ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %r13,%rcx
  lea    D19(%rip),%rdi
  mov    %r14,%r8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  mov    0xc(%rsp),%r11d
  mov    0x10(%rsp),%ecx
  mov    0x14(%rsp),%r10d
  mov    %edi,%eax
  and    %r11d,%eax
  and    %ecx,%eax
  and    %r10d,%eax
  cmp    $0xffffffff,%eax
  je     L17
  lea    0x1(%r10),%edx
  mov    %edi,%r8d
  lea    0x5304(%r10),%r9d
  mov    %edx,%esi
  sub    %r11d,%r8d
  sub    %edi,%esi
  sub    %ecx,%edi
  jmp    L31
  cs nopw 0x0(%rax,%rax,1)
L52:
  add    $0x1,%edx
  add    $0x1,%esi
  cmp    %r9d,%edx
  jg     L30
L31:
  lea    (%r8,%rsi,1),%ecx
  movslq %ecx,%rax
  mov    %ecx,%r11d
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  add    %ecx,%eax
  sar    $0x4,%eax
  sub    %r11d,%eax
  lea    (%rdi,%rsi,1),%r11d
  imul   $0x1c,%eax,%eax
  mov    %r11d,%r15d
  sar    $0x1f,%r15d
  sub    %eax,%ecx
  movslq %r11d,%rax
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x23,%rax
  sub    %r15d,%eax
  mov    %eax,%r15d
  shl    $0x5,%r15d
  add    %r15d,%eax
  sub    %eax,%r11d
  movslq %esi,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  or     %r11d,%ecx
  mov    %esi,%r11d
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  add    %esi,%eax
  sar    $0x4,%eax
  sub    %r11d,%eax
  mov    %esi,%r11d
  imul   $0x17,%eax,%eax
  sub    %eax,%r11d
  or     %ecx,%r11d
  jne    L52
L30:
  mov    %ebx,%esi
  sub    %r10d,%edx
  lea    D10(%rip),%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  jmp    L37
L17:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D19
data_type: string
name: parse_data
arguments: label: D10
data_type: string

tool: D19: %d %d %d %d

tool: D10: Case %d: the next triple peak occurs in %d days.


assistant: ```c
int main() {
  int p;
  int e;
  int i;
  int d;
  int count = 1;
  int k;
  while (1) {
    scanf("%d %d %d %d", &p, &e, &i, &d);
    if (p == -1 && e == -1 && i == -1 && d == -1) {
      break;
    }
    for (k = d + 1; k <= 21252 + d; k++) {
      if ((k - p) % 23 == 0 && (k - e) % 28 == 0 && (k - i) % 33 == 0)
        break;
    }
    printf("Case %d: the next triple peak occurs in %d days.\n", count, k - d);
    count++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%r14
  lea    0x10(%rsp),%r13
  lea    0xc(%rsp),%r12
  lea    0x8(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L58:
  mov    %r13,%rcx
  lea    D13(%rip),%rdi
  mov    %r14,%r8
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  mov    0xc(%rsp),%r11d
  mov    0x10(%rsp),%ecx
  mov    0x14(%rsp),%r10d
  mov    %edi,%eax
  and    %r11d,%eax
  and    %ecx,%eax
  and    %r10d,%eax
  cmp    $0xffffffff,%eax
  je     L44
  lea    0x1(%r10),%edx
  mov    %edi,%r8d
  lea    0x5304(%r10),%r9d
  mov    %edx,%esi
  sub    %r11d,%r8d
  sub    %edi,%esi
  sub    %ecx,%edi
  jmp    L34
  cs nopw 0x0(%rax,%rax,1)
L42:
  add    $0x1,%edx
  add    $0x1,%esi
  cmp    %r9d,%edx
  jg     L45
L34:
  lea    (%r8,%rsi,1),%ecx
  movslq %ecx,%rax
  mov    %ecx,%r11d
  imul   $0xffffffff92492493,%rax,%rax
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  add    %ecx,%eax
  sar    $0x4,%eax
  sub    %r11d,%eax
  lea    (%rdi,%rsi,1),%r11d
  imul   $0x1c,%eax,%eax
  mov    %r11d,%r15d
  sar    $0x1f,%r15d
  sub    %eax,%ecx
  movslq %r11d,%rax
  imul   $0x3e0f83e1,%rax,%rax
  sar    $0x23,%rax
  sub    %r15d,%eax
  mov    %eax,%r15d
  shl    $0x5,%r15d
  add    %r15d,%eax
  sub    %eax,%r11d
  movslq %esi,%rax
  imul   $0xffffffffb21642c9,%rax,%rax
  or     %r11d,%ecx
  mov    %esi,%r11d
  sar    $0x1f,%r11d
  shr    $0x20,%rax
  add    %esi,%eax
  sar    $0x4,%eax
  sub    %r11d,%eax
  mov    %esi,%r11d
  imul   $0x17,%eax,%eax
  sub    %eax,%r11d
  or     %ecx,%r11d
  jne    L42
L45:
  mov    %ebx,%esi
  sub    %r10d,%edx
  lea    D4(%rip),%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  jmp    L58
L44:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
type_infered: string
name: extract_symbolic_data
arguments: tag: D4
type_infered: string

tool: D13: %d %d %d %d

tool: D4: Case %d: the next triple peak occurs in %d days.


assistant: ```c
int main() {
  int p;
  int e;
  int i;
  int d;
  int count = 1;
  int k;
  while (1) {
    scanf("%d %d %d %d", &p, &e, &i, &d);
    if (p == -1 && e == -1 && i == -1 && d == -1) {
      break;
    }
    for (k = d + 1; k <= 21252 + d; k++) {
      if ((k - p) % 23 == 0 && (k - e) % 28 == 0 && (k - i) % 33 == 0)
        break;
    }
    printf("Case %d: the next triple peak occurs in %d days.\n", count, k - d);
    count++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<puts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L25
L19:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L24
  mov    $0xd,%edi
  call   <putch@plt>
L24:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putch@plt>
  addq   $0x1,-0x8(%rbp)
L25:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void puts(const char *a) {
  for (const char *in = a; *in; in++) {
    if (*in == '\n')
      putch('\r');
    putch(*in);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L38
L21:
  pop    %rbx
  ret
L14:
  movsbl (%rbx),%edi
  call   <putch@plt>
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L21
L38:
  cmp    $0xa,%al
  jne    L14
  mov    $0xd,%edi
  call   <putch@plt>
  jmp    L14
```

assistant: ```c
void puts(const char *a) {
  for (const char *in = a; *in; in++) {
    if (*in == '\n')
      putch('\r');
    putch(*in);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<puts>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L39
  push   %rbx
  mov    %rdi,%rbx
  jmp    L19
  nopl   0x0(%rax)
L27:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <putch@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L11
L19:
  cmp    $0xa,%al
  jne    L27
  mov    $0xd,%edi
  call   <putch@plt>
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L11:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L39:
  ret
```

assistant: ```c
void puts(const char *a) {
  for (const char *in = a; *in; in++) {
    if (*in == '\n')
      putch('\r');
    putch(*in);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<puts>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L30
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L42:
  call   <putch@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L39
L30:
  cmp    $0xa,%dil
  jne    L42
  mov    $0xd,%edi
  call   <putch@plt>
  movsbl (%rbx),%edi
  jmp    L42
  nop
L39:
  pop    %rbx
  ret
```

assistant: ```c
void puts(const char *a) {
  for (const char *in = a; *in; in++) {
    if (*in == '\n')
      putch('\r');
    putch(*in);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<psmat4_translation>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <mat4_translation@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
__attribute__((always_)) struct mat4 *psmat4_translation(struct mat4 *result, struct mat4 *v) {
  mat4_translation((float *)result, (float *)v);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<psmat4_translation>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <mat4_translation@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
__attribute__((always_)) struct mat4 *psmat4_translation(struct mat4 *result, struct mat4 *v) {
  mat4_translation((float *)result, (float *)v);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<psmat4_translation>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <mat4_translation@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct mat4 *psmat4_translation(struct mat4 *result, struct mat4 *v) {
  mat4_translation((float *)result, (float *)v);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<psmat4_translation>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  call   <mat4_translation@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
__attribute__((always_)) struct mat4 *psmat4_translation(struct mat4 *result, struct mat4 *v) {
  mat4_translation((float *)result, (float *)v);
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sort_ascending>:
  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),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jge    L15
  mov    $0xffffffff,%eax
  jmp    L10
L15:
  mov    $0x1,%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int sort_ascending(const void *a, const void *b) {
  const int64_t *c = a;
  const int64_t *d = b;
  if (*c < *d)
    return -1;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sort_ascending>:
  endbr64
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int sort_ascending(const void *a, const void *b) {
  const int64_t *c = a;
  const int64_t *d = b;
  if (*c < *d)
    return -1;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sort_ascending>:
  endbr64
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int sort_ascending(const void *a, const void *b) {
  const int64_t *c = a;
  const int64_t *d = b;
  if (*c < *d)
    return -1;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sort_ascending>:
  endbr64
  mov    (%rsi),%rax
  cmp    %rax,(%rdi)
  setge  %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int sort_ascending(const void *a, const void *b) {
  const int64_t *c = a;
  const int64_t *d = b;
  if (*c < *d)
    return -1;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gen_swap_stack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <output_line@plt>
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <output_line@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <output_line@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: string
name: disassemble_data
arguments: data_label: D21
type_guessed: string
name: disassemble_data
arguments: data_label: D14
type_guessed: string

tool: D2: ldu ,s

tool: D21: std ,s

tool: D14: tfr u,d

assistant: ```c
void gen_swap_stack(void) {
  output_line("ldu ,s");
  output_line("std ,s");
  output_line("tfr u,d");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gen_swap_stack>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <output_line@plt>
  lea    D7(%rip),%rdi
  call   <output_line@plt>
  lea    D4(%rip),%rdi
  call   <output_line@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D7
data_type: string
name: disassemble_data
arguments: tag: D4
data_type: string

tool: D0: ldu ,s

tool: D7: std ,s

tool: D4: tfr u,d

assistant: ```c
void gen_swap_stack(void) {
  output_line("ldu ,s");
  output_line("std ,s");
  output_line("tfr u,d");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gen_swap_stack>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <output_line@plt>
  lea    D5(%rip),%rdi
  call   <output_line@plt>
  lea    D25(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <output_line@plt>
```

tool_calls: name: parse_data
arguments: tag: D10
data_type: string
name: parse_data
arguments: tag: D5
data_type: string
name: parse_data
arguments: tag: D25
data_type: string

tool: D10: ldu ,s

tool: D5: std ,s

tool: D25: tfr u,d

assistant: ```c
void gen_swap_stack(void) {
  output_line("ldu ,s");
  output_line("std ,s");
  output_line("tfr u,d");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<gen_swap_stack>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <output_line@plt>
  lea    D16(%rip),%rdi
  call   <output_line@plt>
  lea    D3(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <output_line@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type_infered: string
name: disassemble_data
arguments: data_tag: D16
type_infered: string
name: disassemble_data
arguments: data_tag: D3
type_infered: string

tool: D2: ldu ,s

tool: D16: std ,s

tool: D3: tfr u,d

assistant: ```c
void gen_swap_stack(void) {
  output_line("ldu ,s");
  output_line("std ,s");
  output_line("tfr u,d");
}
```
user: What is the c source code of the assembly code below:

```asm
<tAvg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  lea    0x0(,%rbp,4),%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    (%rbx),%ebp
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
uint64_t tAvg(uint64_t *pTList, int pLen) {
  int i, j, jsteps;
  uint64_t curSample;
  uint64_t *mySamples;
  mySamples = malloc(1 * sizeof(uint64_t));
  memcpy(mySamples, pTList, pLen * sizeof(uint64_t));
  for (i = 0; i < 0; i++) {
    jsteps = 1 / pow(10, i + 1);
    for (j = 0; j < jsteps; j++) {
      curSample = mySamples[j * 10];
      curSample += mySamples[(j * 10) + 1];
      curSample += mySamples[(j * 10) + 2];
      curSample += mySamples[(j * 10) + 3];
      curSample += mySamples[(j * 10) + 4];
      curSample += mySamples[(j * 10) + 5];
      curSample += mySamples[(j * 10) + 6];
      curSample += mySamples[(j * 10) + 7];
      curSample += mySamples[(j * 10) + 8];
      curSample += mySamples[(j * 10) + 9];
      mySamples[j] = curSample / 10;
    }
  }
  curSample = mySamples[0];
  free(mySamples);
  return curSample;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tAvg>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x4,%edi
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  lea    0x0(,%rbx,4),%edx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    0x0(%rbp),%r12d
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t tAvg(uint64_t *pTList, int pLen) {
  int i, j, jsteps;
  uint64_t curSample;
  uint64_t *mySamples;
  mySamples = malloc(1 * sizeof(uint64_t));
  memcpy(mySamples, pTList, pLen * sizeof(uint64_t));
  for (i = 0; i < 0; i++) {
    jsteps = 1 / pow(10, i + 1);
    for (j = 0; j < jsteps; j++) {
      curSample = mySamples[j * 10];
      curSample += mySamples[(j * 10) + 1];
      curSample += mySamples[(j * 10) + 2];
      curSample += mySamples[(j * 10) + 3];
      curSample += mySamples[(j * 10) + 4];
      curSample += mySamples[(j * 10) + 5];
      curSample += mySamples[(j * 10) + 6];
      curSample += mySamples[(j * 10) + 7];
      curSample += mySamples[(j * 10) + 8];
      curSample += mySamples[(j * 10) + 9];
      mySamples[j] = curSample / 10;
    }
  }
  curSample = mySamples[0];
  free(mySamples);
  return curSample;
}
```
user: What is the c source code of the assembly code below?

```asm
<tAvg>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    $0x4,%edi
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  call   <malloc@plt>
  lea    0x0(,%rbx,4),%edx
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    0x0(%rbp),%r12d
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
uint64_t tAvg(uint64_t *pTList, int pLen) {
  int i, j, jsteps;
  uint64_t curSample;
  uint64_t *mySamples;
  mySamples = malloc(1 * sizeof(uint64_t));
  memcpy(mySamples, pTList, pLen * sizeof(uint64_t));
  for (i = 0; i < 0; i++) {
    jsteps = 1 / pow(10, i + 1);
    for (j = 0; j < jsteps; j++) {
      curSample = mySamples[j * 10];
      curSample += mySamples[(j * 10) + 1];
      curSample += mySamples[(j * 10) + 2];
      curSample += mySamples[(j * 10) + 3];
      curSample += mySamples[(j * 10) + 4];
      curSample += mySamples[(j * 10) + 5];
      curSample += mySamples[(j * 10) + 6];
      curSample += mySamples[(j * 10) + 7];
      curSample += mySamples[(j * 10) + 8];
      curSample += mySamples[(j * 10) + 9];
      mySamples[j] = curSample / 10;
    }
  }
  curSample = mySamples[0];
  free(mySamples);
  return curSample;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ifn_array2wk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  shr    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L18
L14:
  mov    -0x8(%rbp),%eax
  lea    0x0(,%rax,4),%rcx
  mov    -0xc(%rbp),%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  shrl   -0x4(%rbp)
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jbe    L14
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ifn_array2wk(float *p2arr, const unsigned int const cuic_Nlwl, const unsigned int const cuic_N, float *a4p2k[]) {
  unsigned int ui_i;
  unsigned int ui_sum;
  unsigned int ui_sd;
  for (ui_i = 0, ui_sum = 0, ui_sd = (cuic_N >> 1); ui_i <= cuic_Nlwl; ui_i += 1) {
    a4p2k[ui_i] = &p2arr[ui_sum];
    ui_sum += ui_sd;
    ui_sd >>= 1;
  }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ifn_array2wk>:
  endbr64
  shr    %edx
  mov    $0x0,%r8d
  mov    $0x0,%eax
L2:
  mov    %eax,%r9d
  mov    %r8d,%r10d
  lea    (%rdi,%r10,4),%r10
  mov    %r10,(%rcx,%r9,8)
  add    %edx,%r8d
  shr    %edx
  add    $0x1,%eax
  cmp    %eax,%esi
  jae    L2
  ret
```

assistant: ```c
void ifn_array2wk(float *p2arr, const unsigned int const cuic_Nlwl, const unsigned int const cuic_N, float *a4p2k[]) {
  unsigned int ui_i;
  unsigned int ui_sum;
  unsigned int ui_sd;
  for (ui_i = 0, ui_sum = 0, ui_sd = (cuic_N >> 1); ui_i <= cuic_Nlwl; ui_i += 1) {
    a4p2k[ui_i] = &p2arr[ui_sum];
    ui_sum += ui_sd;
    ui_sd >>= 1;
  }
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ifn_array2wk>:
  endbr64
  shr    %edx
  xor    %r8d,%r8d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %r8d,%r10d
  mov    %eax,%r9d
  add    $0x1,%eax
  add    %edx,%r8d
  lea    (%rdi,%r10,4),%r10
  shr    %edx
  mov    %r10,(%rcx,%r9,8)
  cmp    %eax,%esi
  jae    L7
  ret
```

assistant: ```c
void ifn_array2wk(float *p2arr, const unsigned int const cuic_Nlwl, const unsigned int const cuic_N, float *a4p2k[]) {
  unsigned int ui_i;
  unsigned int ui_sum;
  unsigned int ui_sd;
  for (ui_i = 0, ui_sum = 0, ui_sd = (cuic_N >> 1); ui_i <= cuic_Nlwl; ui_i += 1) {
    a4p2k[ui_i] = &p2arr[ui_sum];
    ui_sum += ui_sd;
    ui_sd >>= 1;
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<ifn_array2wk>:
  endbr64
  shr    %edx
  xor    %r8d,%r8d
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %r8d,%r10d
  mov    %eax,%r9d
  add    $0x1,%eax
  add    %edx,%r8d
  lea    (%rdi,%r10,4),%r10
  shr    %edx
  mov    %r10,(%rcx,%r9,8)
  cmp    %eax,%esi
  jae    L5
  ret
```

assistant: ```c
void ifn_array2wk(float *p2arr, const unsigned int const cuic_Nlwl, const unsigned int const cuic_N, float *a4p2k[]) {
  unsigned int ui_i;
  unsigned int ui_sum;
  unsigned int ui_sd;
  for (ui_i = 0, ui_sum = 0, ui_sd = (cuic_N >> 1); ui_i <= cuic_Nlwl; ui_i += 1) {
    a4p2k[ui_i] = &p2arr[ui_sum];
    ui_sum += ui_sd;
    ui_sd >>= 1;
  }
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fec_decode>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %rcx,%r12
  mov    %r8,%r13
  mov    %r9,%r14
  movzwl 0x48(%rsp),%r15d
  push   %r15
  call   <reduce@plt>
  mov    %r15d,%r9d
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <resolve@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fec_decode(unsigned int blockSize, unsigned char **data_blocks, unsigned int nr_data_blocks, unsigned char **fec_blocks, unsigned int *fec_block_nos, unsigned int *erased_blocks, unsigned short nr_fec_blocks) {
  reduce(blockSize, data_blocks, nr_data_blocks, fec_blocks, fec_block_nos, erased_blocks, nr_fec_blocks);
  resolve(blockSize, data_blocks, fec_blocks, fec_block_nos, erased_blocks, nr_fec_blocks);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<fec_decode>:
  endbr64
  push   %r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %r9,%rbx
  sub    $0x10,%rsp
  movzwl 0x48(%rsp),%r15d
  push   %r15
  call   <reduce@plt>
  add    $0x18,%rsp
  mov    %r15d,%r9d
  mov    %rbx,%r8
  mov    %r14,%rcx
  pop    %rbx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <resolve@plt>
```

assistant: ```c
void fec_decode(unsigned int blockSize, unsigned char **data_blocks, unsigned int nr_data_blocks, unsigned char **fec_blocks, unsigned int *fec_block_nos, unsigned int *erased_blocks, unsigned short nr_fec_blocks) {
  reduce(blockSize, data_blocks, nr_data_blocks, fec_blocks, fec_block_nos, erased_blocks, nr_fec_blocks);
  resolve(blockSize, data_blocks, fec_blocks, fec_block_nos, erased_blocks, nr_fec_blocks);
}
```