <test1@plt>
  test   %eax,%eax
  je     L109
  mov    $0x0,%eax
  call   <abort@plt>
L109:
  mov    $0x1,%edi
  call   <test1@plt>
  test   %eax,%eax
  je     L71
  mov    $0x0,%eax
  call   <abort@plt>
L71:
  mov    $0xffffffff,%edi
  call   <test1@plt>
  cmp    $0x1,%eax
  je     L30
  mov    $0x0,%eax
  call   <abort@plt>
L30:
  mov    $0x0,%edi
  call   <test2@plt>
  test   %eax,%eax
  je     L100
  mov    $0x0,%eax
  call   <abort@plt>
L100:
  mov    $0x1,%edi
  call   <test2@plt>
  test   %eax,%eax
  je     L10
  mov    $0x0,%eax
  call   <abort@plt>
L10:
  mov    $0xffffffff,%edi
  call   <test2@plt>
  cmp    $0xffffffff,%eax
  je     L99
  mov    $0x0,%eax
  call   <abort@plt>
L99:
  mov    $0x0,%edi
  call   <test3@plt>
  test   %eax,%eax
  je     L1
  mov    $0x0,%eax
  call   <abort@plt>
L1:
  mov    $0x1,%edi
  call   <test3@plt>
  test   %eax,%eax
  je     L95
  mov    $0x0,%eax
  call   <abort@plt>
L95:
  mov    $0xffffffff,%edi
  call   <test3@plt>
  cmp    $0x1,%eax
  je     L67
  mov    $0x0,%eax
  call   <abort@plt>
L67:
  mov    $0x0,%edi
  call   <test4@plt>
  test   %eax,%eax
  je     L63
  mov    $0x0,%eax
  call   <abort@plt>
L63:
  mov    $0x1,%edi
  call   <test4@plt>
  test   %eax,%eax
  je     L57
  mov    $0x0,%eax
  call   <abort@plt>
L57:
  mov    $0xffffffff,%edi
  call   <test4@plt>
  cmp    $0xffffffff,%eax
  je     L58
  mov    $0x0,%eax
  call   <abort@plt>
L58:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (test1(0) != 0)
    abort();
  if (test1(1) != 0)
    abort();
  if (test1(-1) != 1)
    abort();
  if (test2(0) != 0)
    abort();
  if (test2(1) != 0)
    abort();
  if (test2((unsigned int)-1) != -1)
    abort();
  if (test3(0) != 0)
    abort();
  if (test3(1) != 0)
    abort();
  if (test3(-1) != 1)
    abort();
  if (test4(0) != 0)
    abort();
  if (test4(1) != 0)
    abort();
  if (test4((unsigned int)-1) != -1)
    abort();
  return 0;
}
```