ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<sort_desc>:
  endbr64
  mov    $0x0,%r9d
L28:
  test   %esi,%esi
  jle    L11
  lea    0x1(%rsi),%r8d
  mov    $0x1,%eax
  mov    %r9d,%esi
  jmp    L8
L46:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L27
L8:
  mov    -0x4(%rdi,%rax,4),%ecx
  mov    (%rdi,%rax,4),%edx
  cmp    %edx,%ecx
  jge    L46
  mov    %ecx,(%rdi,%rax,4)
  mov    %edx,-0x4(%rdi,%rax,4)
  mov    %eax,%esi
  jmp    L46
L27:
  test   %esi,%esi
  jg     L28
L11:
  ret
```

assistant: ```c
void sort_desc(int vec[], int n) {
  int k;
  do {
    k = 0;
    for (int j = 1; j <= n; j++) {
      if (vec[j - 1] < vec[j]) {
        int temp = vec[j];
        vec[j] = vec[j - 1];
        vec[j - 1] = temp;
        k = j;
      }
    }
    n = k;
  } while (n > 0);
}
```
user: What is the c source code of the assembly code below?

```asm
<sort_desc>:
  endbr64
  nopl   0x0(%rax)
L12:
  test   %esi,%esi
  jle    L7
  lea    0x1(%rsi),%r8d
  mov    $0x1,%eax
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L9:
  mov    -0x4(%rdi,%rax,4),%ecx
  mov    (%rdi,%rax,4),%edx
  cmp    %edx,%ecx
  jge    L30
  mov    %ecx,(%rdi,%rax,4)
  mov    %eax,%esi
  mov    %edx,-0x4(%rdi,%rax,4)
L30:
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L9
  test   %esi,%esi
  jne    L12
L7:
  ret
```

assistant: ```c
void sort_desc(int vec[], int n) {
  int k;
  do {
    k = 0;
    for (int j = 1; j <= n; j++) {
      if (vec[j - 1] < vec[j]) {
        int temp = vec[j];
        vec[j] = vec[j - 1];
        vec[j - 1] = temp;
        k = j;
      }
    }
    n = k;
  } while (n > 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sort_desc>:
  endbr64
  nopl   0x0(%rax)
L7:
  test   %esi,%esi
  jle    L25
  lea    0x1(%rsi),%r8d
  mov    $0x1,%edx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L6:
  movq   -0x4(%rdi,%rdx,4),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%eax
  movd   %xmm2,%ecx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L36
  movq   %xmm1,-0x4(%rdi,%rdx,4)
  mov    %edx,%esi
L36:
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L6
  test   %esi,%esi
  jne    L7
L25:
  ret
```

assistant: ```c
void sort_desc(int vec[], int n) {
  int k;
  do {
    k = 0;
    for (int j = 1; j <= n; j++) {
      if (vec[j - 1] < vec[j]) {
        int temp = vec[j];
        vec[j] = vec[j - 1];
        vec[j - 1] = temp;
        k = j;
      }
    }
    n = k;
  } while (n > 0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cprop_result2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x9,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cprop_result2(int *data) {
  int j;
  j = 9;
  data[0] = j;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cprop_result2>:
  endbr64
  movl   $0x9,(%rdi)
  ret
```

assistant: ```c
void cprop_result2(int *data) {
  int j;
  j = 9;
  data[0] = j;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cprop_result2>:
  endbr64
  movl   $0x9,(%rdi)
  ret
```

assistant: ```c
void cprop_result2(int *data) {
  int j;
  j = 9;
  data[0] = j;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cprop_result2>:
  endbr64
  movl   $0x9,(%rdi)
  ret
```

assistant: ```c
void cprop_result2(int *data) {
  int j;
  j = 9;
  data[0] = j;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x240,%rsp
  mov    %edi,-0x234(%rbp)
  mov    %rsi,-0x240(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,-0x220(%rbp)
  mov    -0x220(%rbp),%rcx
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,-0x22c(%rbp)
  mov    -0x220(%rbp),%rax
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mul    %rdx
  mov    %rdx,%rcx
  shr    $0x3,%rcx
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,-0x228(%rbp)
  mov    -0x220(%rbp),%rax
  shr    $0x2,%rax
  movabs $0x183c977ab2bedd29,%rdx
  mul    %rdx
  mov    %rdx,%rcx
  shr    $0x4,%rcx
  movabs $0x4ec4ec4ec4ec4ec5,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,-0x224(%rbp)
  mov    -0x22c(%rbp),%eax
  add    $0x40,%eax
  mov    %al,-0x218(%rbp)
  mov    -0x228(%rbp),%eax
  add    $0x40,%eax
  mov    %al,-0x217(%rbp)
  mov    -0x224(%rbp),%eax
  add    $0x40,%eax
  mov    %al,-0x216(%rbp)
  movb   $0x0,-0x215(%rbp)
  cmpl   $0x2,-0x234(%rbp)
  je     L20
  mov    $0x1,%eax
  jmp    L17
L20:
  mov    -0x240(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    -0x210(%rbp),%rdx
  lea    -0x218(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <reg_convert@plt>
  mov    -0x240(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    -0x210(%rbp),%rdx
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L17:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_guessed: string

tool: D5: name=%s
code=%s


assistant: ```c
int main(int argc, char *argv[]) {
  char buffer[512], code[8];
  unsigned long t = time(NULL);
  int b1 = t % 26, b2 = (t / 26) % 26, b3 = (t / 26 / 26) % 26;
  code[0] = '@' + b1;
  code[1] = '@' + b2;
  code[2] = '@' + b3;
  code[3] = '\0';
  if (argc != 2)
    return 1;
  reg_convert(argv[1], code, buffer);
  printf("name=%s\ncode=%s\n", argv[1], buffer);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x228,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edi
  call   <time@plt>
  mov    %rax,%rcx
  movabs $0x4ec4ec4ec4ec4ec5,%rdi
  mul    %rdi
  mov    %rdx,%rsi
  shr    $0x3,%rsi
  lea    (%rsi,%rsi,2),%rax
  lea    (%rsi,%rax,4),%rdx
  add    %rdx,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  add    $0x40,%eax
  mov    %al,0x8(%rsp)
  mov    %rsi,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    %rax,%rax
  sub    %rax,%rsi
  add    $0x40,%esi
  mov    %sil,0x9(%rsp)
  shr    $0x2,%rcx
  movabs $0x183c977ab2bedd29,%rdx
  mov    %rcx,%rax
  mul    %rdx
  mov    %rdx,%rcx
  shr    $0x4,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  add    $0x40,%ecx
  mov    %cl,0xa(%rsp)
  movb   $0x0,0xb(%rsp)
  mov    $0x1,%eax
  cmp    $0x2,%ebx
  je     L11
L5:
  mov    0x218(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x228,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  lea    0x10(%rsp),%rbx
  lea    0x8(%rsp),%rsi
  mov    0x8(%rbp),%rdi
  mov    %rbx,%rdx
  call   <reg_convert@plt>
  mov    0x8(%rbp),%rsi
  mov    %rbx,%rdx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L5
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
type_guessed: string

tool: D0: name=%s
code=%s


assistant: ```c
int main(int argc, char *argv[]) {
  char buffer[512], code[8];
  unsigned long t = time(NULL);
  int b1 = t % 26, b2 = (t / 26) % 26, b3 = (t / 26 / 26) % 26;
  code[0] = '@' + b1;
  code[1] = '@' + b2;
  code[2] = '@' + b3;
  code[3] = '\0';
  if (argc != 2)
    return 1;
  reg_convert(argv[1], code, buffer);
  printf("name=%s\ncode=%s\n", argv[1], buffer);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x220,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  movb   $0x0,0xb(%rsp)
  movabs $0x4ec4ec4ec4ec4ec5,%rdi
  mov    %rax,%rcx
  mul    %rdi
  mov    %rdx,%rsi
  shr    $0x3,%rsi
  lea    (%rsi,%rsi,2),%rax
  lea    (%rsi,%rax,4),%rdx
  mov    %rcx,%rax
  shr    $0x2,%rcx
  add    %rdx,%rdx
  sub    %rdx,%rax
  add    $0x40,%eax
  mov    %al,0x8(%rsp)
  mov    %rsi,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  movabs $0x183c977ab2bedd29,%rdx
  add    %rax,%rax
  sub    %rax,%rsi
  mov    %rcx,%rax
  mul    %rdx
  add    $0x40,%esi
  mov    %sil,0x9(%rsp)
  mov    %rdx,%rcx
  shr    $0x4,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    $0x1,%eax
  add    $0x40,%ecx
  mov    %cl,0xa(%rsp)
  cmp    $0x2,%ebx
  je     L1
L2:
  mov    0x218(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x220,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  mov    0x8(%rbp),%rdi
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdx
  call   <reg_convert@plt>
  mov    0x8(%rbp),%rsi
  mov    %r12,%rdx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L2
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D0
type: string

tool: D0: name=%s
code=%s


assistant: ```c
int main(int argc, char *argv[]) {
  char buffer[512], code[8];
  unsigned long t = time(NULL);
  int b1 = t % 26, b2 = (t / 26) % 26, b3 = (t / 26 / 26) % 26;
  code[0] = '@' + b1;
  code[1] = '@' + b2;
  code[2] = '@' + b3;
  code[3] = '\0';
  if (argc != 2)
    return 1;
  reg_convert(argv[1], code, buffer);
  printf("name=%s\ncode=%s\n", argv[1], buffer);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x220,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  call   <time@plt>
  xor    %r8d,%r8d
  movb   $0x0,0xb(%rsp)
  movabs $0x4ec4ec4ec4ec4ec5,%rdi
  mov    %rax,%rcx
  mul    %rdi
  mov    %rdx,%rsi
  shr    $0x3,%rsi
  lea    (%rsi,%rsi,2),%rax
  lea    (%rsi,%rax,4),%rdx
  mov    %rcx,%rax
  shr    $0x2,%rcx
  add    %rdx,%rdx
  sub    %rdx,%rax
  add    $0x40,%eax
  mov    %al,%r8b
  mov    %rsi,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  mov    %r8d,%edx
  add    %rax,%rax
  sub    %rax,%rsi
  lea    0x40(%rsi),%eax
  mov    %al,%dh
  mov    %rcx,%rax
  mov    %dx,0x8(%rsp)
  movabs $0x183c977ab2bedd29,%rdx
  mul    %rdx
  mov    %rdx,%rcx
  shr    $0x4,%rcx
  mov    %rcx,%rax
  mul    %rdi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,2),%rax
  lea    (%rdx,%rax,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    $0x1,%eax
  add    $0x40,%ecx
  mov    %cl,0xa(%rsp)
  cmp    $0x2,%ebx
  je     L11
L17:
  mov    0x218(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x220,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  mov    0x8(%rbp),%rdi
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdx
  call   <reg_convert@plt>
  mov    0x8(%rbp),%rsi
  mov    %r12,%rdx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L17
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
data_type: string

tool: D6: name=%s
code=%s


assistant: ```c
int main(int argc, char *argv[]) {
  char buffer[512], code[8];
  unsigned long t = time(NULL);
  int b1 = t % 26, b2 = (t / 26) % 26, b3 = (t / 26 / 26) % 26;
  code[0] = '@' + b1;
  code[1] = '@' + b2;
  code[2] = '@' + b3;
  code[3] = '\0';
  if (argc != 2)
    return 1;
  reg_convert(argv[1], code, buffer);
  printf("name=%s\ncode=%s\n", argv[1], buffer);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Util_sscanhex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L98
  mov    $0xffffffff,%eax
  jmp    L149
L98:
  movl   $0x0,-0x4(%rbp)
L124:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L28
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L28
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L138
L28:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L15
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x46,%al
  jg     L15
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  sub    $0x37,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L138
L15:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L72
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x66,%al
  jg     L72
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  sub    $0x57,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L138
L72:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L143
  mov    -0x4(%rbp),%eax
  jmp    L149
L143:
  mov    $0xffffffff,%eax
  jmp    L149
L138:
  addq   $0x1,-0x18(%rbp)
  jmp    L124
L149:
  pop    %rbp
  ret
```

assistant: ```c
int Util_sscanhex(const char *s) {
  int result;
  if (*s == '\0')
    return -1;
  result = 0;
  for (;;) {
    if (*s >= '0' && *s <= '9')
      result = 16 * result + *s - '0';
    else if (*s >= 'A' && *s <= 'F')
      result = 16 * result + *s - 'A' + 10;
    else if (*s >= 'a' && *s <= 'f')
      result = 16 * result + *s - 'a' + 10;
    else if (*s == '\0')
      return result;
    else
      return -1;
    s++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Util_sscanhex>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L4
  mov    $0x0,%eax
  jmp    L17
L55:
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x30(%rax,%rdx,1),%eax
L30:
  add    $0x1,%rdi
L17:
  movzbl (%rdi),%edx
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L55
  lea    -0x41(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L35
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x37(%rax,%rdx,1),%eax
  jmp    L30
L35:
  lea    -0x61(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L11
  shl    $0x4,%eax
  movsbl %dl,%edx
  lea    -0x57(%rax,%rdx,1),%eax
  jmp    L30
L11:
  test   %dl,%dl
  mov    $0xffffffff,%edx
  cmovne %edx,%eax
  ret
L4:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int Util_sscanhex(const char *s) {
  int result;
  if (*s == '\0')
    return -1;
  result = 0;
  for (;;) {
    if (*s >= '0' && *s <= '9')
      result = 16 * result + *s - '0';
    else if (*s >= 'A' && *s <= 'F')
      result = 16 * result + *s - 'A' + 10;
    else if (*s >= 'a' && *s <= 'f')
      result = 16 * result + *s - 'a' + 10;
    else if (*s == '\0')
      return result;
    else
      return -1;
    s++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Util_sscanhex>:
  endbr64
  movsbl (%rdi),%edx
  test   %dl,%dl
  je     L2
  xor    %eax,%eax
  jmp    L69
  nop
L25:
  shl    $0x4,%eax
  lea    -0x30(%rax,%rdx,1),%eax
L20:
  movsbl 0x1(%rdi),%edx
  add    $0x1,%rdi
L69:
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L25
  lea    -0x41(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L52
  shl    $0x4,%eax
  lea    -0x37(%rax,%rdx,1),%eax
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L52:
  lea    -0x61(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L11
  shl    $0x4,%eax
  lea    -0x57(%rax,%rdx,1),%eax
  jmp    L20
  nopl   0x0(%rax)
L11:
  test   %dl,%dl
  mov    $0xffffffff,%edx
  cmovne %edx,%eax
  ret
L2:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int Util_sscanhex(const char *s) {
  int result;
  if (*s == '\0')
    return -1;
  result = 0;
  for (;;) {
    if (*s >= '0' && *s <= '9')
      result = 16 * result + *s - '0';
    else if (*s >= 'A' && *s <= 'F')
      result = 16 * result + *s - 'A' + 10;
    else if (*s >= 'a' && *s <= 'f')
      result = 16 * result + *s - 'a' + 10;
    else if (*s == '\0')
      return result;
    else
      return -1;
    s++;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Util_sscanhex>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L5
  xor    %eax,%eax
  jmp    L48
  nopl   (%rax)
L10:
  shl    $0x4,%eax
  lea    -0x30(%rax,%rdx,1),%eax
L11:
  add    $0x1,%rdi
L48:
  movsbl (%rdi),%edx
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L10
  lea    -0x41(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L47
  shl    $0x4,%eax
  lea    -0x37(%rax,%rdx,1),%eax
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L47:
  lea    -0x61(%rdx),%ecx
  cmp    $0x5,%cl
  ja     L63
  shl    $0x4,%eax
  lea    -0x57(%rax,%rdx,1),%eax
  jmp    L11
  nopl   0x0(%rax)
L63:
  test   %dl,%dl
  mov    $0xffffffff,%edx
  cmovne %edx,%eax
  ret
L5:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int Util_sscanhex(const char *s) {
  int result;
  if (*s == '\0')
    return -1;
  result = 0;
  for (;;) {
    if (*s >= '0' && *s <= '9')
      result = 16 * result + *s - '0';
    else if (*s >= 'A' && *s <= 'F')
      result = 16 * result + *s - 'A' + 10;
    else if (*s >= 'a' && *s <= 'f')
      result = 16 * result + *s - 'a' + 10;
    else if (*s == '\0')
      return result;
    else
      return -1;
    s++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string

tool: D1: ls -l

assistant: ```c
int main(void) {
  system("ls -l");
  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
  lea    D5(%rip),%rdi
  call   <system@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D5
type_infered: string

tool: D5: ls -l

assistant: ```c
int main(void) {
  system("ls -l");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <system@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
data_type: string

tool: D5: ls -l

assistant: ```c
int main(void) {
  system("ls -l");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <system@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_infered: string

tool: D2: ls -l

assistant: ```c
int main(void) {
  system("ls -l");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L30
L1:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L14
L35:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D28(%rip),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L14:
  cmpl   $0x2,-0x14(%rbp)
  jle    L35
  addl   $0x1,-0x18(%rbp)
L30:
  cmpl   $0x6,-0x18(%rbp)
  jle    L1
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D28
data_type: string
name: disassemble_data
arguments: label: D23
data_type: string
name: disassemble_data
arguments: label: D13
data_type: string

tool: D28: dapxsumkbn

tool: D23: # c::%s:len=%d


tool: D13: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dapxsumkbn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D2(%rip),%r15
  jmp    L25
L24:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L0
L25:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L27:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L27
  jmp    L24
L0:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_infered: string
name: retrieve_data
arguments: data_tag: D16
type_infered: string
name: retrieve_data
arguments: data_tag: D27
type_infered: string

tool: D2: dapxsumkbn

tool: D16: # c::%s:len=%d


tool: D27: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dapxsumkbn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```