ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x6,0xc(%rsp)
  movl   $0xa,0x10(%rsp)
  movl   $0xf,0x14(%rsp)
  movl   $0x15,0x18(%rsp)
  movl   $0x1c,0x1c(%rsp)
  movl   $0x24,0x20(%rsp)
  mov    %rsp,%rbp
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x3,%edi
  call   <binsearch1@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x3,%edi
  call   <binsearch2@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x15,%edi
  call   <binsearch1@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x15,%edi
  call   <binsearch2@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <binsearch1@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %rbp,%rsi
  mov    $0x5,%edi
  call   <binsearch2@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d


assistant: ```c
int main() {
  int w[] = {0, 1, 3, 6, 10, 15, 21, 28, 36};
  int m = sizeof(w) / sizeof(int);
  int i1, i2;
  int j1, j2;
  int k1, k2;
  i1 = binsearch1(3, w, m);
  printf("%d\n", i1);
  i2 = binsearch2(3, w, m);
  printf("%d\n", i2);
  j1 = binsearch1(21, w, m);
  printf("%d\n", j1);
  j2 = binsearch2(21, w, m);
  printf("%d\n", j2);
  k1 = binsearch1(5, w, m);
  printf("%d\n", k1);
  k2 = binsearch2(5, w, m);
  printf("%d\n", k2);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x9,%edx
  mov    $0x3,%edi
  push   %rbp
  lea    D1(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x100000000,%rax
  mov    %rsp,%r12
  movl   $0x24,0x20(%rsp)
  mov    %r12,%rsi
  mov    %rax,(%rsp)
  movabs $0x600000003,%rax
  mov    %rax,0x8(%rsp)
  movabs $0xf0000000a,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x1c00000015,%rax
  mov    %rax,0x18(%rsp)
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x3,%edi
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x15,%edi
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x15,%edi
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x5,%edi
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    $0x5,%edi
  mov    $0x9,%edx
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string

tool: D1: %d


assistant: ```c
int main() {
  int w[] = {0, 1, 3, 6, 10, 15, 21, 28, 36};
  int m = sizeof(w) / sizeof(int);
  int i1, i2;
  int j1, j2;
  int k1, k2;
  i1 = binsearch1(3, w, m);
  printf("%d\n", i1);
  i2 = binsearch2(3, w, m);
  printf("%d\n", i2);
  j1 = binsearch1(21, w, m);
  printf("%d\n", j1);
  j2 = binsearch2(21, w, m);
  printf("%d\n", j2);
  k1 = binsearch1(5, w, m);
  printf("%d\n", k1);
  k2 = binsearch2(5, w, m);
  printf("%d\n", k2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x9,%edx
  mov    $0x3,%edi
  push   %rbp
  lea    D12(%rip),%rbp
  sub    $0x38,%rsp
  movdqa D8(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  movl   $0x24,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D15(%rip),%xmm0
  mov    %r12,%rsi
  movaps %xmm0,0x10(%rsp)
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x3,%edi
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x15,%edi
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x15,%edi
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x9,%edx
  mov    %r12,%rsi
  mov    $0x5,%edi
  call   <binsearch1@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    $0x5,%edi
  mov    $0x9,%edx
  call   <binsearch2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D12
type_infered: string
name: retrieve_data
arguments: label: D8
type_infered: byte[16]
name: retrieve_data
arguments: label: D15
type_infered: byte[16]

tool: D12: %d


tool: {"D8": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: {"D15": [0x0a,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x15,0x00,0x00,0x00,0x1c,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int w[] = {0, 1, 3, 6, 10, 15, 21, 28, 36};
  int m = sizeof(w) / sizeof(int);
  int i1, i2;
  int j1, j2;
  int k1, k2;
  i1 = binsearch1(3, w, m);
  printf("%d\n", i1);
  i2 = binsearch2(3, w, m);
  printf("%d\n", i2);
  j1 = binsearch1(21, w, m);
  printf("%d\n", j1);
  j2 = binsearch2(21, w, m);
  printf("%d\n", j2);
  k1 = binsearch1(5, w, m);
  printf("%d\n", k1);
  k2 = binsearch2(5, w, m);
  printf("%d\n", k2);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L19
L12:
  addq   $0x1,-0x10(%rbp)
L19:
  cmpq   $0x1fff,-0x10(%rbp)
  ja     L1
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
L1:
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *_strdup(const char *s) {
  size_t l = 0;
  while (l < 8192 && s[l]) {
    l++;
  }
  char *r = malloc(l + 1);
  strncpy(r, s, l);
  r[l] = '\0';
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x0,%ebx
L14:
  cmpb   $0x0,0x0(%rbp,%rbx,1)
  je     L19
  add    $0x1,%rbx
  cmp    $0x2000,%rbx
  jne    L14
L19:
  lea    0x1(%rbx),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_strdup(const char *s) {
  size_t l = 0;
  while (l < 8192 && s[l]) {
    l++;
  }
  char *r = malloc(l + 1);
  strncpy(r, s, l);
  r[l] = '\0';
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  jmp    L7
  nop
L29:
  cmp    $0x2000,%rdi
  je     L8
L7:
  movzbl 0x0(%rbp,%rdi,1),%eax
  mov    %rdi,%rbx
  add    $0x1,%rdi
  test   %al,%al
  jne    L29
L34:
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  mov    $0x2000,%ebx
  mov    $0x2001,%edi
  jmp    L34
```

assistant: ```c
char *_strdup(const char *s) {
  size_t l = 0;
  while (l < 8192 && s[l]) {
    l++;
  }
  char *r = malloc(l + 1);
  strncpy(r, s, l);
  r[l] = '\0';
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  jmp    L19
  nop
L18:
  cmp    $0x2000,%rdi
  je     L20
L19:
  movzbl 0x0(%rbp,%rdi,1),%eax
  mov    %rdi,%rbx
  add    $0x1,%rdi
  test   %al,%al
  jne    L18
L33:
  call   <malloc@plt>
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L20:
  mov    $0x2000,%ebx
  mov    $0x2001,%edi
  jmp    L33
```

assistant: ```c
char *_strdup(const char *s) {
  size_t l = 0;
  while (l < 8192 && s[l]) {
    l++;
  }
  char *r = malloc(l + 1);
  strncpy(r, s, l);
  r[l] = '\0';
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L162
  mov    $0x1,%eax
  jmp    L5
L162:
  mov    $0x0,%edx
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L101
  mov    $0x1,%eax
  jmp    L5
L101:
  mov    $0x2,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L79
  mov    $0x1,%eax
  jmp    L5
L79:
  mov    $0x2,%edx
  mov    $0xfffffffd,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L164
  mov    $0x1,%eax
  jmp    L5
L164:
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L138
  mov    $0x1,%eax
  jmp    L5
L138:
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L117
  mov    $0x1,%eax
  jmp    L5
L117:
  mov    $0x0,%edx
  mov    $0x1,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L97
  mov    $0x1,%eax
  jmp    L5
L97:
  mov    $0x0,%edx
  mov    $0x2,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L85
  mov    $0x1,%eax
  jmp    L5
L85:
  mov    $0x0,%edx
  mov    $0xfffffffe,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L56
  mov    $0x1,%eax
  jmp    L5
L56:
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L140
  mov    $0x1,%eax
  jmp    L5
L140:
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L26
  mov    $0x1,%eax
  jmp    L5
L26:
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (test__modsi3(0, 1, 0))
    return 1;
  if (test__modsi3(0, -1, 0))
    return 1;
  if (test__modsi3(5, 3, 2))
    return 1;
  if (test__modsi3(5, -3, 2))
    return 1;
  if (test__modsi3(-5, 3, -2))
    return 1;
  if (test__modsi3(-5, -3, -2))
    return 1;
  if (test__modsi3(0x80000000, 1, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, -2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 3, -2))
    return 1;
  if (test__modsi3(0x80000000, -3, -2))
    return 1;
  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    $0x0,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L27
L19:
  add    $0x8,%rsp
  ret
L27:
  mov    $0xffffffff,%esi
  mov    $0x0,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0x2,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0x2,%edx
  mov    $0xfffffffd,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0x1,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0x2,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0xfffffffe,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L19
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L19
```

assistant: ```c
int main() {
  if (test__modsi3(0, 1, 0))
    return 1;
  if (test__modsi3(0, -1, 0))
    return 1;
  if (test__modsi3(5, 3, 2))
    return 1;
  if (test__modsi3(5, -3, 2))
    return 1;
  if (test__modsi3(-5, 3, -2))
    return 1;
  if (test__modsi3(-5, -3, -2))
    return 1;
  if (test__modsi3(0x80000000, 1, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, -2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 3, -2))
    return 1;
  if (test__modsi3(0x80000000, -3, -2))
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L89
L80:
  mov    $0x1,%eax
L81:
  add    $0x8,%rsp
  ret
L89:
  xor    %edx,%edx
  or     $0xffffffff,%esi
  xor    %edi,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0x2,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0x2,%edx
  mov    $0xfffffffd,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  xor    %edx,%edx
  mov    $0x1,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  xor    %edx,%edx
  mov    $0x2,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  xor    %edx,%edx
  mov    $0xfffffffe,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L80
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L81
```

assistant: ```c
int main() {
  if (test__modsi3(0, 1, 0))
    return 1;
  if (test__modsi3(0, -1, 0))
    return 1;
  if (test__modsi3(5, 3, 2))
    return 1;
  if (test__modsi3(5, -3, 2))
    return 1;
  if (test__modsi3(-5, 3, -2))
    return 1;
  if (test__modsi3(-5, -3, -2))
    return 1;
  if (test__modsi3(0x80000000, 1, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, -2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 3, -2))
    return 1;
  if (test__modsi3(0x80000000, -3, -2))
    return 1;
  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
  xor    %edx,%edx
  xor    %edi,%edi
  mov    $0x1,%esi
  call   <test__modsi3@plt>
  test   %rax,%rax
  je     L9
L61:
  mov    $0x1,%eax
L17:
  add    $0x8,%rsp
  ret
L9:
  xor    %edx,%edx
  or     $0xffffffff,%esi
  xor    %edi,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0x2,%edx
  mov    $0x3,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0x2,%edx
  mov    $0xfffffffd,%esi
  mov    $0x5,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0xfffffffb,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  xor    %edx,%edx
  mov    $0x1,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  xor    %edx,%edx
  mov    $0x2,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  xor    %edx,%edx
  mov    $0xfffffffe,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0xfffffffe,%edx
  mov    $0x3,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  jne    L61
  mov    $0xfffffffe,%edx
  mov    $0xfffffffd,%esi
  mov    $0x80000000,%edi
  call   <test__modsi3@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L17
```

assistant: ```c
int main() {
  if (test__modsi3(0, 1, 0))
    return 1;
  if (test__modsi3(0, -1, 0))
    return 1;
  if (test__modsi3(5, 3, 2))
    return 1;
  if (test__modsi3(5, -3, 2))
    return 1;
  if (test__modsi3(-5, 3, -2))
    return 1;
  if (test__modsi3(-5, -3, -2))
    return 1;
  if (test__modsi3(0x80000000, 1, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, -2, 0x0))
    return 1;
  if (test__modsi3(0x80000000, 3, -2))
    return 1;
  if (test__modsi3(0x80000000, -3, -2))
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<wcscmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L22
L13:
  addq   $0x8,-0x8(%rbp)
  addq   $0x8,-0x10(%rbp)
L22:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L1
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L13
L1:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int wcscmp(const wchar_t *s1, const wchar_t *s2) {
  while ((*s1) && (*s1 == *s2)) {
    ++s1;
    ++s2;
  }
  return (*(wchar_t *)s1 - *(wchar_t *)s2);
}
```
user: What is the c source code of the assembly code below:

```asm
<wcscmp>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  je     L9
L3:
  cmp    %rax,(%rsi)
  jne    L9
  add    $0x8,%rdi
  add    $0x8,%rsi
  mov    (%rdi),%rax
  test   %rax,%rax
  jne    L3
L9:
  sub    (%rsi),%eax
  ret
```

assistant: ```c
int wcscmp(const wchar_t *s1, const wchar_t *s2) {
  while ((*s1) && (*s1 == *s2)) {
    ++s1;
    ++s2;
  }
  return (*(wchar_t *)s1 - *(wchar_t *)s2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<wcscmp>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  jne    L38
  jmp    L31
  xchg   %ax,%ax
L30:
  mov    0x8(%rdi),%rax
  add    $0x8,%rdi
  lea    0x8(%rsi),%rdx
  test   %rax,%rax
  je     L25
  mov    %rdx,%rsi
L38:
  mov    (%rsi),%rdx
  cmp    %rax,%rdx
  je     L30
L47:
  sub    %edx,%eax
  ret
  nop
L25:
  mov    0x8(%rsi),%rdx
  sub    %edx,%eax
  ret
L31:
  mov    (%rsi),%rdx
  jmp    L47
```

assistant: ```c
int wcscmp(const wchar_t *s1, const wchar_t *s2) {
  while ((*s1) && (*s1 == *s2)) {
    ++s1;
    ++s2;
  }
  return (*(wchar_t *)s1 - *(wchar_t *)s2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wcscmp>:
  endbr64
  mov    (%rdi),%rax
  test   %rax,%rax
  jne    L17
  jmp    L44
  xchg   %ax,%ax
L48:
  mov    0x8(%rdi),%rax
  add    $0x8,%rdi
  lea    0x8(%rsi),%rdx
  test   %rax,%rax
  je     L13
  mov    %rdx,%rsi
L17:
  mov    (%rsi),%rdx
  cmp    %rax,%rdx
  je     L48
L1:
  sub    %edx,%eax
  ret
  nop
L13:
  mov    0x8(%rsi),%rdx
  sub    %edx,%eax
  ret
L44:
  mov    (%rsi),%rdx
  jmp    L1
```

assistant: ```c
int wcscmp(const wchar_t *s1, const wchar_t *s2) {
  while ((*s1) && (*s1 == *s2)) {
    ++s1;
    ++s2;
  }
  return (*(wchar_t *)s1 - *(wchar_t *)s2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L12
L15:
  mov    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L15
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int factorial(int n) {
  int j = 1, fact = 1;
  for (j = 1; j <= n; j++)
    fact = fact * j;
  return fact;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  test   %edi,%edi
  jle    L22
  add    $0x1,%edi
  mov    $0x1,%edx
  mov    $0x1,%eax
L2:
  imul   %eax,%edx
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L2
L20:
  mov    %edx,%eax
  ret
L22:
  mov    $0x1,%edx
  jmp    L20
```

assistant: ```c
int factorial(int n) {
  int j = 1, fact = 1;
  for (j = 1; j <= n; j++)
    fact = fact * j;
  return fact;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<factorial>:
  endbr64
  test   %edi,%edi
  jle    L7
  add    $0x1,%edi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L17:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L17
  mov    %r8d,%eax
  ret
  nop
L7:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int factorial(int n) {
  int j = 1, fact = 1;
  for (j = 1; j <= n; j++)
    fact = fact * j;
  return fact;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factorial>:
  endbr64
  test   %edi,%edi
  jle    L19
  add    $0x1,%edi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  cs nopw 0x0(%rax,%rax,1)
L6:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L6
  mov    %r8d,%eax
  ret
  nop
L19:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int factorial(int n) {
  int j = 1, fact = 1;
  for (j = 1; j <= n; j++)
    fact = fact * j;
  return fact;
}
```