ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test_append_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  lea    -0xc000(%rsp),%r11
L11:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L11
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc020(%rbp),%rax
  lea    D58(%rip),%rdx
  lea    D61(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0xc020(%rbp),%rax
  lea    D23(%rip),%rdx
  lea    D61(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0x8020(%rbp),%rcx
  lea    -0xc020(%rbp),%rdx
  lea    -0x4020(%rbp),%rax
  lea    D32(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    -0xc020(%rbp),%rdx
  lea    -0x8020(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <append_string@plt>
  lea    -0xc020(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  lea    -0x4020(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x0,-0xc024(%rbp)
  jmp    L22
L10:
  mov    -0xc024(%rbp),%eax
  cltq
  movzbl -0xc020(%rbp,%rax,1),%edx
  mov    -0xc024(%rbp),%eax
  cltq
  movzbl -0x4020(%rbp,%rax,1),%eax
  cmp    %al,%dl
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  addl   $0x1,-0xc024(%rbp)
L22:
  lea    -0xc020(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc024(%rbp)
  jl     L10
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
data_type: string
name: disassemble_data
arguments: data_label: D58
data_type: string
name: disassemble_data
arguments: data_label: D61
data_type: string
name: disassemble_data
arguments: data_label: D23
data_type: string
name: disassemble_data
arguments: data_label: D32
data_type: string
name: disassemble_data
arguments: data_label: D1
data_type: string

tool: D9: test_append_string...

tool: D58: string1

tool: D61: %s

tool: D23: string2

tool: D32: %s
%s

tool: D1: Ok


assistant: ```c
void test_append_string() {
  char string1[(4096 * 4)];
  char string2[(4096 * 4)];
  char expected[(4096 * 4)];
  int i;
  printf("test_append_string...");
  sprintf(string1, "%s", "string1");
  sprintf(string1, "%s", "string2");
  sprintf(expected, "%s\n%s", string1, string2);
  append_string(string2, string1);
  assert(strlen(string1) == strlen(expected));
  for (i = 0; i < strlen(string1); i++) {
    assert(string1[i] == expected[i]);
  }
  printf("Ok\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_append_string>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc000(%rsp),%r11
L27:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L27
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc008(%rsp)
  xor    %eax,%eax
  lea    D55(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  lea    D27(%rip),%rdx
  lea    D12(%rip),%rbp
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    D48(%rip),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  lea    0x4000(%rsp),%r12
  lea    0x8000(%rsp),%rbp
  mov    %r12,%rcx
  mov    %rbx,%rdx
  lea    D35(%rip),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <append_string@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %eax,%ebx
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%ebx
  mov    %rsp,%rbp
  jmp    L11
L10:
  movzbl 0x8000(%rsp,%rbx,1),%eax
  cmp    %al,0x0(%rbp,%rbx,1)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x1,%rbx
L11:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L10
  lea    D50(%rip),%rdi
  call   <printf@plt>
  mov    0xc008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0xc010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D55
type_guessed: string
name: read_data
arguments: data_label: D27
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string
name: read_data
arguments: data_label: D48
type_guessed: string
name: read_data
arguments: data_label: D35
type_guessed: string
name: read_data
arguments: data_label: D50
type_guessed: string

tool: D55: test_append_string...

tool: D27: string1

tool: D12: %s

tool: D48: string2

tool: D35: %s
%s

tool: D50: Ok


assistant: ```c
void test_append_string() {
  char string1[(4096 * 4)];
  char string2[(4096 * 4)];
  char expected[(4096 * 4)];
  int i;
  printf("test_append_string...");
  sprintf(string1, "%s", "string1");
  sprintf(string1, "%s", "string2");
  sprintf(expected, "%s\n%s", string1, string2);
  append_string(string2, string1);
  assert(strlen(string1) == strlen(expected));
  for (i = 0; i < strlen(string1); i++) {
    assert(string1[i] == expected[i]);
  }
  printf("Ok\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_append_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc000(%rsp),%r11
L6:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L6
  sub    $0x18,%rsp
  lea    D15(%rip),%r12
  lea    D49(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xc008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x4000(%rsp),%r13
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D35(%rip),%rdx
  call   <sprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D51(%rip),%rdx
  lea    0x8000(%rsp),%r12
  call   <sprintf@plt>
  mov    %r13,%rcx
  mov    %rbp,%rdx
  mov    %r12,%rdi
  lea    D7(%rip),%rsi
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <append_string@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  xor    %edi,%edi
  cmp    %eax,%ebx
  sete   %dil
  xor    %ebx,%ebx
  call   <assert@plt>
  jmp    L21
  nopl   (%rax)
L22:
  movzbl (%r12,%rbx,1),%eax
  xor    %edi,%edi
  cmp    %al,0x0(%rbp,%rbx,1)
  sete   %dil
  add    $0x1,%rbx
  call   <assert@plt>
L21:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L22
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0xc008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0xc018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D15
data_type: string
name: retrieve_data
arguments: tag: D49
data_type: string
name: retrieve_data
arguments: tag: D35
data_type: string
name: retrieve_data
arguments: tag: D51
data_type: string
name: retrieve_data
arguments: tag: D7
data_type: string
name: retrieve_data
arguments: tag: D24
data_type: string

tool: D15: %s

tool: D49: test_append_string...

tool: D35: string1

tool: D51: string2

tool: D7: %s
%s

tool: D24: Ok


assistant: ```c
void test_append_string() {
  char string1[(4096 * 4)];
  char string2[(4096 * 4)];
  char expected[(4096 * 4)];
  int i;
  printf("test_append_string...");
  sprintf(string1, "%s", "string1");
  sprintf(string1, "%s", "string2");
  sprintf(expected, "%s\n%s", string1, string2);
  append_string(string2, string1);
  assert(strlen(string1) == strlen(expected));
  for (i = 0; i < strlen(string1); i++) {
    assert(string1[i] == expected[i]);
  }
  printf("Ok\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_append_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xc000(%rsp),%r11
L20:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L20
  sub    $0x18,%rsp
  lea    D14(%rip),%r12
  lea    D45(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0xc008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x4000(%rsp),%r13
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D38(%rip),%rdx
  call   <sprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    D59(%rip),%rdx
  lea    0x8000(%rsp),%r12
  call   <sprintf@plt>
  mov    %r13,%rcx
  mov    %rbp,%rdx
  mov    %r12,%rdi
  lea    D4(%rip),%rsi
  xor    %eax,%eax
  call   <sprintf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <append_string@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  xor    %edi,%edi
  cmp    %eax,%ebx
  sete   %dil
  xor    %ebx,%ebx
  call   <assert@plt>
  jmp    L36
  nopl   (%rax)
L4:
  movzbl (%r12,%rbx,1),%eax
  xor    %edi,%edi
  cmp    %al,0x0(%rbp,%rbx,1)
  sete   %dil
  add    $0x1,%rbx
  call   <assert@plt>
L36:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L4
  lea    D57(%rip),%rdi
  call   <printf@plt>
  mov    0xc008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0xc018,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D14
type_infered: string
name: unpack_data
arguments: data_tag: D45
type_infered: string
name: unpack_data
arguments: data_tag: D38
type_infered: string
name: unpack_data
arguments: data_tag: D59
type_infered: string
name: unpack_data
arguments: data_tag: D4
type_infered: string
name: unpack_data
arguments: data_tag: D57
type_infered: string

tool: D14: %s

tool: D45: test_append_string...

tool: D38: string1

tool: D59: string2

tool: D4: %s
%s

tool: D57: Ok


assistant: ```c
void test_append_string() {
  char string1[(4096 * 4)];
  char string2[(4096 * 4)];
  char expected[(4096 * 4)];
  int i;
  printf("test_append_string...");
  sprintf(string1, "%s", "string1");
  sprintf(string1, "%s", "string2");
  sprintf(expected, "%s\n%s", string1, string2);
  append_string(string2, string1);
  assert(strlen(string1) == strlen(expected));
  for (i = 0; i < strlen(string1); i++) {
    assert(string1[i] == expected[i]);
  }
  printf("Ok\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <_uname@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L7
L13:
  mov    $0x2,%eax
L7:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret = 0;
  (void)argc;
  ret += _uname(argv[0]);
  return (ret == 0) ? 0 : 2;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <_uname@plt>
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  add    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret = 0;
  (void)argc;
  ret += _uname(argv[0]);
  return (ret == 0) ? 0 : 2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <_uname@plt>
  test   %eax,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  add    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret = 0;
  (void)argc;
  ret += _uname(argv[0]);
  return (ret == 0) ? 0 : 2;
}
```
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
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  call   <_uname@plt>
  test   %eax,%eax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  add    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret = 0;
  (void)argc;
  ret += _uname(argv[0]);
  return (ret == 0) ? 0 : 2;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<simple_add_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L33
L13:
  movl   $0x0,-0x4(%rbp)
  jmp    L37
L10:
  mov    -0x8(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   -0x28(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  imul   0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  imul   0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L37:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L10
  addl   $0x1,-0x8(%rbp)
L33:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L13
  nop
  pop    %rbp
  ret
```

assistant: ```c
void simple_add_float(int m, int n, float *a, int a_row_stride, int a_col_stride, float *b, int b_row_stride, int b_col_stride) {
  int i, j;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      b[(i)*b_row_stride + (j)*b_col_stride] += a[(i)*a_row_stride + (j)*a_col_stride];
    }
  }
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<simple_add_float>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x30(%rsp),%r14d
  test   %edi,%edi
  jle    L23
  mov    %edi,%ebx
  mov    %rdx,%rbp
  mov    %ecx,%r12d
  mov    %r9,%r13
  movslq 0x38(%rsp),%r10
  shl    $0x2,%r10
  movslq %r8d,%r8
  lea    0x0(,%r8,4),%rdi
  mov    $0x0,%r9d
  mov    $0x0,%r11d
  mov    $0x0,%r8d
  jmp    L11
L37:
  movslq %r11d,%rax
  lea    0x0(%r13,%rax,4),%rax
  movslq %r9d,%rdx
  lea    0x0(%rbp,%rdx,4),%rcx
  mov    $0x0,%edx
L0:
  movss  (%rax),%xmm0
  addss  (%rcx),%xmm0
  movss  %xmm0,(%rax)
  add    $0x1,%edx
  add    %r10,%rax
  add    %rdi,%rcx
  cmp    %edx,%esi
  jne    L0
L49:
  add    $0x1,%r8d
  add    %r14d,%r11d
  add    %r12d,%r9d
  cmp    %r8d,%ebx
  je     L23
L11:
  test   %esi,%esi
  jg     L37
  jmp    L49
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void simple_add_float(int m, int n, float *a, int a_row_stride, int a_col_stride, float *b, int b_row_stride, int b_col_stride) {
  int i, j;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      b[(i)*b_row_stride + (j)*b_col_stride] += a[(i)*a_row_stride + (j)*a_col_stride];
    }
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<simple_add_float>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x30(%rsp),%r14d
  test   %edi,%edi
  jle    L7
  movslq 0x38(%rsp),%r10
  movslq %r8d,%r8
  mov    %edi,%ebx
  mov    %r9,%r13
  mov    %rdx,%rbp
  mov    %ecx,%r12d
  xor    %r9d,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r8,4),%rdi
  shl    $0x2,%r10
  xor    %r8d,%r8d
  nopl   (%rax)
L36:
  test   %esi,%esi
  jle    L10
  movslq %r9d,%rdx
  movslq %r11d,%rax
  lea    0x0(%rbp,%rdx,4),%rcx
  lea    0x0(%r13,%rax,4),%rax
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L24:
  movss  (%rax),%xmm0
  addss  (%rcx),%xmm0
  add    $0x1,%edx
  add    %rdi,%rcx
  movss  %xmm0,(%rax)
  add    %r10,%rax
  cmp    %edx,%esi
  jne    L24
L10:
  add    $0x1,%r8d
  add    %r14d,%r11d
  add    %r12d,%r9d
  cmp    %r8d,%ebx
  jne    L36
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void simple_add_float(int m, int n, float *a, int a_row_stride, int a_col_stride, float *b, int b_row_stride, int b_col_stride) {
  int i, j;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      b[(i)*b_row_stride + (j)*b_col_stride] += a[(i)*a_row_stride + (j)*a_col_stride];
    }
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<partition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xa(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L11
L29:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0xa(%rbp)
  jle    L18
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
L18:
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L29
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int partition(char *array, int low, int high) {
  char pivot, temp;
  int pindex, i;
  pivot = array[high];
  pindex = low;
  for (i = low; i < high; ++i) {
    if (array[i] < pivot) {
      temp = array[i];
      array[i] = array[pindex];
      array[pindex] = temp;
      ++pindex;
    }
  }
  temp = array[high];
  array[high] = array[pindex];
  array[pindex] = temp;
  return (pindex);
}
```
user: What is the c source code of the assembly code below:

```asm
<partition>:
  endbr64
  mov    %esi,%eax
  movslq %edx,%r11
  add    %rdi,%r11
  movzbl (%r11),%r9d
  cmp    %esi,%edx
  jle    L20
  movslq %esi,%rsi
  lea    (%rdi,%rsi,1),%rcx
  lea    0x1(%rdi,%rsi,1),%r8
  sub    %eax,%edx
  lea    -0x1(%rdx),%edx
  add    %rdx,%r8
  jmp    L30
L39:
  add    $0x1,%rcx
  cmp    %r8,%rcx
  je     L20
L30:
  movzbl (%rcx),%esi
  cmp    %r9b,%sil
  jge    L39
  movslq %eax,%rdx
  add    %rdi,%rdx
  movzbl (%rdx),%r10d
  mov    %r10b,(%rcx)
  mov    %sil,(%rdx)
  add    $0x1,%eax
  jmp    L39
L20:
  movzbl (%r11),%edx
  movslq %eax,%rcx
  add    %rcx,%rdi
  movzbl (%rdi),%ecx
  mov    %cl,(%r11)
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
int partition(char *array, int low, int high) {
  char pivot, temp;
  int pindex, i;
  pivot = array[high];
  pindex = low;
  for (i = low; i < high; ++i) {
    if (array[i] < pivot) {
      temp = array[i];
      array[i] = array[pindex];
      array[pindex] = temp;
      ++pindex;
    }
  }
  temp = array[high];
  array[high] = array[pindex];
  array[pindex] = temp;
  return (pindex);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<partition>:
  endbr64
  movslq %edx,%r11
  mov    %esi,%eax
  add    %rdi,%r11
  movzbl (%r11),%r8d
  cmp    %esi,%edx
  jle    L1
  movslq %esi,%rsi
  sub    %eax,%edx
  lea    0x1(%rdi,%rsi,1),%r9
  sub    $0x1,%edx
  lea    (%rdi,%rsi,1),%rcx
  add    %rdx,%r9
  nopl   0x0(%rax,%rax,1)
L22:
  movzbl (%rcx),%esi
  cmp    %r8b,%sil
  jge    L29
  movslq %eax,%rdx
  add    $0x1,%eax
  add    %rdi,%rdx
  movzbl (%rdx),%r10d
  mov    %r10b,(%rcx)
  mov    %sil,(%rdx)
L29:
  add    $0x1,%rcx
  cmp    %r9,%rcx
  jne    L22
  movzbl (%r11),%r8d
L1:
  movslq %eax,%rdx
  add    %rdx,%rdi
  movzbl (%rdi),%edx
  mov    %dl,(%r11)
  mov    %r8b,(%rdi)
  ret
```

assistant: ```c
int partition(char *array, int low, int high) {
  char pivot, temp;
  int pindex, i;
  pivot = array[high];
  pindex = low;
  for (i = low; i < high; ++i) {
    if (array[i] < pivot) {
      temp = array[i];
      array[i] = array[pindex];
      array[pindex] = temp;
      ++pindex;
    }
  }
  temp = array[high];
  array[high] = array[pindex];
  array[pindex] = temp;
  return (pindex);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<partition>:
  endbr64
  movslq %edx,%r11
  mov    %esi,%eax
  add    %rdi,%r11
  movzbl (%r11),%r8d
  cmp    %esi,%edx
  jle    L26
  movslq %esi,%rsi
  sub    %eax,%edx
  lea    0x1(%rdi,%rsi,1),%r9
  sub    $0x1,%edx
  lea    (%rdi,%rsi,1),%rcx
  add    %rdx,%r9
  nopl   0x0(%rax,%rax,1)
L6:
  movzbl (%rcx),%esi
  cmp    %r8b,%sil
  jge    L0
  movslq %eax,%rdx
  add    $0x1,%eax
  add    %rdi,%rdx
  movzbl (%rdx),%r10d
  mov    %r10b,(%rcx)
  mov    %sil,(%rdx)
L0:
  add    $0x1,%rcx
  cmp    %r9,%rcx
  jne    L6
  movzbl (%r11),%r8d
L26:
  movslq %eax,%rdx
  add    %rdx,%rdi
  movzbl (%rdi),%edx
  mov    %dl,(%r11)
  mov    %r8b,(%rdi)
  ret
```

assistant: ```c
int partition(char *array, int low, int high) {
  char pivot, temp;
  int pindex, i;
  pivot = array[high];
  pindex = low;
  for (i = low; i < high; ++i) {
    if (array[i] < pivot) {
      temp = array[i];
      array[i] = array[pindex];
      array[pindex] = temp;
      ++pindex;
    }
  }
  temp = array[high];
  array[high] = array[pindex];
  array[pindex] = temp;
  return (pindex);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<checkAlive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L36
L14:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L38
  mov    $0x1,%eax
  jmp    L32
L38:
  addl   $0x1,-0x4(%rbp)
L36:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L14
  mov    $0x0,%eax
L32:
  pop    %rbp
  ret
```

assistant: ```c
int checkAlive(int *current, int length) {
  for (int i = 0; i < length - 1; i++)
    if (*(current + i) != 0)
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<checkAlive>:
  endbr64
  cmp    $0x1,%esi
  jle    L32
  mov    %rdi,%rax
  lea    -0x2(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
L40:
  mov    (%rax),%edx
  test   %edx,%edx
  jne    L35
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L40
  jmp    L44
L32:
  mov    $0x0,%edx
  jmp    L44
L35:
  mov    $0x1,%edx
L44:
  mov    %edx,%eax
  ret
```

assistant: ```c
int checkAlive(int *current, int length) {
  for (int i = 0; i < length - 1; i++)
    if (*(current + i) != 0)
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<checkAlive>:
  endbr64
  cmp    $0x1,%esi
  jle    L4
  lea    -0x2(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L12:
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  je     L37
L39:
  mov    (%rdi),%eax
  test   %eax,%eax
  je     L12
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L37:
  ret
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int checkAlive(int *current, int length) {
  for (int i = 0; i < length - 1; i++)
    if (*(current + i) != 0)
      return 1;
  return 0;
}
```