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

```asm
<HardwareInit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <SetupFPU@plt>
  mov    $0x1,%edi
  call   <SetupClock@plt>
  mov    $0x0,%eax
  call   <SetupStdio@plt>
  mov    $0x0,%eax
  call   <main@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void HardwareInit() {
  SetupFPU();
  SetupClock(1);
  SetupStdio();
  main();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<HardwareInit>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <SetupFPU@plt>
  mov    $0x1,%edi
  call   <SetupClock@plt>
  xor    %eax,%eax
  call   <SetupStdio@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10a0 <main@plt>
```

assistant: ```c
void HardwareInit() {
  SetupFPU();
  SetupClock(1);
  SetupStdio();
  main();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HardwareInit>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <SetupFPU@plt>
  mov    $0x1,%edi
  call   <SetupClock@plt>
  xor    %eax,%eax
  call   <SetupStdio@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10a0 <main@plt>
```

assistant: ```c
void HardwareInit() {
  SetupFPU();
  SetupClock(1);
  SetupStdio();
  main();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ckstrpbrk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L103
  cmpq   $0x0,-0x20(%rbp)
  jne    L28
L103:
  mov    $0x0,%eax
  jmp    L41
L28:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L13
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
L13:
  mov    $0x0,%eax
  jmp    L41
L48:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L71:
  movzbl -0x9(%rbp),%eax
  cmp    -0xa(%rbp),%al
  jne    L6
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  jmp    L41
L6:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x0,-0x9(%rbp)
  jne    L71
L17:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0xa(%rbp)
  cmpb   $0x0,-0xa(%rbp)
  jne    L48
  mov    $0x0,%eax
L41:
  pop    %rbp
  ret
```

assistant: ```c
char *ckstrpbrk(s1, s2)
char *s1, *s2;
{
  char c1, c2, *s3;
  if (!s1 || !s2)
    return (0L);
  if (!*s1 || !*s2)
    return (0L);
  while ((c1 = *s1++)) {
    s3 = s2;
    while ((c2 = *s3++)) {
      if (c2 == c1)
        return (s1 - 1);
    }
  }
  return (0L);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ckstrpbrk>:
  endbr64
  test   %rdi,%rdi
  je     L48
  test   %rsi,%rsi
  je     L48
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L58
  movzbl (%rsi),%r9d
  test   %r9b,%r9b
  je     L33
L93:
  mov    %rdi,%r8
  lea    0x1(%rsi),%rdx
  mov    %r9d,%eax
L27:
  cmp    %al,%cl
  je     L8
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%eax
  test   %al,%al
  jne    L27
  add    $0x1,%rdi
  movzbl (%rdi),%ecx
  test   %cl,%cl
  jne    L93
  mov    $0x0,%r8d
  jmp    L8
L48:
  mov    $0x0,%r8d
L8:
  mov    %r8,%rax
  ret
L58:
  mov    $0x0,%r8d
  jmp    L8
L33:
  mov    $0x0,%r8d
  jmp    L8
```

assistant: ```c
char *ckstrpbrk(s1, s2)
char *s1, *s2;
{
  char c1, c2, *s3;
  if (!s1 || !s2)
    return (0L);
  if (!*s1 || !*s2)
    return (0L);
  while ((c1 = *s1++)) {
    s3 = s2;
    while ((c2 = *s3++)) {
      if (c2 == c1)
        return (s1 - 1);
    }
  }
  return (0L);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ckstrpbrk>:
  endbr64
  test   %rdi,%rdi
  je     L58
  test   %rsi,%rsi
  je     L58
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L58
  movzbl (%rsi),%r9d
  test   %r9b,%r9b
  je     L58
  add    $0x1,%rsi
L45:
  mov    %rdi,%r8
  mov    %r9d,%eax
  mov    %rsi,%rdx
  nopl   0x0(%rax,%rax,1)
L38:
  cmp    %al,%cl
  je     L44
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  jne    L38
  movzbl 0x1(%rdi),%ecx
  add    $0x1,%rdi
  test   %cl,%cl
  jne    L45
  nopl   0x0(%rax,%rax,1)
L58:
  xor    %r8d,%r8d
L44:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *ckstrpbrk(s1, s2)
char *s1, *s2;
{
  char c1, c2, *s3;
  if (!s1 || !s2)
    return (0L);
  if (!*s1 || !*s2)
    return (0L);
  while ((c1 = *s1++)) {
    s3 = s2;
    while ((c2 = *s3++)) {
      if (c2 == c1)
        return (s1 - 1);
    }
  }
  return (0L);
}
```
user: What is the c source code of the assembly code below?

```asm
<ckstrpbrk>:
  endbr64
  test   %rdi,%rdi
  je     L9
  test   %rsi,%rsi
  je     L9
  movzbl (%rdi),%ecx
  test   %cl,%cl
  je     L9
  movzbl (%rsi),%r9d
  test   %r9b,%r9b
  je     L9
  add    $0x1,%rsi
L51:
  mov    %rdi,%r8
  mov    %r9d,%eax
  mov    %rsi,%rdx
  nopl   0x0(%rax,%rax,1)
L64:
  cmp    %al,%cl
  je     L55
  movzbl (%rdx),%eax
  add    $0x1,%rdx
  test   %al,%al
  jne    L64
  movzbl 0x1(%rdi),%ecx
  add    $0x1,%rdi
  test   %cl,%cl
  jne    L51
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %r8d,%r8d
L55:
  mov    %r8,%rax
  ret
```

assistant: ```c
char *ckstrpbrk(s1, s2)
char *s1, *s2;
{
  char c1, c2, *s3;
  if (!s1 || !s2)
    return (0L);
  if (!*s1 || !*s2)
    return (0L);
  while ((c1 = *s1++)) {
    s3 = s2;
    while ((c2 = *s3++)) {
      if (c2 == c1)
        return (s1 - 1);
    }
  }
  return (0L);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  jg     L5
  mov    $0x1,%eax
  jmp    L1
L5:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <fib>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <fib>
  add    %ebx,%eax
L1:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fib(n) {
  if (n <= 2)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: What is the c source code of the assembly code below:

```asm
<fib>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x2,%edi
  jg     L5
  ret
L5:
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  mov    $0x0,%eax
  call   <fib>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  mov    $0x0,%eax
  call   <fib>
  add    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fib(n) {
  if (n <= 2)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fib>:
  endbr64
  push   %r15
  mov    $0x1,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,0x4(%rsp)
  movl   $0x0,0x8(%rsp)
  cmp    $0x2,%edi
  jle    L37
L120:
  mov    0x4(%rsp),%eax
  movl   $0x0,0xc(%rsp)
  sub    $0x1,%eax
  mov    %eax,(%rsp)
  cmp    $0x2,%eax
  je     L207
L13:
  mov    (%rsp),%eax
  movl   $0x0,0x10(%rsp)
  lea    -0x1(%rax),%r9d
  cmp    $0x2,%r9d
  je     L31
L252:
  lea    -0x1(%r9),%esi
  xor    %r8d,%r8d
  cmp    $0x2,%esi
  je     L205
L253:
  lea    -0x1(%rsi),%ebp
  xor    %ecx,%ecx
  cmp    $0x2,%ebp
  je     L215
L149:
  lea    -0x1(%rbp),%ebx
  xor    %r14d,%r14d
  mov    %ebx,%r10d
  cmp    $0x2,%ebx
  je     L180
L9:
  lea    -0x1(%r10),%edx
  xor    %ebx,%ebx
  cmp    $0x2,%edx
  je     L1
L248:
  lea    -0x1(%rdx),%r12d
  xor    %r15d,%r15d
  mov    %ebx,%r11d
  cmp    $0x2,%r12d
  je     L40
L112:
  lea    -0x1(%r12),%r13d
  xor    %ebx,%ebx
  cmp    $0x2,%r13d
  je     L62
L93:
  lea    -0x1(%r13),%edi
  xor    %eax,%eax
  mov    %edx,0x2c(%rsp)
  sub    $0x2,%r13d
  mov    %r10d,0x28(%rsp)
  mov    %r11d,0x24(%rsp)
  mov    %ecx,0x20(%rsp)
  mov    %r8d,0x1c(%rsp)
  mov    %esi,0x18(%rsp)
  mov    %r9d,0x14(%rsp)
  call   <fib>
  mov    0x14(%rsp),%r9d
  mov    0x18(%rsp),%esi
  add    %eax,%ebx
  cmp    $0x2,%r13d
  mov    0x1c(%rsp),%r8d
  mov    0x20(%rsp),%ecx
  mov    0x24(%rsp),%r11d
  mov    0x28(%rsp),%r10d
  mov    0x2c(%rsp),%edx
  jg     L93
  add    $0x1,%ebx
L106:
  sub    $0x2,%r12d
  add    %ebx,%r15d
  cmp    $0x2,%r12d
  jg     L112
  mov    %r11d,%ebx
  add    $0x1,%r15d
L174:
  sub    $0x2,%edx
  add    %r15d,%ebx
  cmp    $0x2,%edx
  jg     L248
  add    $0x1,%ebx
L97:
  sub    $0x2,%r10d
  add    %ebx,%r14d
  cmp    $0x2,%r10d
  jg     L9
  add    $0x1,%r14d
L246:
  sub    $0x2,%ebp
  add    %r14d,%ecx
  cmp    $0x2,%ebp
  jg     L149
  add    $0x1,%ecx
L10:
  sub    $0x2,%esi
  add    %ecx,%r8d
  cmp    $0x2,%esi
  jg     L253
  add    $0x1,%r8d
L30:
  sub    $0x2,%r9d
  add    %r8d,0x10(%rsp)
  cmp    $0x2,%r9d
  jg     L252
  mov    0x10(%rsp),%eax
  add    $0x1,%eax
L182:
  subl   $0x2,(%rsp)
  mov    (%rsp),%edi
  add    %eax,0xc(%rsp)
  cmp    $0x2,%edi
  jg     L13
  mov    0xc(%rsp),%eax
  add    $0x1,%eax
L240:
  subl   $0x2,0x4(%rsp)
  mov    0x4(%rsp),%edi
  add    %eax,0x8(%rsp)
  cmp    $0x2,%edi
  jg     L120
  mov    0x8(%rsp),%eax
  add    $0x1,%eax
L37:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  mov    $0x1,%ebx
  jmp    L97
  nopl   0x0(%rax)
L40:
  mov    $0x1,%r15d
  jmp    L174
  nopl   0x0(%rax,%rax,1)
L62:
  mov    $0x1,%ebx
  jmp    L106
L180:
  mov    $0x1,%r14d
  jmp    L246
L215:
  mov    $0x1,%ecx
  jmp    L10
L205:
  mov    $0x1,%r8d
  jmp    L30
L31:
  mov    $0x1,%eax
  jmp    L182
L207:
  mov    $0x1,%eax
  jmp    L240
```

assistant: ```c
int fib(n) {
  if (n <= 2)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fib>:
  endbr64
  push   %r15
  mov    $0x1,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,0x4(%rsp)
  movl   $0x0,0x8(%rsp)
  cmp    $0x2,%edi
  jle    L203
L26:
  mov    0x4(%rsp),%eax
  movl   $0x0,0xc(%rsp)
  sub    $0x1,%eax
  mov    %eax,(%rsp)
  cmp    $0x2,%eax
  je     L176
L172:
  mov    (%rsp),%eax
  movl   $0x0,0x10(%rsp)
  lea    -0x1(%rax),%r9d
  cmp    $0x2,%r9d
  je     L72
L182:
  lea    -0x1(%r9),%esi
  xor    %r8d,%r8d
  cmp    $0x2,%esi
  je     L197
L238:
  lea    -0x1(%rsi),%ebp
  xor    %ecx,%ecx
  cmp    $0x2,%ebp
  je     L51
L132:
  lea    -0x1(%rbp),%ebx
  xor    %r14d,%r14d
  mov    %ebx,%r10d
  cmp    $0x2,%ebx
  je     L5
L52:
  lea    -0x1(%r10),%edx
  xor    %ebx,%ebx
  cmp    $0x2,%edx
  je     L42
L39:
  lea    -0x1(%rdx),%r12d
  xor    %r15d,%r15d
  mov    %ebx,%r11d
  cmp    $0x2,%r12d
  je     L21
L34:
  lea    -0x1(%r12),%r13d
  xor    %ebx,%ebx
  cmp    $0x2,%r13d
  je     L192
L131:
  lea    -0x1(%r13),%edi
  xor    %eax,%eax
  mov    %edx,0x2c(%rsp)
  sub    $0x2,%r13d
  mov    %r10d,0x28(%rsp)
  mov    %r11d,0x24(%rsp)
  mov    %ecx,0x20(%rsp)
  mov    %r8d,0x1c(%rsp)
  mov    %esi,0x18(%rsp)
  mov    %r9d,0x14(%rsp)
  call   <fib>
  mov    0x14(%rsp),%r9d
  mov    0x18(%rsp),%esi
  add    %eax,%ebx
  cmp    $0x2,%r13d
  mov    0x1c(%rsp),%r8d
  mov    0x20(%rsp),%ecx
  mov    0x24(%rsp),%r11d
  mov    0x28(%rsp),%r10d
  mov    0x2c(%rsp),%edx
  jg     L131
  add    $0x1,%ebx
L231:
  sub    $0x2,%r12d
  add    %ebx,%r15d
  cmp    $0x2,%r12d
  jg     L34
  mov    %r11d,%ebx
  add    $0x1,%r15d
L85:
  sub    $0x2,%edx
  add    %r15d,%ebx
  cmp    $0x2,%edx
  jg     L39
  add    $0x1,%ebx
L234:
  sub    $0x2,%r10d
  add    %ebx,%r14d
  cmp    $0x2,%r10d
  jg     L52
  add    $0x1,%r14d
L193:
  sub    $0x2,%ebp
  add    %r14d,%ecx
  cmp    $0x2,%ebp
  jg     L132
  add    $0x1,%ecx
L95:
  sub    $0x2,%esi
  add    %ecx,%r8d
  cmp    $0x2,%esi
  jg     L238
  add    $0x1,%r8d
L117:
  sub    $0x2,%r9d
  add    %r8d,0x10(%rsp)
  cmp    $0x2,%r9d
  jg     L182
  mov    0x10(%rsp),%eax
  add    $0x1,%eax
L154:
  subl   $0x2,(%rsp)
  mov    (%rsp),%edi
  add    %eax,0xc(%rsp)
  cmp    $0x2,%edi
  jg     L172
  mov    0xc(%rsp),%eax
  add    $0x1,%eax
L138:
  subl   $0x2,0x4(%rsp)
  mov    0x4(%rsp),%edi
  add    %eax,0x8(%rsp)
  cmp    $0x2,%edi
  jg     L26
  mov    0x8(%rsp),%eax
  add    $0x1,%eax
L203:
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  mov    $0x1,%ebx
  jmp    L234
  nopl   0x0(%rax)
L21:
  mov    $0x1,%r15d
  jmp    L85
  nopl   0x0(%rax,%rax,1)
L192:
  mov    $0x1,%ebx
  jmp    L231
L5:
  mov    $0x1,%r14d
  jmp    L193
L51:
  mov    $0x1,%ecx
  jmp    L95
L197:
  mov    $0x1,%r8d
  jmp    L117
L72:
  mov    $0x1,%eax
  jmp    L154
L176:
  mov    $0x1,%eax
  jmp    L138
```

assistant: ```c
int fib(n) {
  if (n <= 2)
    return 1;
  else
    return fib(n - 1) + fib(n - 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

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

tool: D0: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D6: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D1
datatype: string

tool: D1: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

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

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_guessed: string

tool: D6: skipped


assistant: ```c
int main() {
  printf("skipped\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<base_convert>:
  endbr64
  mov    %esi,%r8d
  lea    -0x2(%rsi),%eax
  mov    $0x0,%r10d
  cmp    $0xe,%eax
  ja     L85
  mov    %rcx,%rsi
  mov    %r8d,%r11d
  sub    $0x1,%rdx
  mov    %rdx,%r9
  je     L34
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %r8d
  lea    D10(%rip),%rax
  movzbl (%rax,%rdx,1),%eax
  mov    %al,(%rcx)
  cmp    %edi,%r8d
  jbe    L24
  mov    $0x1,%r10d
  movb   $0x0,(%rcx,%r10,1)
L85:
  mov    %r10,%rax
  ret
L25:
  mov    %r10,%rcx
  jmp    L2
L24:
  push   %rbp
  push   %rbx
  mov    $0x1,%ecx
  lea    D10(%rip),%rbp
L2:
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %r11d
  mov    %eax,%ebx
  mov    %eax,%edi
  cmp    %rcx,%r9
  je     L63
  lea    0x1(%rcx),%r10
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %r11d
  movzbl 0x0(%rbp,%rdx,1),%eax
  mov    %al,-0x1(%rsi,%r10,1)
  cmp    %ebx,%r8d
  jbe    L25
  mov    %r10,%r8
  shr    %r8
  cmp    $0x1,%r10
  jbe    L60
  lea    (%rsi,%rcx,1),%rdx
  mov    $0x0,%eax
L69:
  movzbl (%rsi,%rax,1),%ecx
  movzbl (%rdx),%edi
  mov    %dil,(%rsi,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %r8,%rax
  jb     L69
L60:
  movb   $0x0,(%rsi,%r10,1)
L53:
  mov    %r10,%rax
  pop    %rbx
  pop    %rbp
  ret
L63:
  mov    $0x0,%r10d
  jmp    L53
L34:
  mov    %rdx,%r10
  jmp    L85
```

tool_calls: name: retrieve_data
arguments: tag: D10
type: string

tool: D10: 0123456789ABCDEF

assistant: ```c
size_t base_convert(unsigned num, int base, size_t size, char buf[size]) {
  static const char *digits = "0123456789ABCDEF";
  if (base < 2 || base > 16) {
    return 0;
  }
  for (size_t length = 0; length < size - 1; num /= base) {
    buf[length++] = digits[num % base];
    if (num >= base) {
      continue;
    }
    for (size_t i = 0; i < length / 2; ++i) {
      char tmp = buf[i];
      buf[i] = buf[length - i - 1];
      buf[length - i - 1] = tmp;
    }
    buf[length] = '\0';
    return length;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<base_convert>:
  endbr64
  mov    %rcx,%r11
  lea    -0x2(%rsi),%ecx
  cmp    $0xe,%ecx
  ja     L73
  sub    $0x1,%rdx
  mov    %rdx,%r8
  je     L73
  mov    %edi,%r10d
  xor    %edx,%edx
  push   %rbx
  lea    D2(%rip),%rbx
  mov    %r10d,%eax
  mov    %esi,%edi
  div    %esi
  movzbl (%rbx,%rdx,1),%eax
  mov    %al,(%r11)
  cmp    %r10d,%esi
  jbe    L1
L78:
  mov    $0x1,%r9d
L51:
  mov    %r9,%rax
  movb   $0x0,(%r11,%r9,1)
  pop    %rbx
  ret
L1:
  mov    $0x1,%ecx
L91:
  mov    %r10d,%eax
  xor    %edx,%edx
  div    %edi
  mov    %eax,%r10d
  cmp    %rcx,%r8
  je     L21
  mov    %r10d,%eax
  xor    %edx,%edx
  lea    0x1(%rcx),%r9
  div    %edi
  movzbl (%rbx,%rdx,1),%esi
  mov    %sil,-0x1(%r11,%r9,1)
  cmp    %r10d,%edi
  jbe    L94
  mov    %r9,%r8
  shr    %r8
  je     L78
  lea    (%r11,%rcx,1),%rax
  mov    %r11,%rdx
  mov    %rax,%rdi
  sub    %r8,%rdi
  jmp    L16
  nopl   0x0(%rax)
L68:
  movzbl (%rax),%esi
L16:
  movzbl (%rdx),%ecx
  sub    $0x1,%rax
  mov    %sil,(%rdx)
  add    $0x1,%rdx
  mov    %cl,0x1(%rax)
  cmp    %rdi,%rax
  jne    L68
  jmp    L51
L21:
  xor    %r9d,%r9d
  pop    %rbx
  mov    %r9,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L94:
  mov    %r9,%rcx
  jmp    L91
  nopl   (%rax)
L73:
  xor    %eax,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
data_type: string

tool: D2: 0123456789ABCDEF

assistant: ```c
size_t base_convert(unsigned num, int base, size_t size, char buf[size]) {
  static const char *digits = "0123456789ABCDEF";
  if (base < 2 || base > 16) {
    return 0;
  }
  for (size_t length = 0; length < size - 1; num /= base) {
    buf[length++] = digits[num % base];
    if (num >= base) {
      continue;
    }
    for (size_t i = 0; i < length / 2; ++i) {
      char tmp = buf[i];
      buf[i] = buf[length - i - 1];
      buf[length - i - 1] = tmp;
    }
    buf[length] = '\0';
    return length;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<base_convert>:
  endbr64
  mov    %rcx,%r10
  lea    -0x2(%rsi),%ecx
  mov    %edi,%r9d
  cmp    $0xe,%ecx
  ja     L51
  mov    %rdx,%rdi
  sub    $0x1,%rdi
  je     L51
  mov    %r9d,%eax
  xor    %edx,%edx
  lea    D9(%rip),%r11
  div    %esi
  movzbl (%r11,%rdx,1),%eax
  mov    %al,(%r10)
  cmp    %r9d,%esi
  jbe    L70
L89:
  mov    $0x1,%r8d
L24:
  movb   $0x0,(%r10,%r8,1)
  mov    %r8,%rax
  ret
  nopl   (%rax)
L70:
  mov    $0x1,%ecx
L44:
  mov    %r9d,%eax
  xor    %edx,%edx
  div    %esi
  mov    %eax,%r9d
  cmp    %rcx,%rdi
  je     L51
  mov    %r9d,%eax
  xor    %edx,%edx
  lea    0x1(%rcx),%r8
  div    %esi
  movzbl (%r11,%rdx,1),%eax
  mov    %al,-0x1(%r10,%r8,1)
  cmp    %r9d,%esi
  jbe    L60
  mov    %r8,%rsi
  shr    %rsi
  je     L89
  lea    (%r10,%rcx,1),%rax
  mov    %r10,%rdx
  mov    %rax,%rdi
  sub    %rsi,%rdi
  nopl   0x0(%rax,%rax,1)
L40:
  movzbl (%rdx),%ecx
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %sil,-0x1(%rdx)
  mov    %cl,0x1(%rax)
  cmp    %rdi,%rax
  jne    L40
  jmp    L24
  nopl   0x0(%rax)
L51:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
  nop
L60:
  mov    %r8,%rcx
  jmp    L44
```

tool_calls: name: disassemble_data
arguments: data_label: D9
data_type: string

tool: D9: 0123456789ABCDEF

assistant: ```c
size_t base_convert(unsigned num, int base, size_t size, char buf[size]) {
  static const char *digits = "0123456789ABCDEF";
  if (base < 2 || base > 16) {
    return 0;
  }
  for (size_t length = 0; length < size - 1; num /= base) {
    buf[length++] = digits[num % base];
    if (num >= base) {
      continue;
    }
    for (size_t i = 0; i < length / 2; ++i) {
      char tmp = buf[i];
      buf[i] = buf[length - i - 1];
      buf[length - i - 1] = tmp;
    }
    buf[length] = '\0';
    return length;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atmc_add64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <compare_and_swap64@plt>
  test   %eax,%eax
  je     L3
  nop
  nop
  leave
  ret
```

assistant: ```c
void atmc_add64(volatile unsigned long long *address, unsigned long long inc) {
  long long old_value;
  long long new_value;
  do {
    old_value = *address;
    new_value = old_value + inc;
  } while (!compare_and_swap64(address, old_value, new_value));
}
```
user: What is the c source code of the assembly code below?

```asm
<atmc_add64>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
L3:
  mov    (%rbx),%rsi
  lea    (%rsi,%rbp,1),%rdx
  mov    %rbx,%rdi
  call   <compare_and_swap64@plt>
  test   %eax,%eax
  je     L3
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void atmc_add64(volatile unsigned long long *address, unsigned long long inc) {
  long long old_value;
  long long new_value;
  do {
    old_value = *address;
    new_value = old_value + inc;
  } while (!compare_and_swap64(address, old_value, new_value));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<atmc_add64>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
L8:
  mov    (%rbx),%rsi
  mov    %rbx,%rdi
  lea    (%rsi,%rbp,1),%rdx
  call   <compare_and_swap64@plt>
  test   %eax,%eax
  je     L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void atmc_add64(volatile unsigned long long *address, unsigned long long inc) {
  long long old_value;
  long long new_value;
  do {
    old_value = *address;
    new_value = old_value + inc;
  } while (!compare_and_swap64(address, old_value, new_value));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<atmc_add64>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
L1:
  mov    (%rbx),%rsi
  mov    %rbx,%rdi
  lea    (%rsi,%rbp,1),%rdx
  call   <compare_and_swap64@plt>
  test   %eax,%eax
  je     L1
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void atmc_add64(volatile unsigned long long *address, unsigned long long inc) {
  long long old_value;
  long long new_value;
  do {
    old_value = *address;
    new_value = old_value + inc;
  } while (!compare_and_swap64(address, old_value, new_value));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<zipmapRawEntryLength>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <zipmapRawKeyLength@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <zipmapRawValueLength@plt>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
unsigned int zipmapRawEntryLength(unsigned char *p) {
  unsigned int l = zipmapRawKeyLength(p);
  return l + zipmapRawValueLength(p + l);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zipmapRawEntryLength>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <zipmapRawKeyLength@plt>
  mov    %eax,%ebp
  mov    %eax,%eax
  lea    (%rbx,%rax,1),%rdi
  call   <zipmapRawValueLength@plt>
  add    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned int zipmapRawEntryLength(unsigned char *p) {
  unsigned int l = zipmapRawKeyLength(p);
  return l + zipmapRawValueLength(p + l);
}
```