ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<VADFrameCalculateZeroCrossingRate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x10(%rbp)
  movw   $0x0,-0x12(%rbp)
  movw   $0x0,-0x14(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L26
L51:
  mov    -0x8(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  test   %ax,%ax
  jle    L10
  mov    $0x1,%eax
  jmp    L49
L10:
  mov    $0xffffffff,%eax
L49:
  mov    %ax,-0x12(%rbp)
  cmpw   $0x0,-0x14(%rbp)
  je     L38
  movzwl -0x12(%rbp),%eax
  cmp    -0x14(%rbp),%ax
  je     L38
  addq   $0x1,-0x10(%rbp)
L38:
  movzwl -0x12(%rbp),%eax
  mov    %ax,-0x14(%rbp)
  addq   $0x1,-0x8(%rbp)
L26:
  mov    -0x8(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L51
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t VADFrameCalculateZeroCrossingRate(const short *frame, size_t frames_count) {
  size_t zcr = 0;
  short sign = 0;
  short last_sign = 0;
  for (size_t i = 0; i < frames_count; i++) {
    sign = (frame[i] > 0) ? 1 : -1;
    if ((last_sign != 0) && (sign != last_sign)) {
      zcr++;
    }
    last_sign = sign;
  }
  return zcr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<VADFrameCalculateZeroCrossingRate>:
  endbr64
  test   %rsi,%rsi
  je     L28
  mov    %rdi,%rdx
  lea    (%rdi,%rsi,2),%rdi
  mov    $0x0,%ecx
  mov    $0x0,%esi
  jmp    L20
L44:
  add    $0x2,%rdx
  cmp    %rdi,%rdx
  je     L28
  mov    %eax,%ecx
L20:
  cmpw   $0x0,(%rdx)
  setg   %al
  movzbl %al,%eax
  lea    -0x1(%rax,%rax,1),%eax
  test   %cx,%cx
  je     L44
  cmp    %cx,%ax
  setne  %cl
  cmp    $0x1,%cl
  sbb    $0xffffffffffffffff,%rsi
  jmp    L44
L28:
  mov    %rsi,%rax
  ret
```

assistant: ```c
size_t VADFrameCalculateZeroCrossingRate(const short *frame, size_t frames_count) {
  size_t zcr = 0;
  short sign = 0;
  short last_sign = 0;
  for (size_t i = 0; i < frames_count; i++) {
    sign = (frame[i] > 0) ? 1 : -1;
    if ((last_sign != 0) && (sign != last_sign)) {
      zcr++;
    }
    last_sign = sign;
  }
  return zcr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<VADFrameCalculateZeroCrossingRate>:
  endbr64
  test   %rsi,%rsi
  je     L12
  lea    (%rdi,%rsi,2),%rcx
  xor    %edx,%edx
  xor    %r8d,%r8d
  jmp    L30
  nopl   0x0(%rax)
L17:
  mov    %eax,%edx
L30:
  xor    %eax,%eax
  cmpw   $0x0,(%rdi)
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  test   %dx,%dx
  je     L0
  cmp    %dx,%ax
  setne  %dl
  cmp    $0x1,%dl
  sbb    $0xffffffffffffffff,%r8
L0:
  add    $0x2,%rdi
  cmp    %rcx,%rdi
  jne    L17
  mov    %r8,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t VADFrameCalculateZeroCrossingRate(const short *frame, size_t frames_count) {
  size_t zcr = 0;
  short sign = 0;
  short last_sign = 0;
  for (size_t i = 0; i < frames_count; i++) {
    sign = (frame[i] > 0) ? 1 : -1;
    if ((last_sign != 0) && (sign != last_sign)) {
      zcr++;
    }
    last_sign = sign;
  }
  return zcr;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<VADFrameCalculateZeroCrossingRate>:
  endbr64
  test   %rsi,%rsi
  je     L1
  lea    (%rdi,%rsi,2),%rcx
  xor    %edx,%edx
  xor    %r8d,%r8d
  jmp    L4
  nopl   0x0(%rax)
L23:
  mov    %eax,%edx
L4:
  xor    %eax,%eax
  cmpw   $0x0,(%rdi)
  setg   %al
  lea    -0x1(%rax,%rax,1),%eax
  test   %dx,%dx
  je     L10
  cmp    %ax,%dx
  setne  %dl
  cmp    $0x1,%dl
  sbb    $0xffffffffffffffff,%r8
L10:
  add    $0x2,%rdi
  cmp    %rdi,%rcx
  jne    L23
  mov    %r8,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
size_t VADFrameCalculateZeroCrossingRate(const short *frame, size_t frames_count) {
  size_t zcr = 0;
  short sign = 0;
  short last_sign = 0;
  for (size_t i = 0; i < frames_count; i++) {
    sign = (frame[i] > 0) ? 1 : -1;
    if ((last_sign != 0) && (sign != last_sign)) {
      zcr++;
    }
    last_sign = sign;
  }
  return zcr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  movl   $0x0,-0x2c(%rbp)
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L4
L16:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    -0x2c(%rbp),%eax
  cltq
  add    $0x2,%rax
  add    %rax,-0x28(%rbp)
L4:
  mov    -0x28(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L16
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  test   %eax,%eax
  jne    L24
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  jmp    L36
L24:
  mov    -0x28(%rbp),%rax
L36:
  leave
  ret
```

assistant: ```c
char *get_basename(const char *path) {
  char *basename;
  char *copy;
  int l = 0;
  char *ssc;
  basename = malloc(500);
  strcpy(basename, path);
  while (ssc = strstr(basename, "/")) {
    l = strlen(ssc) + 1;
    copy = basename;
    basename = &basename[strlen(basename) - l + 2];
  }
  if (!strcmp(basename, "")) {
    char *temp = malloc(500);
    size_t length = (strlen(copy) - 1);
    strncpy(temp, copy, length);
    temp[length] = '\0';
    return temp;
  }
  return basename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_basename>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L28
L1:
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    $0x1,%ebx
  sub    %ebx,%eax
  cltq
  mov    %rbp,%rbx
  lea    0x2(%rbp,%rax,1),%rbp
L28:
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L1
  cmpb   $0x0,0x0(%rbp)
  je     L25
L32:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%r12d
  movslq %r12d,%r12
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strncpy@plt>
  movb   $0x0,0x0(%rbp,%r12,1)
  jmp    L32
```

assistant: ```c
char *get_basename(const char *path) {
  char *basename;
  char *copy;
  int l = 0;
  char *ssc;
  basename = malloc(500);
  strcpy(basename, path);
  while (ssc = strstr(basename, "/")) {
    l = strlen(ssc) + 1;
    copy = basename;
    basename = &basename[strlen(basename) - l + 2];
  }
  if (!strcmp(basename, "")) {
    char *temp = malloc(500);
    size_t length = (strlen(copy) - 1);
    strncpy(temp, copy, length);
    temp[length] = '\0';
    return temp;
  }
  return basename;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_basename>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    $0x1f4,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L3:
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %r12,%rbp
  mov    %eax,%ebx
  call   <strlen@plt>
  add    $0x1,%ebx
  sub    %ebx,%eax
  cltq
  lea    0x2(%r12,%rax,1),%r12
L5:
  mov    %r12,%rdi
  mov    $0x2f,%esi
  call   <strchr@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L3
  cmpb   $0x0,(%r12)
  je     L9
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdx
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *get_basename(const char *path) {
  char *basename;
  char *copy;
  int l = 0;
  char *ssc;
  basename = malloc(500);
  strcpy(basename, path);
  while (ssc = strstr(basename, "/")) {
    l = strlen(ssc) + 1;
    copy = basename;
    basename = &basename[strlen(basename) - l + 2];
  }
  if (!strcmp(basename, "")) {
    char *temp = malloc(500);
    size_t length = (strlen(copy) - 1);
    strncpy(temp, copy, length);
    temp[length] = '\0';
    return temp;
  }
  return basename;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_basename>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    $0x1f4,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L2:
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %r12,%rbp
  mov    %eax,%ebx
  call   <strlen@plt>
  add    $0x1,%ebx
  sub    %ebx,%eax
  cltq
  lea    0x2(%r12,%rax,1),%r12
L9:
  mov    %r12,%rdi
  mov    $0x2f,%esi
  call   <strchr@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L2
  cmpb   $0x0,(%r12)
  je     L19
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0x1f4,%edi
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  mov    %rbx,%rdx
  call   <strncpy@plt>
  movb   $0x0,(%r12,%rbx,1)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *get_basename(const char *path) {
  char *basename;
  char *copy;
  int l = 0;
  char *ssc;
  basename = malloc(500);
  strcpy(basename, path);
  while (ssc = strstr(basename, "/")) {
    l = strlen(ssc) + 1;
    copy = basename;
    basename = &basename[strlen(basename) - l + 2];
  }
  if (!strcmp(basename, "")) {
    char *temp = malloc(500);
    size_t length = (strlen(copy) - 1);
    strncpy(temp, copy, length);
    temp[length] = '\0';
    return temp;
  }
  return basename;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_create_new_session>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <setsid@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jns    L9
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L18
L9:
  mov    $0x0,%eax
L18:
  leave
  ret
```

assistant: ```c
int _create_new_session(void) {
  pid_t sid;
  sid = setsid();
  if (sid < 0)
    return -errno;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_create_new_session>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <setsid@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  js     L2
L17:
  add    $0x8,%rsp
  ret
L2:
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  neg    %eax
  jmp    L17
```

assistant: ```c
int _create_new_session(void) {
  pid_t sid;
  sid = setsid();
  if (sid < 0)
    return -errno;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_create_new_session>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <setsid@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  js     L3
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L3:
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  neg    %eax
  ret
```

assistant: ```c
int _create_new_session(void) {
  pid_t sid;
  sid = setsid();
  if (sid < 0)
    return -errno;
  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
<_create_new_session>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <setsid@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  js     L5
  add    $0x8,%rsp
  ret
  xchg   %ax,%ax
L5:
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  add    $0x8,%rsp
  neg    %eax
  ret
```

assistant: ```c
int _create_new_session(void) {
  pid_t sid;
  sid = setsid();
  if (sid < 0)
    return -errno;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<vec2_normalize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  movss  0x4(%rax),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0xc(%rbp),%xmm0
  movaps %xmm0,%xmm1
  mulss  %xmm0,%xmm1
  movss  -0x8(%rbp),%xmm0
  mulss  %xmm0,%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comiss %xmm1,%xmm0
  ja     L17
  jmp    L11
L17:
  mov    -0x4(%rbp),%eax
  movd   %eax,%xmm0
  call   <sqrtf@plt>
  mov    %eax,%ecx
  mov    $0x1,%eax
  cltd
  idiv   %ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0x4(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mulss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
L11:
  nop
  leave
  ret
```

assistant: ```c
void vec2_normalize(float *dst) {
  float x = dst[0], y = dst[1];
  float len = x * x + y * y;
  if (len > 0) {
    len = 1 / sqrtf(len);
    dst[0] = dst[0] * len;
    dst[1] = dst[1] * len;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<vec2_normalize>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  comiss D4(%rip),%xmm0
  ja     L0
  ret
L0:
  push   %rbx
  mov    %rdi,%rbx
  call   <sqrtf@plt>
  mov    %eax,%ecx
  mov    $0x1,%eax
  cltd
  idiv   %ecx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movaps %xmm0,%xmm1
  mulss  (%rbx),%xmm1
  movss  %xmm1,(%rbx)
  mulss  0x4(%rbx),%xmm0
  movss  %xmm0,0x4(%rbx)
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
datatype: f32

tool: D4: 0

assistant: ```c
void vec2_normalize(float *dst) {
  float x = dst[0], y = dst[1];
  float len = x * x + y * y;
  if (len > 0) {
    len = 1 / sqrtf(len);
    dst[0] = dst[0] * len;
    dst[1] = dst[1] * len;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vec2_normalize>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  comiss D0(%rip),%xmm0
  ja     L9
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  push   %rbx
  mov    %rdi,%rbx
  call   <sqrtf@plt>
  pxor   %xmm0,%xmm0
  movss  (%rbx),%xmm1
  mov    %eax,%r8d
  mov    $0x1,%eax
  cltd
  idiv   %r8d
  cvtsi2ss %eax,%xmm0
  mulss  %xmm0,%xmm1
  mulss  0x4(%rbx),%xmm0
  movss  %xmm1,(%rbx)
  movss  %xmm0,0x4(%rbx)
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type: f32

tool: D0: 0

assistant: ```c
void vec2_normalize(float *dst) {
  float x = dst[0], y = dst[1];
  float len = x * x + y * y;
  if (len > 0) {
    len = 1 / sqrtf(len);
    dst[0] = dst[0] * len;
    dst[1] = dst[1] * len;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vec2_normalize>:
  endbr64
  movss  (%rdi),%xmm0
  movss  0x4(%rdi),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  comiss D5(%rip),%xmm0
  ja     L5
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  push   %rbx
  mov    %rdi,%rbx
  call   <sqrtf@plt>
  pxor   %xmm1,%xmm1
  movq   (%rbx),%xmm0
  mov    %eax,%r8d
  mov    $0x1,%eax
  cltd
  idiv   %r8d
  cvtsi2ss %eax,%xmm1
  shufps $0xe0,%xmm1,%xmm1
  mulps  %xmm1,%xmm0
  movlps %xmm0,(%rbx)
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_guessed: f32

tool: D5: 0

assistant: ```c
void vec2_normalize(float *dst) {
  float x = dst[0], y = dst[1];
  float len = x * x + y * y;
  if (len > 0) {
    len = 1 / sqrtf(len);
    dst[0] = dst[0] * len;
    dst[1] = dst[1] * len;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<red>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movsd  %xmm0,-0x38(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L9
L38:
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L39
L18:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L43
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  movsd  -0x38(%rbp),%xmm0
  movsd  D14(%rip),%xmm3
  movapd %xmm0,%xmm2
  divsd  %xmm3,%xmm2
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm3
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  sub    $0x8,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm0,%xmm3
  mov    -0x8(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm0,%xmm3
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    $0x1,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm3,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm4
  movsd  D14(%rip),%xmm3
  mulsd  %xmm4,%xmm3
  subsd  %xmm3,%xmm0
  mulsd  %xmm2,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
L43:
  addl   $0x1,-0x4(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L18
  addl   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L38
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D14
type: f64

tool: D14: 4

assistant: ```c
void red(double **u_previous, double **u_current, int min_x, int max_x, int min_y, int max_y, double omega) {
  int i, j;
  for (i = min_x; i < max_x; i++)
    for (j = min_y; j < max_y; j++)
      if ((i + j) % 2 == 0)
        u_current[i][j] = u_previous[i][j] + (omega / 4.0) * (u_previous[i - 1][j] + u_previous[i][j - 1] + u_previous[i + 1][j] + u_previous[i][j + 1] - 4 * u_previous[i][j]);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<red>:
  endbr64
  cmp    %ecx,%edx
  jge    L33
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  mov    %rsi,%r11
  mov    %ecx,%ebx
  movslq %edx,%rdi
  lea    (%rdx,%r9,1),%esi
  lea    (%rax,%rdi,8),%rcx
  movslq %r8d,%rax
  lea    0x8(,%rax,8),%rbp
  movsd  D10(%rip),%xmm5
  movsd  D17(%rip),%xmm4
  jmp    L44
L9:
  add    $0x1,%edx
  add    $0x8,%rax
  cmp    %esi,%edx
  je     L67
L56:
  test   $0x1,%dl
  jne    L9
  mov    (%rcx),%r12
  movsd  -0x8(%r12,%rax,1),%xmm2
  mov    (%r11,%rdi,8),%r10
  mov    -0x8(%rcx),%r14
  movsd  -0x8(%r14,%rax,1),%xmm1
  addsd  -0x10(%r12,%rax,1),%xmm1
  mov    0x8(%rcx),%r14
  addsd  -0x8(%r14,%rax,1),%xmm1
  addsd  (%r12,%rax,1),%xmm1
  movapd %xmm2,%xmm3
  mulsd  %xmm5,%xmm3
  subsd  %xmm3,%xmm1
  movapd %xmm0,%xmm3
  mulsd  %xmm4,%xmm3
  mulsd  %xmm3,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,-0x8(%r10,%rax,1)
  jmp    L9
L67:
  add    $0x1,%rdi
  add    $0x1,%esi
  add    $0x8,%rcx
  cmp    %edi,%ebx
  jle    L25
L44:
  lea    (%r8,%rdi,1),%edx
  mov    %rbp,%rax
  cmp    %r9d,%r8d
  jl     L56
  jmp    L67
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L33:
  ret
```

tool_calls: name: retrieve_data
arguments: label: D10
type_infered: f64
name: retrieve_data
arguments: label: D17
type_infered: byte[8]

tool: D10: 4

tool: {"D17": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

assistant: ```c
void red(double **u_previous, double **u_current, int min_x, int max_x, int min_y, int max_y, double omega) {
  int i, j;
  for (i = min_x; i < max_x; i++)
    for (j = min_y; j < max_y; j++)
      if ((i + j) % 2 == 0)
        u_current[i][j] = u_previous[i][j] + (omega / 4.0) * (u_previous[i - 1][j] + u_previous[i][j - 1] + u_previous[i + 1][j] + u_previous[i][j + 1] - 4 * u_previous[i][j]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<red>:
  endbr64
  cmp    %ecx,%edx
  jge    L28
  mov    %rdi,%rax
  movslq %edx,%rdi
  push   %r14
  movsd  D15(%rip),%xmm5
  push   %r12
  movsd  D18(%rip),%xmm4
  mov    %rsi,%r11
  lea    (%rdx,%r9,1),%esi
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  lea    (%rax,%rdi,8),%rcx
  movslq %r8d,%rax
  lea    0x8(,%rax,8),%rbp
L30:
  lea    (%r8,%rdi,1),%edx
  mov    %rbp,%rax
  cmp    %r9d,%r8d
  jge    L13
  nopl   0x0(%rax)
L3:
  test   $0x1,%dl
  jne    L8
  mov    (%rcx),%r12
  mov    -0x8(%rcx),%r14
  mov    (%r11,%rdi,8),%r10
  movsd  -0x8(%r12,%rax,1),%xmm2
  movsd  -0x8(%r14,%rax,1),%xmm1
  mov    0x8(%rcx),%r14
  addsd  -0x10(%r12,%rax,1),%xmm1
  movapd %xmm2,%xmm3
  mulsd  %xmm5,%xmm3
  addsd  -0x8(%r14,%rax,1),%xmm1
  addsd  (%r12,%rax,1),%xmm1
  subsd  %xmm3,%xmm1
  movapd %xmm0,%xmm3
  mulsd  %xmm4,%xmm3
  mulsd  %xmm3,%xmm1
  addsd  %xmm2,%xmm1
  movsd  %xmm1,-0x8(%r10,%rax,1)
L8:
  add    $0x1,%edx
  add    $0x8,%rax
  cmp    %esi,%edx
  jne    L3
L13:
  add    $0x1,%rdi
  add    $0x1,%esi
  add    $0x8,%rcx
  cmp    %edi,%ebx
  jg     L30
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L28:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D15
type_infered: f64
name: disassemble_data
arguments: data_tag: D18
type_infered: byte[8]

tool: D15: 4

tool: {"D18": [0x00,0x00,0x00,0x00,0x00,0x00,0xd0,0x3f]}

assistant: ```c
void red(double **u_previous, double **u_current, int min_x, int max_x, int min_y, int max_y, double omega) {
  int i, j;
  for (i = min_x; i < max_x; i++)
    for (j = min_y; j < max_y; j++)
      if ((i + j) % 2 == 0)
        u_current[i][j] = u_previous[i][j] + (omega / 4.0) * (u_previous[i - 1][j] + u_previous[i][j - 1] + u_previous[i + 1][j] + u_previous[i][j + 1] - 4 * u_previous[i][j]);
}
```