ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<copy_recursive_exec>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    $0x1,%edi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x1c(%rsp),%rdx
  call   <real_xstat@plt>
  cmp    $0xffffffff,%eax
  je     L43
L34:
  xor    %r12d,%r12d
L36:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L67
  add    $0x38,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L43:
  mov    $0x2f,%esi
  mov    %r12,%rdi
  call   <strrchr@plt>
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  mov    %rax,%r13
  call   <strrchr@plt>
  lea    0x10(%rsp),%rdx
  mov    %r12,%rsi
  mov    $0x1,%edi
  mov    %rax,%rbx
  call   <real_xstat@plt>
  cmp    $0xffffffff,%eax
  je     L69
  movb   $0x0,0x0(%r13)
  mov    %r12,%rdi
  mov    %rbp,%rsi
  movb   $0x0,(%rbx)
  call   <copy_recursive_exec>
  movb   $0x2f,0x0(%r13)
  movb   $0x2f,(%rbx)
  mov    0x18(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  je     L34
  mov    $0x41c9,%esi
  mov    %rbp,%rdi
  call   <real_mkdir@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L34
  mov    0x10(%rsp),%rax
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  mov    %rax,0x8(%rsp)
  call   <utime@plt>
  jmp    L36
L69:
  mov    $0xfffffffe,%r12d
  jmp    L36
L67:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int copy_recursive_exec(const char *oldpath, const char *cachepath) {
  char *path_bn;
  char *cache_bn;
  char *cpath = (char *)cachepath;
  char *ppath = (char *)oldpath;
  struct stat pathstat;
  struct stat cachepathstat;
  struct utimbuf dtime;
  if (real_xstat(1, cpath, &cachepathstat) == -1) {
    path_bn = strrchr(ppath, '/');
    cache_bn = strrchr(cpath, '/');
    if (real_xstat(1, ppath, &pathstat) == -1) {
      return -2;
    }
    *path_bn = '\0';
    *cache_bn = '\0';
    copy_recursive_exec(ppath, cpath);
    *path_bn = '/';
    *cache_bn = '/';
    if (S_ISDIR(pathstat.st_mode)) {
      if (!real_mkdir(cpath, 040711)) {
        dtime.actime = pathstat.st_atime;
        dtime.modtime = pathstat.st_mtime;
        utime(cpath, &dtime);
      }
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<memset>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    %esi,%ecx
  mov    %rdx,%rax
  mov    -0x10(%rbp),%rbx
  test   %rax,%rax
  je     L20
  add    $0x1,%rax
  jmp    L3
L1:
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %ecx,%esi
  mov    %sil,(%rdx)
L3:
  sub    $0x1,%rax
  test   %rax,%rax
  jne    L1
L20:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void *memset(void *dest, register const int c, register size_t n) {
  register char *s = dest;
  if (n > 0) {
    n++;
    while (--n > 0)
      *s++ = c;
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L5
  add    %rdi,%rdx
  mov    %rdi,%rcx
L11:
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rcx,%rdx
  jne    L11
L5:
  ret
```

assistant: ```c
void *memset(void *dest, register const int c, register size_t n) {
  register char *s = dest;
  if (n > 0) {
    n++;
    while (--n > 0)
      *s++ = c;
  }
  return dest;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L8
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *dest, register const int c, register size_t n) {
  register char *s = dest;
  if (n > 0) {
    n++;
    while (--n > 0)
      *s++ = c;
  }
  return dest;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L3
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *dest, register const int c, register size_t n) {
  register char *s = dest;
  if (n > 0) {
    n++;
    while (--n > 0)
      *s++ = c;
  }
  return dest;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L84:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L84
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D30(%rip),%rdi
  call   <scanf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%ecx
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  lea    D30(%rip),%r12
  test   %ecx,%ecx
  jg     L105
L107:
  lea    0x1a0(%rsp),%r8
  lea    0x10(%rsp),%rsi
  mov    $0x1,%edi
  lea    -0x1(%rcx),%r9d
  jmp    L12
L74:
  lea    0x10(%rsp,%r14,4),%rbp
  mov    $0x0,%ebx
L91:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L91
L123:
  add    $0x1,%r13d
  mov    0xc(%rsp),%ecx
  add    $0x64,%r14
  cmp    %r13d,%ecx
  jle    L107
L105:
  cmpl   $0x0,0xc(%rsp)
  jg     L74
  jmp    L123
L8:
  add    $0x1,%rdi
  add    $0x194,%r8
  add    $0x190,%rsi
L12:
  lea    -0x1(%rdi),%eax
  cmp    %eax,%r9d
  jle    L44
  cmp    %edi,%ecx
  jle    L8
  mov    %r8,%rdx
  mov    %rdi,%rax
L43:
  mov    (%rdx),%ebx
  cmp    %ebx,(%rsi,%rax,4)
  jne    L94
  add    $0x1,%rax
  add    $0x190,%rdx
  cmp    %eax,%ecx
  jg     L43
  jmp    L8
L44:
  lea    D16(%rip),%rdi
  call   <printf@plt>
L20:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L96
  mov    $0x0,%eax
  add    $0x9c60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L94:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L96:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D28
type: string
name: disassemble_data
arguments: data_tag: D30
type: string
name: disassemble_data
arguments: data_tag: D35
type: string
name: disassemble_data
arguments: data_tag: D16
type: string
name: disassemble_data
arguments: data_tag: D11
type: string

tool: D28: Vnesete dimenzija na kvadratna matrica: 


tool: D30: %d

tool: D35: Vnesete gi elementite: 


tool: D16: Matricata e SIMETRICHNA vo odnos na glavnata dijagonala


tool: D11: Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala


assistant: ```c
int main() {
  int a[100][100], n, i, j, simetrichna = 1;
  printf("Vnesete dimenzija na kvadratna matrica: \n");
  scanf("%d", &n);
  printf("Vnesete gi elementite: \n");
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j)
      scanf("%d", &a[i][j]);
  for (i = 0; i < n - 1; ++i) {
    for (j = i + 1; j < n; ++j)
      if (a[i][j] != a[j][i]) {
        simetrichna = 0;
        break;
      }
    if (!simetrichna)
      break;
  }
  if (simetrichna)
    printf("Matricata e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  else
    printf("Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  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
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L72:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L72
  sub    $0xc68,%rsp
  lea    D4(%rip),%rdi
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  lea    D15(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L15
  nopl   0x0(%rax)
L41:
  lea    0x0(%rbp,%r14,4),%r15
  xor    %ebx,%ebx
  nop
L23:
  mov    %r15,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  cmp    %ebx,%ecx
  jg     L23
  add    $0x1,%r13d
  add    $0x64,%r14
  cmp    %r13d,%ecx
  jg     L41
  cmp    $0x1,%ecx
  jle    L15
  lea    -0x2(%rcx),%r8d
  lea    0x1a0(%rsp),%rdi
  mov    $0x1,%esi
  add    $0x2,%r8
  nopl   0x0(%rax)
L67:
  cmp    %esi,%ecx
  jle    L73
  mov    %rdi,%rdx
  mov    %rsi,%rax
  jmp    L89
  nopl   0x0(%rax)
L18:
  add    $0x1,%rax
  add    $0x190,%rdx
  cmp    %eax,%ecx
  jle    L73
L89:
  mov    (%rdx),%ebx
  cmp    %ebx,0x0(%rbp,%rax,4)
  je     L18
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L19
  nopl   (%rax)
L73:
  add    $0x1,%rsi
  add    $0x194,%rdi
  add    $0x190,%rbp
  cmp    %r8,%rsi
  jne    L67
L15:
  lea    D23(%rip),%rdi
  call   <printf@plt>
L19:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  add    $0x9c68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
datatype: string
name: retrieve_data
arguments: data_tag: D15
datatype: string
name: retrieve_data
arguments: data_tag: D8
datatype: string
name: retrieve_data
arguments: data_tag: D17
datatype: string
name: retrieve_data
arguments: data_tag: D23
datatype: string

tool: D4: Vnesete dimenzija na kvadratna matrica: 


tool: D15: %d

tool: D8: Vnesete gi elementite: 


tool: D17: Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala


tool: D23: Matricata e SIMETRICHNA vo odnos na glavnata dijagonala


assistant: ```c
int main() {
  int a[100][100], n, i, j, simetrichna = 1;
  printf("Vnesete dimenzija na kvadratna matrica: \n");
  scanf("%d", &n);
  printf("Vnesete gi elementite: \n");
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j)
      scanf("%d", &a[i][j]);
  for (i = 0; i < n - 1; ++i) {
    for (j = i + 1; j < n; ++j)
      if (a[i][j] != a[j][i]) {
        simetrichna = 0;
        break;
      }
    if (!simetrichna)
      break;
  }
  if (simetrichna)
    printf("Matricata e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  else
    printf("Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  return 0;
}
```
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
  lea    -0x9000(%rsp),%r11
L66:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L66
  sub    $0xc68,%rsp
  lea    D7(%rip),%rdi
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  lea    D34(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L109
L62:
  lea    0x0(%rbp,%r13,4),%r15
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L98:
  mov    %r15,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%ecx
  cmp    %ebx,%ecx
  jg     L98
  add    $0x1,%r14d
  cmp    %r14d,%ecx
  jle    L20
  add    $0x64,%r13
  jmp    L62
  nopl   0x0(%rax)
L74:
  add    $0x1,%rsi
  add    $0x194,%rdi
  add    $0x190,%rbp
  cmp    %r8,%rsi
  jne    L11
L109:
  lea    D18(%rip),%rdi
  call   <printf@plt>
L129:
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L112
  add    $0x9c68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  cmp    $0x1,%ecx
  jle    L109
  lea    -0x2(%rcx),%r8d
  lea    0x1a0(%rsp),%rdi
  mov    $0x1,%esi
  add    $0x2,%r8
  nopl   0x0(%rax)
L11:
  cmp    %esi,%ecx
  jle    L74
  mov    %rdi,%rdx
  mov    %rsi,%rax
  jmp    L110
  nopl   0x0(%rax)
L27:
  add    $0x1,%rax
  add    $0x190,%rdx
  cmp    %eax,%ecx
  jle    L74
L110:
  mov    (%rdx),%ebx
  cmp    %ebx,0x0(%rbp,%rax,4)
  je     L27
  lea    D38(%rip),%rdi
  call   <printf@plt>
  jmp    L129
L112:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
data_type: string
name: parse_data
arguments: tag: D34
data_type: string
name: parse_data
arguments: tag: D29
data_type: string
name: parse_data
arguments: tag: D18
data_type: string
name: parse_data
arguments: tag: D38
data_type: string

tool: D7: Vnesete dimenzija na kvadratna matrica: 


tool: D34: %d

tool: D29: Vnesete gi elementite: 


tool: D18: Matricata e SIMETRICHNA vo odnos na glavnata dijagonala


tool: D38: Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala


assistant: ```c
int main() {
  int a[100][100], n, i, j, simetrichna = 1;
  printf("Vnesete dimenzija na kvadratna matrica: \n");
  scanf("%d", &n);
  printf("Vnesete gi elementite: \n");
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j)
      scanf("%d", &a[i][j]);
  for (i = 0; i < n - 1; ++i) {
    for (j = i + 1; j < n; ++j)
      if (a[i][j] != a[j][i]) {
        simetrichna = 0;
        break;
      }
    if (!simetrichna)
      break;
  }
  if (simetrichna)
    printf("Matricata e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  else
    printf("Matricata ne e SIMETRICHNA vo odnos na glavnata dijagonala\n");
  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
<invert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    $0xffffffff,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  not    %eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  not    %eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    -0x1c(%rbp),%eax
  mov    $0xffffffff,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  or     %edx,%eax
  not    %eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  or     -0x4(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int invert(int x, unsigned int p, unsigned int n) {
  int a;
  int i;
  a = x & ~(~(~0 << n) << (p - n + 1));
  i = (~((x >> (p - n + 1)) & ~(~0 << n)) & ~(~0 << n)) << (p - n + 1);
  x = a | i;
  return x;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<invert>:
  endbr64
  mov    $0xffffffff,%r8d
  mov    %edx,%ecx
  shl    %cl,%r8d
  add    $0x1,%esi
  sub    %edx,%esi
  mov    %esi,%ecx
  mov    %r8d,%eax
  not    %eax
  shl    %cl,%eax
  not    %eax
  and    %edi,%eax
  sar    %cl,%edi
  or     %r8d,%edi
  not    %edi
  shl    %cl,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
int invert(int x, unsigned int p, unsigned int n) {
  int a;
  int i;
  a = x & ~(~(~0 << n) << (p - n + 1));
  i = (~((x >> (p - n + 1)) & ~(~0 << n)) & ~(~0 << n)) << (p - n + 1);
  x = a | i;
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<invert>:
  endbr64
  mov    %edx,%ecx
  mov    $0xffffffff,%r8d
  add    $0x1,%esi
  shl    %cl,%r8d
  sub    %edx,%esi
  mov    %r8d,%eax
  mov    %esi,%ecx
  not    %eax
  shl    %cl,%eax
  not    %eax
  and    %edi,%eax
  sar    %cl,%edi
  or     %r8d,%edi
  not    %edi
  shl    %cl,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
int invert(int x, unsigned int p, unsigned int n) {
  int a;
  int i;
  a = x & ~(~(~0 << n) << (p - n + 1));
  i = (~((x >> (p - n + 1)) & ~(~0 << n)) & ~(~0 << n)) << (p - n + 1);
  x = a | i;
  return x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<invert>:
  endbr64
  mov    %edx,%ecx
  mov    $0xffffffff,%r8d
  add    $0x1,%esi
  shl    %cl,%r8d
  sub    %edx,%esi
  mov    %r8d,%eax
  mov    %esi,%ecx
  not    %eax
  shl    %cl,%eax
  not    %eax
  and    %edi,%eax
  sar    %cl,%edi
  or     %r8d,%edi
  not    %edi
  shl    %cl,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
int invert(int x, unsigned int p, unsigned int n) {
  int a;
  int i;
  a = x & ~(~(~0 << n) << (p - n + 1));
  i = (~((x >> (p - n + 1)) & ~(~0 << n)) & ~(~0 << n)) << (p - n + 1);
  x = a | i;
  return x;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<arc4_seed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <arc4_seed_urandom@plt>
  test   %rax,%rax
  jne    L6
  movl   $0x1,-0x4(%rbp)
L6:
  mov    $0x0,%eax
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  test   %rax,%rax
  jne    L1
  movl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x0,-0x4(%rbp)
  je     L21
  mov    $0x0,%eax
  jmp    L9
L21:
  mov    $0xffffffff,%eax
L9:
  leave
  ret
```

assistant: ```c
int arc4_seed(void) {
  int ok = 0;
  if (0 == arc4_seed_urandom())
    ok = 1;
  if (0 == arc4_seed_proc_sys_kernel_random_uuid())
    ok = 1;
  return ok ? 0 : -1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<arc4_seed>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <arc4_seed_urandom@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  test   %rax,%rax
  sete   %al
  test   %rbx,%rbx
  sete   %dl
  or     %edx,%eax
  movzbl %al,%eax
  sub    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int arc4_seed(void) {
  int ok = 0;
  if (0 == arc4_seed_urandom())
    ok = 1;
  if (0 == arc4_seed_proc_sys_kernel_random_uuid())
    ok = 1;
  return ok ? 0 : -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<arc4_seed>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <arc4_seed_urandom@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  je     L4
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  neg    %rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int arc4_seed(void) {
  int ok = 0;
  if (0 == arc4_seed_urandom())
    ok = 1;
  if (0 == arc4_seed_proc_sys_kernel_random_uuid())
    ok = 1;
  return ok ? 0 : -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<arc4_seed>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <arc4_seed_urandom@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  je     L4
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  neg    %rax
  sbb    %eax,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  call   <arc4_seed_proc_sys_kernel_random_uuid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int arc4_seed(void) {
  int ok = 0;
  if (0 == arc4_seed_urandom())
    ok = 1;
  if (0 == arc4_seed_proc_sys_kernel_random_uuid())
    ok = 1;
  return ok ? 0 : -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<curl_slist_append>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <check_expected@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <check_expected@plt>
  mov    $0x0,%eax
  call   <mock@plt>
  leave
  ret
```

assistant: ```c
struct curl_slist *curl_slist_append(struct curl_slist *list, const char *string) {
  check_expected(list);
  check_expected(string);
  return (struct curl_slist *)mock();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<curl_slist_append>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <check_expected@plt>
  mov    %rbx,%rdi
  call   <check_expected@plt>
  mov    $0x0,%eax
  call   <mock@plt>
  pop    %rbx
  ret
```

assistant: ```c
struct curl_slist *curl_slist_append(struct curl_slist *list, const char *string) {
  check_expected(list);
  check_expected(string);
  return (struct curl_slist *)mock();
}
```
user: What is the c source code of the assembly code below?

```asm
<curl_slist_append>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <check_expected@plt>
  mov    %rbp,%rdi
  call   <check_expected@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <mock@plt>
```

assistant: ```c
struct curl_slist *curl_slist_append(struct curl_slist *list, const char *string) {
  check_expected(list);
  check_expected(string);
  return (struct curl_slist *)mock();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<curl_slist_append>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <check_expected@plt>
  mov    %rbp,%rdi
  call   <check_expected@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <mock@plt>
```

assistant: ```c
struct curl_slist *curl_slist_append(struct curl_slist *list, const char *string) {
  check_expected(list);
  check_expected(string);
  return (struct curl_slist *)mock();
}
```
user: Translate the below assembly code into C programming language:

```asm
<insertionsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L43
L32:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L48
L0:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x8(%rbp)
L48:
  cmpl   $0x0,-0x8(%rbp)
  jle    L4
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jl     L0
L4:
  addl   $0x1,-0xc(%rbp)
L43:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L32
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void insertionsort(int a[], int length) {
  int i, j, tmp;
  for (i = 0; i < length; i++)
    for (j = i; j > 0 && a[j] < a[j - 1]; j--) {
      tmp = a[j];
      a[j] = a[j - 1];
      a[j - 1] = tmp;
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<insertionsort>:
  endbr64
  test   %esi,%esi
  jle    L18
  lea    0x4(%rdi),%r9
  mov    $0x0,%r8d
  jmp    L14
L49:
  add    $0x4,%r9
L14:
  add    $0x1,%r8d
  cmp    %r8d,%esi
  je     L18
  test   %r8d,%r8d
  jle    L49
  mov    %r9,%rax
L8:
  mov    (%rax),%ecx
  mov    -0x4(%rax),%edx
  cmp    %edx,%ecx
  jge    L49
  mov    %edx,(%rax)
  mov    %ecx,-0x4(%rax)
  sub    $0x4,%rax
  cmp    %rdi,%rax
  jne    L8
  jmp    L49
L18:
  ret
```

assistant: ```c
void insertionsort(int a[], int length) {
  int i, j, tmp;
  for (i = 0; i < length; i++)
    for (j = i; j > 0 && a[j] < a[j - 1]; j--) {
      tmp = a[j];
      a[j] = a[j - 1];
      a[j - 1] = tmp;
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<insertionsort>:
  endbr64
  test   %esi,%esi
  jle    L20
  xor    %r8d,%r8d
  lea    0x4(%rdi),%r9
  add    $0x1,%r8d
  cmp    %r8d,%esi
  je     L20
  nopl   0x0(%rax,%rax,1)
L31:
  mov    %r9,%rax
  jmp    L48
  nopl   (%rax)
L8:
  mov    %edx,(%rax)
  sub    $0x4,%rax
  mov    %ecx,(%rax)
  cmp    %rax,%rdi
  je     L6
L48:
  mov    (%rax),%ecx
  mov    -0x4(%rax),%edx
  cmp    %edx,%ecx
  jl     L8
L6:
  add    $0x1,%r8d
  add    $0x4,%r9
  cmp    %r8d,%esi
  jne    L31
L20:
  ret
```

assistant: ```c
void insertionsort(int a[], int length) {
  int i, j, tmp;
  for (i = 0; i < length; i++)
    for (j = i; j > 0 && a[j] < a[j - 1]; j--) {
      tmp = a[j];
      a[j] = a[j - 1];
      a[j - 1] = tmp;
    }
}
```