ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  mov    $0x18,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %rsp,%rbx
  lea    0x640(%rsp),%r12
  mov    %rbx,%rbp
L14:
  mov    $0x78,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  add    $0x8,%rbp
  cmp    %r12,%rbp
  jne    L14
  mov    $0x18,%edi
  call   <malloc@plt>
L7:
  mov    (%rbx),%rdi
  call   <free@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L7
  mov    $0x18,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %r13,%rsi
  lea    D0(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x658,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D0
type_infered: string

tool: D0: %p


assistant: ```c
int main() {
  void *m1 = malloc(24);
  int t = 0;
  void *ms[200];
  for (t = 0; t < 200; t++)
    ms[t] = malloc(120);
  malloc(24);
  for (t = 0; t < 200; t++)
    free(ms[t]);
  void *m2 = malloc(24);
  printf("%p\n", m1);
  printf("%p\n", m2);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x18,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x640(%rsp),%r12
  call   <malloc@plt>
  mov    %rbx,%rbp
  mov    %rax,%r13
  nop
L1:
  mov    $0x78,%edi
  add    $0x8,%rbp
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmp    %r12,%rbp
  jne    L1
  mov    $0x18,%edi
  call   <malloc@plt>
  nopl   0x0(%rax)
L24:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %r12,%rbx
  jne    L24
  mov    $0x18,%edi
  lea    D9(%rip),%r12
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x658,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
datatype: string

tool: D9: %p


assistant: ```c
int main() {
  void *m1 = malloc(24);
  int t = 0;
  void *ms[200];
  for (t = 0; t < 200; t++)
    ms[t] = malloc(120);
  malloc(24);
  for (t = 0; t < 200; t++)
    free(ms[t]);
  void *m2 = malloc(24);
  printf("%p\n", m1);
  printf("%p\n", m2);
  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   %r13
  mov    $0x18,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x658,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x648(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x640(%rsp),%r12
  call   <malloc@plt>
  mov    %rbx,%rbp
  mov    %rax,%r13
  nop
L26:
  mov    $0x78,%edi
  add    $0x8,%rbp
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmp    %r12,%rbp
  jne    L26
  mov    $0x18,%edi
  call   <malloc@plt>
  nopl   0x0(%rax)
L16:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %r12,%rbx
  jne    L16
  mov    $0x18,%edi
  lea    D5(%rip),%r12
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x648(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x658,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: %p


assistant: ```c
int main() {
  void *m1 = malloc(24);
  int t = 0;
  void *ms[200];
  for (t = 0; t < 200; t++)
    ms[t] = malloc(120);
  malloc(24);
  for (t = 0; t < 200; t++)
    free(ms[t]);
  void *m2 = malloc(24);
  printf("%p\n", m1);
  printf("%p\n", m2);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<make_devname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2f,(%rax)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *make_devname(const char *path, const char *name) {
  int plen = strlen(path);
  int nlen = strlen(name);
  char *devname = malloc(plen + nlen + 2);
  strcpy(devname, path);
  devname[plen] = '/';
  strcpy(devname + plen + 1, name);
  return devname;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<make_devname>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x2(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movslq %ebx,%rbx
  movb   $0x2f,0x0(%rbp,%rbx,1)
  lea    0x1(%rbp,%rbx,1),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *make_devname(const char *path, const char *name) {
  int plen = strlen(path);
  int nlen = strlen(name);
  char *devname = malloc(plen + nlen + 2);
  strcpy(devname, path);
  devname[plen] = '/';
  strcpy(devname + plen + 1, name);
  return devname;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<make_devname>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  lea    0x2(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  movb   $0x2f,(%r12,%rbx,1)
  lea    0x1(%r12,%rbx,1),%rdi
  mov    %rbp,%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *make_devname(const char *path, const char *name) {
  int plen = strlen(path);
  int nlen = strlen(name);
  char *devname = malloc(plen + nlen + 2);
  strcpy(devname, path);
  devname[plen] = '/';
  strcpy(devname + plen + 1, name);
  return devname;
}
```
user: What is the c source code of the assembly code below?

```asm
<make_devname>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  call   <strlen@plt>
  lea    0x2(%rbx,%rax,1),%edi
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  movb   $0x2f,(%r12,%rbx,1)
  lea    0x1(%r12,%rbx,1),%rdi
  mov    %rbp,%rsi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *make_devname(const char *path, const char *name) {
  int plen = strlen(path);
  int nlen = strlen(name);
  char *devname = malloc(plen + nlen + 2);
  strcpy(devname, path);
  devname[plen] = '/';
  strcpy(devname + plen + 1, name);
  return devname;
}
```
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)
  movl   $0x8,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  cmpl   $0x1,-0x14(%rbp)
  jle    L6
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <BackTrack@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type: string

tool: D3: %d -%s- 


assistant: ```c
int main(int argc, char **argv) {
  int n = 8;
  printf("%d -%s- \n", argc, argv[1]);
  if (argc > 1) {
    n = atoi(argv[1]);
  }
  BackTrack(n);
  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    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    0x8(%rsi),%rdx
  mov    %edi,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edi
  cmp    $0x1,%ebx
  jg     L8
L13:
  call   <BackTrack@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
  jmp    L13
```

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

tool: D6: %d -%s- 


assistant: ```c
int main(int argc, char **argv) {
  int n = 8;
  printf("%d -%s- \n", argc, argv[1]);
  if (argc > 1) {
    n = atoi(argv[1]);
  }
  BackTrack(n);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdx
  mov    %edi,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edi
  cmp    $0x1,%ebx
  jle    L5
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L5:
  call   <BackTrack@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D4: %d -%s- 


assistant: ```c
int main(int argc, char **argv) {
  int n = 8;
  printf("%d -%s- \n", argc, argv[1]);
  if (argc > 1) {
    n = atoi(argv[1]);
  }
  BackTrack(n);
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdx
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edi
  cmp    $0x1,%ebx
  jle    L5
  mov    0x8(%rbp),%rdi
  call   <atoi@plt>
  mov    %eax,%edi
L5:
  call   <BackTrack@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
data_type: string

tool: D8: %d -%s- 


assistant: ```c
int main(int argc, char **argv) {
  int n = 8;
  printf("%d -%s- \n", argc, argv[1]);
  if (argc > 1) {
    n = atoi(argv[1]);
  }
  BackTrack(n);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x375f00,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
datatype: string

tool: D7: 10!=%d /n

assistant: ```c
int main() {
  int i;
  i = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10;
  printf("10!=%d /n", i);
  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
  mov    $0x375f00,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D0: 10!=%d /n

assistant: ```c
int main() {
  int i;
  i = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10;
  printf("10!=%d /n", i);
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x375f00,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: 10!=%d /n

assistant: ```c
int main() {
  int i;
  i = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10;
  printf("10!=%d /n", i);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x375f00,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
datatype: string

tool: D3: 10!=%d /n

assistant: ```c
int main() {
  int i;
  i = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10;
  printf("10!=%d /n", i);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hazer_parse_fraction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movq   $0x1,-0x20(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  lea    -0x28(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoull@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  sub    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L26
L3:
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x20(%rbp)
L26:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  test   %rax,%rax
  jne    L3
  mov    -0x40(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
uint64_t hazer_parse_fraction(const char *string, uint64_t *denominatorp) {
  unsigned long long numerator = 0;
  unsigned long long denominator = 1;
  char *end = (char *)0;
  size_t length = 0;
  numerator = strtoull(string, &end, 10);
  length = end - string;
  while ((length--) > 0) {
    denominator = denominator * 10;
  }
  *denominatorp = denominator;
  return numerator;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hazer_parse_fraction>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtoull@plt>
  mov    (%rsp),%rdx
  sub    %rbx,%rdx
  je     L13
  lea    -0x1(%rdx),%rcx
  mov    $0x1,%edx
L16:
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  sub    $0x1,%rcx
  cmp    $0xffffffffffffffff,%rcx
  jne    L16
L28:
  mov    %rdx,0x0(%rbp)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L27
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  mov    $0x1,%edx
  jmp    L28
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t hazer_parse_fraction(const char *string, uint64_t *denominatorp) {
  unsigned long long numerator = 0;
  unsigned long long denominator = 1;
  char *end = (char *)0;
  size_t length = 0;
  numerator = strtoull(string, &end, 10);
  length = end - string;
  while ((length--) > 0) {
    denominator = denominator * 10;
  }
  *denominatorp = denominator;
  return numerator;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hazer_parse_fraction>:
  endbr64
  push   %rbp
  mov    $0xa,%edx
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <strtoull@plt>
  mov    (%rsp),%rdx
  sub    %rbx,%rdx
  je     L23
  lea    -0x1(%rdx),%rcx
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L35:
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  sub    $0x1,%rcx
  jae    L35
L11:
  mov    %rdx,0x0(%rbp)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L26
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  mov    $0x1,%edx
  jmp    L11
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t hazer_parse_fraction(const char *string, uint64_t *denominatorp) {
  unsigned long long numerator = 0;
  unsigned long long denominator = 1;
  char *end = (char *)0;
  size_t length = 0;
  numerator = strtoull(string, &end, 10);
  length = end - string;
  while ((length--) > 0) {
    denominator = denominator * 10;
  }
  *denominatorp = denominator;
  return numerator;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<hazer_parse_fraction>:
  endbr64
  push   %rbp
  mov    $0xa,%edx
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   $0x0,(%rsp)
  call   <strtoull@plt>
  mov    (%rsp),%rdx
  sub    %rbx,%rdx
  je     L32
  lea    -0x1(%rdx),%rcx
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L34:
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  sub    $0x1,%rcx
  jae    L34
L24:
  mov    %rdx,0x0(%rbp)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L38
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  mov    $0x1,%edx
  jmp    L24
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint64_t hazer_parse_fraction(const char *string, uint64_t *denominatorp) {
  unsigned long long numerator = 0;
  unsigned long long denominator = 1;
  char *end = (char *)0;
  size_t length = 0;
  numerator = strtoull(string, &end, 10);
  length = end - string;
  while ((length--) > 0) {
    denominator = denominator * 10;
  }
  *denominatorp = denominator;
  return numerator;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jsr_flush_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x0,%eax
  call   <beq@plt>
  test   %rax,%rax
  jne    L2
  mov    $0x0,%eax
  call   <jsr_hplot@plt>
  mov    $0x0,%eax
  call   <clc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <adc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <adc@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <dec@plt>
  jmp    L4
L2:
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void jsr_flush_line() {
flush_loop:
  lda(1, 0xf4);
  if (beq())
    goto end_flush;
  jsr_hplot();
  clc();
  lda(1, 0xf1);
  adc(0, 1);
  sta(1, 0xf1);
  lda(1, 0xf2);
  adc(0, 0);
  sta(1, 0xf2);
  dec(1, 0xf4);
  goto flush_loop;
end_flush:;
}
```
user: What is the c source code of the assembly code below?

```asm
<jsr_flush_line>:
  endbr64
  sub    $0x8,%rsp
  jmp    L10
L17:
  call   <jsr_hplot@plt>
  mov    $0x0,%eax
  call   <clc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <adc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <adc@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <dec@plt>
L10:
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x0,%eax
  call   <beq@plt>
  test   %rax,%rax
  je     L17
  add    $0x8,%rsp
  ret
```

assistant: ```c
void jsr_flush_line() {
flush_loop:
  lda(1, 0xf4);
  if (beq())
    goto end_flush;
  jsr_hplot();
  clc();
  lda(1, 0xf1);
  adc(0, 1);
  sta(1, 0xf1);
  lda(1, 0xf2);
  adc(0, 0);
  sta(1, 0xf2);
  dec(1, 0xf4);
  goto flush_loop;
end_flush:;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<jsr_flush_line>:
  endbr64
  sub    $0x8,%rsp
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L14:
  call   <jsr_hplot@plt>
  xor    %eax,%eax
  call   <clc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x1,%esi
  xor    %edi,%edi
  call   <adc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <adc@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <dec@plt>
L13:
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  xor    %eax,%eax
  call   <beq@plt>
  test   %rax,%rax
  je     L14
  add    $0x8,%rsp
  ret
```

assistant: ```c
void jsr_flush_line() {
flush_loop:
  lda(1, 0xf4);
  if (beq())
    goto end_flush;
  jsr_hplot();
  clc();
  lda(1, 0xf1);
  adc(0, 1);
  sta(1, 0xf1);
  lda(1, 0xf2);
  adc(0, 0);
  sta(1, 0xf2);
  dec(1, 0xf4);
  goto flush_loop;
end_flush:;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<jsr_flush_line>:
  endbr64
  sub    $0x8,%rsp
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L0:
  call   <jsr_hplot@plt>
  xor    %eax,%eax
  call   <clc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  mov    $0x1,%esi
  xor    %edi,%edi
  call   <adc@plt>
  mov    $0xf1,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <adc@plt>
  mov    $0xf2,%esi
  mov    $0x1,%edi
  call   <sta@plt>
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <dec@plt>
L11:
  mov    $0xf4,%esi
  mov    $0x1,%edi
  call   <lda@plt>
  xor    %eax,%eax
  call   <beq@plt>
  test   %rax,%rax
  je     L0
  add    $0x8,%rsp
  ret
```

assistant: ```c
void jsr_flush_line() {
flush_loop:
  lda(1, 0xf4);
  if (beq())
    goto end_flush;
  jsr_hplot();
  clc();
  lda(1, 0xf1);
  adc(0, 1);
  sta(1, 0xf1);
  lda(1, 0xf2);
  adc(0, 0);
  sta(1, 0xf2);
  dec(1, 0xf4);
  goto flush_loop;
end_flush:;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<maximum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L7
  mov    -0x4(%rbp),%eax
  jmp    L17
L7:
  mov    -0x8(%rbp),%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int maximum(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<maximum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maximum(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<maximum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maximum(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maximum>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovge %edi,%eax
  ret
```

assistant: ```c
int maximum(int a, int b) {
  if (a > b)
    return a;
  else
    return b;
}
```