ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<calc_step_size>:
  endbr64
  push   %rbx
  movapd %xmm0,%xmm1
  sub    $0x10,%rsp
  test   %rdi,%rdi
  js     L27
  pxor   %xmm0,%xmm0
  cvtsi2sd %rdi,%xmm0
L12:
  divsd  D17(%rip),%xmm0
  divsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <log@plt>
  mov    %eax,%ebx
  mov    D47(%rip),%rax
  movq   %rax,%xmm0
  call   <log@plt>
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    %eax,%edi
  call   <floor@plt>
  mov    %eax,%edi
  mov    D47(%rip),%rax
  movq   %rax,%xmm0
  call   <pow@plt>
  movsd  0x8(%rsp),%xmm1
  comisd %xmm1,%xmm0
  jae    L21
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  comisd %xmm1,%xmm2
  jae    L49
  movsd  D6(%rip),%xmm2
  mulsd  %xmm0,%xmm2
  comisd %xmm1,%xmm2
  jae    L49
  mulsd  D47(%rip),%xmm0
L21:
  add    $0x10,%rsp
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %rdi,%rax
  and    $0x1,%edi
  pxor   %xmm0,%xmm0
  shr    %rax
  or     %rdi,%rax
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm0
  jmp    L12
  xchg   %ax,%ax
L49:
  add    $0x10,%rsp
  movapd %xmm2,%xmm0
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D17
data_type: f64
name: retrieve_data
arguments: tag: D47
data_type: f64
name: retrieve_data
arguments: tag: D6
data_type: byte[8]

tool: D17: 14

tool: D47: 10

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x14,0x40]}

assistant: ```c
double calc_step_size(double range, size_t graph_height) {
  const unsigned int factors[] = {1, 2, 5};
  unsigned int i;
  double min_step = range / (graph_height / 14.0);
  double mag = pow(10.0, floor(log(min_step) / log(10.0)));
  for (i = 0; i < 3; ++i)
    if (mag * factors[i] >= min_step)
      return mag * factors[i];
  return mag * 10;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L1
  mov    $0x1,%eax
  jmp    L9
L1:
  mov    -0x10(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
L9:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc != 3)
    return 1;
  return strcmp(argv[1], argv[2]);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x3,%edi
  jne    L7
  sub    $0x8,%rsp
  mov    0x10(%rsi),%rax
  mov    0x8(%rsi),%rdi
  mov    %rax,%rsi
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
L7:
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc != 3)
    return 1;
  return strcmp(argv[1], argv[2]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  cmp    $0x3,%edi
  jne    L3
  mov    0x10(%rsi),%r8
  mov    0x8(%rsi),%rdi
  mov    %r8,%rsi
  jmp    1050 <strcmp@plt>
L3:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc != 3)
    return 1;
  return strcmp(argv[1], argv[2]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  cmp    $0x3,%edi
  jne    L6
  mov    0x10(%rsi),%r8
  mov    0x8(%rsi),%rdi
  mov    %r8,%rsi
  jmp    1050 <strcmp@plt>
L6:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc != 3)
    return 1;
  return strcmp(argv[1], argv[2]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<intCmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int intCmp(const void *s1, const void *s2) {
  int *ip1 = (int *)s1;
  int *ip2 = (int *)s2;
  return *ip1 - *ip2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<intCmp>:
  endbr64
  mov    (%rdi),%eax
  sub    (%rsi),%eax
  ret
```

assistant: ```c
int intCmp(const void *s1, const void *s2) {
  int *ip1 = (int *)s1;
  int *ip2 = (int *)s2;
  return *ip1 - *ip2;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<intCmp>:
  endbr64
  mov    (%rdi),%eax
  sub    (%rsi),%eax
  ret
```

assistant: ```c
int intCmp(const void *s1, const void *s2) {
  int *ip1 = (int *)s1;
  int *ip2 = (int *)s2;
  return *ip1 - *ip2;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<yy_1_BlockContent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void yy_1_BlockContent(char *yytext, int yyleng) {
  ;
  ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<yy_1_BlockContent>:
  endbr64
  ret
```

assistant: ```c
void yy_1_BlockContent(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<yy_1_BlockContent>:
  endbr64
  ret
```

assistant: ```c
void yy_1_BlockContent(char *yytext, int yyleng) {
  ;
  ;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<yy_1_BlockContent>:
  endbr64
  ret
```

assistant: ```c
void yy_1_BlockContent(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rk31sdk_get_sdio_wifi_voltage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0xbb8,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int rk31sdk_get_sdio_wifi_voltage(void) {
  int voltage;
  voltage = 3000;
  return voltage;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rk31sdk_get_sdio_wifi_voltage>:
  endbr64
  mov    $0xbb8,%eax
  ret
```

assistant: ```c
int rk31sdk_get_sdio_wifi_voltage(void) {
  int voltage;
  voltage = 3000;
  return voltage;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rk31sdk_get_sdio_wifi_voltage>:
  endbr64
  mov    $0xbb8,%eax
  ret
```

assistant: ```c
int rk31sdk_get_sdio_wifi_voltage(void) {
  int voltage;
  voltage = 3000;
  return voltage;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rk31sdk_get_sdio_wifi_voltage>:
  endbr64
  mov    $0xbb8,%eax
  ret
```

assistant: ```c
int rk31sdk_get_sdio_wifi_voltage(void) {
  int voltage;
  voltage = 3000;
  return voltage;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<refLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D9(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L16
  addq   $0x4,-0x8(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  lea    D15(%rip),%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  mov    $0x1,%eax
  jmp    L9
L16:
  mov    $0x0,%eax
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_guessed: string
name: read_data
arguments: data_tag: D15
type_guessed: string

tool: D9: REF:

tool: D15: <div>REF: <a href="%s">%s</a></div>


assistant: ```c
int refLine(char *line, FILE *out) {
  char *ref;
  ref = strstr(line, "REF:");
  if (ref) {
    ref += 4;
    fprintf(out, "<div>REF: <a href=\"%s\">%s</a></div>\n", ref, ref);
    return 1;
  } else {
    return 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<refLine>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  lea    D5(%rip),%rsi
  call   <strstr@plt>
  mov    $0x0,%edx
  test   %rax,%rax
  je     L6
  lea    0x4(%rax),%rdx
  mov    %rdx,%rcx
  lea    D14(%rip),%rsi
  mov    %rbx,%rdi
  call   <fprintf@plt>
  mov    $0x1,%edx
L6:
  mov    %edx,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string
name: retrieve_data
arguments: data_tag: D14
data_type: string

tool: D5: REF:

tool: D14: <div>REF: <a href="%s">%s</a></div>


assistant: ```c
int refLine(char *line, FILE *out) {
  char *ref;
  ref = strstr(line, "REF:");
  if (ref) {
    ref += 4;
    fprintf(out, "<div>REF: <a href=\"%s\">%s</a></div>\n", ref, ref);
    return 1;
  } else {
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<refLine>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    D6(%rip),%rsi
  call   <strstr@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L1
  lea    0x4(%rax),%rdx
  lea    D16(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rdx,%rcx
  call   <fprintf@plt>
  mov    $0x1,%r8d
L1:
  mov    %r8d,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string
name: unpack_data
arguments: data_tag: D16
datatype: string

tool: D6: REF:

tool: D16: <div>REF: <a href="%s">%s</a></div>


assistant: ```c
int refLine(char *line, FILE *out) {
  char *ref;
  ref = strstr(line, "REF:");
  if (ref) {
    ref += 4;
    fprintf(out, "<div>REF: <a href=\"%s\">%s</a></div>\n", ref, ref);
    return 1;
  } else {
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<refLine>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  lea    D9(%rip),%rsi
  call   <strstr@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L4
  lea    0x4(%rax),%rdx
  lea    D15(%rip),%rsi
  mov    %rbp,%rdi
  mov    %rdx,%rcx
  call   <fprintf@plt>
  mov    $0x1,%r8d
L4:
  mov    %r8d,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string
name: disassemble_data
arguments: label: D15
type_guessed: string

tool: D9: REF:

tool: D15: <div>REF: <a href="%s">%s</a></div>


assistant: ```c
int refLine(char *line, FILE *out) {
  char *ref;
  ref = strstr(line, "REF:");
  if (ref) {
    ref += 4;
    fprintf(out, "<div>REF: <a href=\"%s\">%s</a></div>\n", ref, ref);
    return 1;
  } else {
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<array_to_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    $0x8,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L38
L18:
  mov    -0x20(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L39
L26:
  mov    -0x10(%rbp),%eax
  imul   -0x20(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L39:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L26
  addl   $0x1,-0x10(%rbp)
L38:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int **array_to_matrix(int *m, int rows, int cols) {
  int i, j;
  int **r;
  r = (int **)calloc(rows, sizeof(int *));
  for (i = 0; i < rows; i++) {
    r[i] = (int *)calloc(cols, sizeof(int));
    for (j = 0; j < cols; j++)
      r[i][j] = m[i * cols + j];
  }
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<array_to_matrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %esi,%ebx
  mov    %edx,%r12d
  mov    $0x8,%esi
  mov    %ebx,%edi
  call   <calloc@plt>
  mov    %rax,0x8(%rsp)
  test   %ebx,%ebx
  jle    L24
  mov    %rax,%rbp
  mov    %ebx,%ebx
  lea    (%rax,%rbx,8),%r15
  lea    -0x1(%r12),%eax
  lea    0x4(,%rax,4),%rbx
  mov    $0x0,%r13d
L26:
  mov    $0x4,%esi
  mov    %r12d,%edi
  call   <calloc@plt>
  mov    %rbp,%rsi
  mov    %rax,0x0(%rbp)
  test   %r12d,%r12d
  jle    L7
  movslq %r13d,%rax
  lea    (%r14,%rax,4),%rdi
  mov    $0x0,%eax
L21:
  mov    (%rdi,%rax,1),%ecx
  mov    (%rsi),%rdx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L21
L7:
  add    $0x8,%rbp
  add    %r12d,%r13d
  cmp    %r15,%rbp
  jne    L26
L24:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **array_to_matrix(int *m, int rows, int cols) {
  int i, j;
  int **r;
  r = (int **)calloc(rows, sizeof(int *));
  for (i = 0; i < rows; i++) {
    r[i] = (int *)calloc(cols, sizeof(int));
    for (j = 0; j < cols; j++)
      r[i][j] = m[i * cols + j];
  }
  return r;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<array_to_matrix>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  mov    $0x8,%esi
  mov    %ebx,%edi
  sub    $0x18,%rsp
  call   <calloc@plt>
  mov    %rax,0x8(%rsp)
  test   %ebx,%ebx
  jle    L34
  lea    (%rax,%rbx,8),%r12
  lea    -0x1(%r15),%ebx
  mov    %rax,%r14
  xor    %r13d,%r13d
  add    $0x1,%rbx
  nopl   0x0(%rax)
L27:
  mov    $0x4,%esi
  mov    %r15d,%edi
  call   <calloc@plt>
  mov    %rax,(%r14)
  test   %r15d,%r15d
  jle    L21
  movslq %r13d,%rdx
  lea    (%rax,%rbx,4),%rsi
  lea    0x0(%rbp,%rdx,4),%rcx
  sub    %rax,%rcx
  nopl   0x0(%rax)
L1:
  mov    (%rcx,%rax,1),%edx
  add    $0x4,%rax
  mov    %edx,-0x4(%rax)
  cmp    %rsi,%rax
  jne    L1
L21:
  add    $0x8,%r14
  add    %r15d,%r13d
  cmp    %r12,%r14
  jne    L27
L34:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int **array_to_matrix(int *m, int rows, int cols) {
  int i, j;
  int **r;
  r = (int **)calloc(rows, sizeof(int *));
  for (i = 0; i < rows; i++) {
    r[i] = (int *)calloc(cols, sizeof(int));
    for (j = 0; j < cols; j++)
      r[i][j] = m[i * cols + j];
  }
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<array_to_matrix>:
  endbr64
  push   %r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    $0x8,%esi
  mov    %ebx,%edi
  sub    $0x48,%rsp
  call   <calloc@plt>
  mov    %rax,0x38(%rsp)
  test   %ebx,%ebx
  jle    L18
  mov    %rax,%r13
  lea    (%rax,%rbx,8),%rax
  mov    %r14d,%ebx
  xor    %r15d,%r15d
  mov    %rax,(%rsp)
  mov    %r14d,%eax
  and    $0xfffffffc,%ebx
  shr    $0x2,%eax
  lea    -0x1(%rax),%ebp
  movslq %ebx,%rax
  shl    $0x2,%rax
  add    $0x1,%rbp
  mov    %rax,0x10(%rsp)
  lea    0x1(%rbx),%eax
  shl    $0x4,%rbp
  mov    %eax,0x20(%rsp)
  cltq
  shl    $0x2,%rax
  mov    %rax,0x18(%rsp)
  lea    0x2(%rbx),%eax
  mov    %eax,0x24(%rsp)
  cltq
  shl    $0x2,%rax
  mov    %rax,0x28(%rsp)
  lea    -0x1(%r14),%eax
  mov    %eax,0xc(%rsp)
  lea    0x4(,%rax,4),%rax
  mov    %rax,0x30(%rsp)
  mov    %rbp,%rax
  mov    %r14d,%ebp
  mov    %r13,%r14
  mov    %r15d,%r13d
  mov    %rax,%r15
  nopl   0x0(%rax)
L52:
  mov    $0x4,%esi
  mov    %ebp,%edi
  call   <calloc@plt>
  mov    %rax,(%r14)
  mov    %rax,%rdx
  test   %ebp,%ebp
  jle    L28
  movslq %r13d,%rax
  lea    0x4(,%rax,4),%rsi
  mov    %rdx,%rax
  lea    (%r12,%rsi,1),%r10
  sub    %r10,%rax
  cmp    $0x8,%rax
  jbe    L106
  cmpl   $0x2,0xc(%rsp)
  jbe    L106
  mov    %r12,%r11
  lea    (%r15,%rdx,1),%r10
  mov    %rdx,%rax
  sub    %rdx,%r11
  lea    -0x4(%r11,%rsi,1),%rsi
  cs nopw 0x0(%rax,%rax,1)
L82:
  movdqu (%rax,%rsi,1),%xmm0
  add    $0x10,%rax
  movups %xmm0,-0x10(%rax)
  cmp    %r10,%rax
  jne    L82
  cmp    %ebp,%ebx
  je     L28
  lea    0x0(%r13,%rbx,1),%eax
  mov    0x10(%rsp),%rdi
  cltq
  mov    (%r12,%rax,4),%eax
  mov    %eax,(%rdx,%rdi,1)
  mov    0x20(%rsp),%eax
  cmp    %eax,%ebp
  jle    L28
  add    %r13d,%eax
  mov    0x18(%rsp),%rdi
  cltq
  mov    (%r12,%rax,4),%eax
  mov    %eax,(%rdx,%rdi,1)
  mov    0x24(%rsp),%eax
  cmp    %eax,%ebp
  jle    L28
  add    %r13d,%eax
  mov    0x28(%rsp),%rcx
  cltq
  mov    (%r12,%rax,4),%eax
  mov    %eax,(%rdx,%rcx,1)
L28:
  add    $0x8,%r14
  add    %ebp,%r13d
  cmp    (%rsp),%r14
  jne    L52
L18:
  mov    0x38(%rsp),%rax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L106:
  mov    0x30(%rsp),%rdi
  sub    $0x4,%rsi
  mov    %rdx,%rax
  sub    %rdx,%rsi
  lea    (%rdx,%rdi,1),%r10
  lea    (%r12,%rsi,1),%rdx
  nopw   0x0(%rax,%rax,1)
L47:
  mov    (%rdx,%rax,1),%ecx
  add    $0x4,%rax
  mov    %ecx,-0x4(%rax)
  cmp    %r10,%rax
  jne    L47
  jmp    L28
```

assistant: ```c
int **array_to_matrix(int *m, int rows, int cols) {
  int i, j;
  int **r;
  r = (int **)calloc(rows, sizeof(int *));
  for (i = 0; i < rows; i++) {
    r[i] = (int *)calloc(cols, sizeof(int));
    for (j = 0; j < cols; j++)
      r[i][j] = m[i * cols + j];
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L36:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L36
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x3e8,%edi
  call   <srand@plt>
  movl   $0x0,-0x9c58(%rbp)
  jmp    L21
L8:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x418505d3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0xa,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xfa1,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x9c58(%rbp),%eax
  cltq
  mov    %edx,-0x9c50(%rbp,%rax,4)
  addl   $0x1,-0x9c58(%rbp)
L21:
  cmpl   $0x1387,-0x9c58(%rbp)
  jle    L8
  lea    -0x9c50(%rbp),%rcx
  lea    -0x4e30(%rbp),%rax
  mov    $0x4e20,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x0,-0x9c54(%rbp)
  jmp    L0
L31:
  lea    -0x9c50(%rbp),%rax
  mov    $0x1388,%esi
  mov    %rax,%rdi
  call   <bubbleSort@plt>
  lea    -0x4e30(%rbp),%rcx
  lea    -0x9c50(%rbp),%rax
  mov    $0x4e20,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  addl   $0x1,-0x9c54(%rbp)
L0:
  cmpl   $0x12b,-0x9c54(%rbp)
  jle    L31
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L52
  call   <__stack_chk_fail@plt>
L52:
  leave
  ret
```

assistant: ```c
int main() {
  srand(1000);
  int mainObject[5000], backupObject[5000];
  for (int i = 0; i < 5000; i++) {
    mainObject[i] = rand() % 4001;
  }
  memcpy(backupObject, mainObject, sizeof(int) * 5000);
  for (int i = 0; i < 300; i++) {
    bubbleSort(mainObject, 5000);
    memcpy(mainObject, backupObject, sizeof(int) * 5000);
  }
  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
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L35:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L35
  sub    $0xc58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    $0x3e8,%edi
  call   <srand@plt>
  mov    %rsp,%rbx
  lea    0x4e20(%rsp),%rbp
L25:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x418505d3,%rdx,%rdx
  sar    $0x2a,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xfa1,%edx,%edx
  sub    %edx,%eax
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L25
  mov    %rsp,%rsi
  lea    0x4e20(%rsp),%rdi
  mov    $0x4e20,%edx
  call   <memcpy@plt>
  mov    $0x12c,%ebx
  mov    %rsp,%rbp
L29:
  mov    $0x1388,%esi
  mov    %rbp,%rdi
  call   <bubbleSort@plt>
  lea    0x4e20(%rsp),%rsi
  mov    $0x4e20,%edx
  mov    %rbp,%rdi
  call   <memcpy@plt>
  sub    $0x1,%ebx
  jne    L29
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x9c58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  srand(1000);
  int mainObject[5000], backupObject[5000];
  for (int i = 0; i < 5000; i++) {
    mainObject[i] = rand() % 4001;
  }
  memcpy(backupObject, mainObject, sizeof(int) * 5000);
  for (int i = 0; i < 300; i++) {
    bubbleSort(mainObject, 5000);
    memcpy(mainObject, backupObject, sizeof(int) * 5000);
  }
  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
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L34:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L34
  sub    $0xc50,%rsp
  mov    $0x3e8,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x4e20(%rsp),%rbp
  call   <srand@plt>
  mov    %rbx,%r12
  nopl   0x0(%rax,%rax,1)
L36:
  xor    %eax,%eax
  add    $0x4,%r12
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x418505d3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x2a,%rdx
  sub    %ecx,%edx
  imul   $0xfa1,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r12)
  cmp    %r12,%rbp
  jne    L36
  mov    $0x4e20,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    $0x12c,%r12d
  call   <memcpy@plt>
L26:
  mov    $0x1388,%esi
  mov    %rbx,%rdi
  call   <bubbleSort@plt>
  mov    $0x4e20,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  sub    $0x1,%r12d
  jne    L26
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x9c50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  srand(1000);
  int mainObject[5000], backupObject[5000];
  for (int i = 0; i < 5000; i++) {
    mainObject[i] = rand() % 4001;
  }
  memcpy(backupObject, mainObject, sizeof(int) * 5000);
  for (int i = 0; i < 300; i++) {
    bubbleSort(mainObject, 5000);
    memcpy(mainObject, backupObject, sizeof(int) * 5000);
  }
  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
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L29:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L29
  sub    $0xc50,%rsp
  mov    $0x3e8,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x4e20(%rsp),%rbp
  call   <srand@plt>
  mov    %rbx,%r12
  nopl   0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  add    $0x4,%r12
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x418505d3,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x2a,%rdx
  sub    %ecx,%edx
  imul   $0xfa1,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r12)
  cmp    %rbp,%r12
  jne    L18
  mov    $0x4e20,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    $0x12c,%r12d
  call   <memcpy@plt>
L13:
  mov    $0x1388,%esi
  mov    %rbx,%rdi
  call   <bubbleSort@plt>
  mov    $0x4e20,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  sub    $0x1,%r12d
  jne    L13
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x9c50,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  srand(1000);
  int mainObject[5000], backupObject[5000];
  for (int i = 0; i < 5000; i++) {
    mainObject[i] = rand() % 4001;
  }
  memcpy(backupObject, mainObject, sizeof(int) * 5000);
  for (int i = 0; i < 300; i++) {
    bubbleSort(mainObject, 5000);
    memcpy(mainObject, backupObject, sizeof(int) * 5000);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<TimerHalWaitForL3SClockBeingActive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
L4:
  mov    $0x44e00004,%eax
  mov    (%rax),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void TimerHalWaitForL3SClockBeingActive(void) {
  while (!((*((volatile unsigned int *)(((0x44E00000) + 0) + (0x4)))) & (0x00000008u)))
    ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<TimerHalWaitForL3SClockBeingActive>:
  endbr64
L2:
  mov    0x44e00004,%eax
  test   $0x8,%al
  je     L2
  ret
```

assistant: ```c
void TimerHalWaitForL3SClockBeingActive(void) {
  while (!((*((volatile unsigned int *)(((0x44E00000) + 0) + (0x4)))) & (0x00000008u)))
    ;
}
```