ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<_add2>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  sar    $0x10,%edi
  sar    $0x10,%esi
  add    %esi,%edi
  shl    $0x10,%edi
  movzwl %ax,%eax
  or     %edi,%eax
  ret
```

assistant: ```c
int _add2(int src1, int src2) {
  short s1[2], s2[2], r[2];
  int result;
  *((int *)s1) = src1;
  *((int *)s2) = src2;
  r[0] = s1[0] + s2[0];
  r[1] = s1[1] + s2[1];
  result = *(int *)r;
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_add2>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  sar    $0x10,%edi
  sar    $0x10,%esi
  add    %esi,%edi
  movzwl %ax,%eax
  shl    $0x10,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
int _add2(int src1, int src2) {
  short s1[2], s2[2], r[2];
  int result;
  *((int *)s1) = src1;
  *((int *)s2) = src2;
  r[0] = s1[0] + s2[0];
  r[1] = s1[1] + s2[1];
  result = *(int *)r;
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<_add2>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  sar    $0x10,%edi
  sar    $0x10,%esi
  add    %esi,%edi
  movzwl %ax,%eax
  shl    $0x10,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
int _add2(int src1, int src2) {
  short s1[2], s2[2], r[2];
  int result;
  *((int *)s1) = src1;
  *((int *)s2) = src2;
  r[0] = s1[0] + s2[0];
  r[1] = s1[1] + s2[1];
  result = *(int *)r;
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<prune_origin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L78
L9:
  cmpq   $0x0,-0x20(%rbp)
  je     L24
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ns_samename@plt>
  cmp    $0x1,%eax
  jne    L24
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  seta   %al
  movzbl %al,%ecx
  mov    %rdx,%rax
  sub    %rcx,%rax
  jmp    L6
L58:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L107
  addq   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L95
  jmp    L78
L107:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L95
  addq   $0x1,-0x18(%rbp)
  jmp    L78
L95:
  addq   $0x1,-0x18(%rbp)
L24:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L58
L78:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
L6:
  leave
  ret
```

assistant: ```c
size_t prune_origin(const char *name, const char *origin) {
  const char *oname = name;
  while (*name != '\0') {
    if (origin != NULL && ns_samename(name, origin) == 1)
      return (name - oname - (name > oname));
    while (*name != '\0') {
      if (*name == '\\') {
        name++;
        if (*name == '\0')
          break;
      } else if (*name == '.') {
        name++;
        break;
      }
      name++;
    }
  }
  return (name - oname);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<prune_origin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  cmpb   $0x0,(%rdi)
  je     L114
  mov    %rsi,%rbp
  mov    %rdi,%rbx
  jmp    L17
L126:
  cmp    %rbx,%r12
  mov    %rbx,%rax
  sbb    %r12,%rax
  jmp    L42
L55:
  lea    0x1(%rbx),%rax
  cmpb   $0x0,0x1(%rbx)
  je     L75
L62:
  lea    0x1(%rax),%rbx
  movzbl 0x1(%rax),%eax
  test   %al,%al
  je     L109
L122:
  cmp    $0x5c,%al
  je     L55
  cmp    $0x2e,%al
  je     L71
  mov    %rbx,%rax
  jmp    L62
L71:
  add    $0x1,%rbx
L6:
  cmpb   $0x0,(%rbx)
  je     L109
L17:
  test   %rbp,%rbp
  je     L25
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ns_samename@plt>
  cmp    $0x1,%eax
  je     L126
L25:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L122
L109:
  mov    %rbx,%rax
  sub    %r12,%rax
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L75:
  mov    %rax,%rbx
  jmp    L6
L114:
  mov    %rdi,%rbx
  jmp    L109
```

assistant: ```c
size_t prune_origin(const char *name, const char *origin) {
  const char *oname = name;
  while (*name != '\0') {
    if (origin != NULL && ns_samename(name, origin) == 1)
      return (name - oname - (name > oname));
    while (*name != '\0') {
      if (*name == '\\') {
        name++;
        if (*name == '\0')
          break;
      } else if (*name == '.') {
        name++;
        break;
      }
      name++;
    }
  }
  return (name - oname);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prune_origin>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L38
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopw   0x0(%rax,%rax,1)
L101:
  test   %rbp,%rbp
  je     L44
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ns_samename@plt>
  cmp    $0x1,%eax
  je     L109
L44:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L36
  jmp    L115
  xchg   %ax,%ax
L114:
  cmp    $0x2e,%al
  je     L22
  mov    %rbx,%rdx
L21:
  movzbl 0x1(%rdx),%eax
  lea    0x1(%rdx),%rbx
  test   %al,%al
  je     L115
L36:
  lea    0x1(%rbx),%rdx
  cmp    $0x5c,%al
  jne    L114
  cmpb   $0x0,0x1(%rbx)
  jne    L21
L22:
  cmpb   $0x0,0x1(%rbx)
  je     L94
  mov    %rdx,%rbx
  jmp    L101
  nopl   0x0(%rax)
L115:
  mov    %rbx,%rax
  sub    %r12,%rax
L76:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L109:
  cmp    %rbx,%r12
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sbb    %r12,%rax
  pop    %r12
  ret
L38:
  xor    %eax,%eax
  ret
L94:
  mov    %rdx,%rax
  sub    %r12,%rax
  jmp    L76
```

assistant: ```c
size_t prune_origin(const char *name, const char *origin) {
  const char *oname = name;
  while (*name != '\0') {
    if (origin != NULL && ns_samename(name, origin) == 1)
      return (name - oname - (name > oname));
    while (*name != '\0') {
      if (*name == '\\') {
        name++;
        if (*name == '\0')
          break;
      } else if (*name == '.') {
        name++;
        break;
      }
      name++;
    }
  }
  return (name - oname);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<prune_origin>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L71
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax)
L45:
  test   %rbp,%rbp
  je     L38
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <ns_samename@plt>
  cmp    $0x1,%eax
  je     L21
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L22
L38:
  mov    %rbx,%rax
  jmp    L16
  nopl   0x0(%rax)
L7:
  cmp    $0x2e,%dl
  je     L96
  mov    %rax,%rbx
L75:
  movzbl 0x1(%rbx),%edx
  lea    0x1(%rbx),%rax
  test   %dl,%dl
  je     L84
L16:
  lea    0x1(%rax),%rbx
  cmp    $0x5c,%dl
  jne    L7
  cmpb   $0x0,0x1(%rax)
  jne    L75
L96:
  movzbl 0x1(%rax),%edx
  test   %dl,%dl
  jne    L45
  mov    %rbx,%rax
  sub    %r12,%rax
  jmp    L49
  nopl   0x0(%rax,%rax,1)
L22:
  mov    %rbx,%rax
  nopl   0x0(%rax,%rax,1)
L84:
  sub    %r12,%rax
L49:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  cmp    %rbx,%r12
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sbb    %r12,%rax
  pop    %r12
  ret
L71:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t prune_origin(const char *name, const char *origin) {
  const char *oname = name;
  while (*name != '\0') {
    if (origin != NULL && ns_samename(name, origin) == 1)
      return (name - oname - (name > oname));
    while (*name != '\0') {
      if (*name == '\\') {
        name++;
        if (*name == '\0')
          break;
      } else if (*name == '.') {
        name++;
        break;
      }
      name++;
    }
  }
  return (name - oname);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<netmask>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L8
  mov    $0x0,%eax
  jmp    L10
L8:
  mov    $0x20,%eax
  sub    -0x4(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  neg    %eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
in_addr_t netmask(int prefix) {
  if (prefix == 0)
    return (~((in_addr_t)-1));
  else
    return (~((1 << (32 - prefix)) - 1));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<netmask>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L3
  mov    $0x20,%ecx
  sub    %edi,%ecx
  mov    $0xffffffff,%eax
  shl    %cl,%eax
L3:
  ret
```

assistant: ```c
in_addr_t netmask(int prefix) {
  if (prefix == 0)
    return (~((in_addr_t)-1));
  else
    return (~((1 << (32 - prefix)) - 1));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<netmask>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L0
  mov    $0x20,%ecx
  mov    $0xffffffff,%eax
  sub    %edi,%ecx
  shl    %cl,%eax
L0:
  ret
```

assistant: ```c
in_addr_t netmask(int prefix) {
  if (prefix == 0)
    return (~((in_addr_t)-1));
  else
    return (~((1 << (32 - prefix)) - 1));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<netmask>:
  endbr64
  mov    %edi,%eax
  test   %edi,%edi
  je     L7
  mov    $0x20,%ecx
  mov    $0xffffffff,%eax
  sub    %edi,%ecx
  shl    %cl,%eax
L7:
  ret
```

assistant: ```c
in_addr_t netmask(int prefix) {
  if (prefix == 0)
    return (~((in_addr_t)-1));
  else
    return (~((1 << (32 - prefix)) - 1));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  jmp    L19
L4:
  mov    -0x14(%rbp),%eax
  cltq
  neg    %rax
  add    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bar@plt>
L19:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jbe    L4
  nop
  nop
  leave
  ret
```

assistant: ```c
void foo(char *left, char *rite, int element) {
  while (left <= rite) {
    rite -= element;
    bar(left, rite, element);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<foo>:
  endbr64
  cmp    %rdi,%rsi
  jb     L14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  movslq %edx,%r13
L15:
  sub    %r13,%rbx
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <bar@plt>
  cmp    %rbx,%rbp
  jbe    L15
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  ret
```

assistant: ```c
void foo(char *left, char *rite, int element) {
  while (left <= rite) {
    rite -= element;
    bar(left, rite, element);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %r13,%r12
  neg    %r13
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %rdi,%rsi
  jb     L5
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L4:
  add    %r13,%rbx
  mov    %r12d,%edx
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  call   <bar@plt>
  cmp    %rbx,%rbp
  jbe    L4
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void foo(char *left, char *rite, int element) {
  while (left <= rite) {
    rite -= element;
    bar(left, rite, element);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  push   %r13
  movslq %edx,%r13
  push   %r12
  mov    %r13,%r12
  neg    %r13
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %rdi,%rsi
  jb     L1
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L3:
  add    %r13,%rbx
  mov    %r12d,%edx
  mov    %rbp,%rdi
  mov    %rbx,%rsi
  call   <bar@plt>
  cmp    %rbx,%rbp
  jbe    L3
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void foo(char *left, char *rite, int element) {
  while (left <= rite) {
    rite -= element;
    bar(left, rite, element);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<graph_id>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  js     L23
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L23
  cmpl   $0x0,-0xc(%rbp)
  js     L23
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L28
L23:
  mov    $0xffffffff,%eax
  jmp    L7
L28:
  mov    -0xc(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
int graph_id(int width, int height, int x, int y) {
  if (y < 0 || y >= height || x < 0 || x >= width) {
    return -1;
  }
  return (x * width) + y;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<graph_id>:
  endbr64
  mov    %ecx,%eax
  or     %edx,%eax
  js     L2
  cmp    %esi,%ecx
  jge    L2
  cmp    %edi,%edx
  jge    L20
  imul   %edi,%edx
  lea    (%rdx,%rcx,1),%eax
  ret
L2:
  mov    $0xffffffff,%eax
  ret
L20:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int graph_id(int width, int height, int x, int y) {
  if (y < 0 || y >= height || x < 0 || x >= width) {
    return -1;
  }
  return (x * width) + y;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<graph_id>:
  endbr64
  mov    %ecx,%eax
  or     %edx,%eax
  js     L8
  cmp    %esi,%ecx
  jge    L8
  cmp    %edi,%edx
  jge    L8
  imul   %edi,%edx
  lea    (%rdx,%rcx,1),%eax
  ret
  nopl   0x0(%rax)
L8:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int graph_id(int width, int height, int x, int y) {
  if (y < 0 || y >= height || x < 0 || x >= width) {
    return -1;
  }
  return (x * width) + y;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<graph_id>:
  endbr64
  mov    %ecx,%eax
  or     %edx,%eax
  js     L4
  cmp    %esi,%ecx
  jge    L4
  cmp    %edi,%edx
  jge    L4
  imul   %edi,%edx
  lea    (%rdx,%rcx,1),%eax
  ret
  nopl   0x0(%rax)
L4:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int graph_id(int width, int height, int x, int y) {
  if (y < 0 || y >= height || x < 0 || x >= width) {
    return -1;
  }
  return (x * width) + y;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<h32_to_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <htonl@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void h32_to_bytes(char *bytes, uint32_t value) {
  value = htonl(value);
  *(uint32_t *)bytes = value;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<h32_to_bytes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  call   <htonl@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void h32_to_bytes(char *bytes, uint32_t value) {
  value = htonl(value);
  *(uint32_t *)bytes = value;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<h32_to_bytes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  call   <htonl@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void h32_to_bytes(char *bytes, uint32_t value) {
  value = htonl(value);
  *(uint32_t *)bytes = value;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<h32_to_bytes>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  call   <htonl@plt>
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void h32_to_bytes(char *bytes, uint32_t value) {
  value = htonl(value);
  *(uint32_t *)bytes = value;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<kernel>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rdx
  mov    %rsi,%r8
  mov    $0x400,%r15d
  mov    $0x2000,%r14d
  movq   $0x0,-0x8(%rsp)
  mov    $0xfffffffffffff000,%r13
L3:
  mov    $0xf8,%ecx
  sub    %r13,%rcx
  mov    $0x0,%r12d
  mov    $0x100,%ebp
  mov    $0xffffffffffffff80,%r11
  mov    $0x20,%ebx
  mov    -0x8(%rsp),%rax
  lea    0x0(%r13,%rax,1),%r10
  add    %rdx,%r10
  lea    0x0(%r13,%r14,1),%r9
  add    %rdx,%r9
L8:
  add    $0x20,%ebx
  lea    -0x78(%rcx),%rax
  lea    (%r11,%rbp,1),%rdi
  add    %rdx,%rdi
  lea    (%r11,%r12,1),%rsi
  add    %rdx,%rsi
L15:
  movss  0x4(%rdx,%rax,1),%xmm0
  addss  0x8(%rdx,%rax,1),%xmm0
  addss  (%rdx,%rax,1),%xmm0
  addss  0x4(%rdi,%rax,1),%xmm0
  addss  0x4(%rsi,%rax,1),%xmm0
  addss  0x4(%r10,%rax,1),%xmm0
  addss  0x4(%r9,%rax,1),%xmm0
  movss  %xmm0,0x4(%r8,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L15
  add    $0xffffffffffffff80,%r11
  sub    $0xffffffffffffff80,%rbp
  sub    $0xffffffffffffff80,%r12
  sub    $0xffffffffffffff80,%rcx
  cmp    $0x3e0,%ebx
  jne    L8
  sub    $0x1000,%r13
  addq   $0x1000,-0x8(%rsp)
  add    $0x1000,%r14
  add    $0x400,%r15
  cmp    $0x7c00,%r15
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void kernel(float *g1, float *g2) {
  int x, y, z;
  int halo_width = 1;
  for (z = halo_width; z < 32 - halo_width; ++z) {
    for (y = halo_width; y < 32 - halo_width; ++y) {
      for (x = halo_width; x < 32 - halo_width; ++x) {
        float v = g1[((x) + (y)*32 + (z)*32 * 32)] + g1[((x + 1) + (y)*32 + (z)*32 * 32)] + g1[((x - 1) + (y)*32 + (z)*32 * 32)] + g1[((x) + (y + 1) * 32 + (z)*32 * 32)] + g1[((x) + (y - 1) * 32 + (z)*32 * 32)] + g1[((x) + (y)*32 + (z - 1) * 32 * 32)] + g1[((x) + (y)*32 + (z + 1) * 32 * 32)];
        g2[((x) + (y)*32 + (z)*32 * 32)] = v;
      }
    }
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kernel>:
  endbr64
  push   %r15
  mov    %rdi,%rdx
  mov    %rsi,%r10
  mov    $0x2000,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    $0xfffffffffffff000,%r13
  push   %r12
  push   %rbp
  push   %rbx
  movq   $0x400,-0x8(%rsp)
L8:
  mov    $0xf8,%edi
  lea    0x0(%r13,%r14,1),%r9
  lea    (%r15,%r13,1),%r8
  mov    $0xffffffffffffff80,%r11
  sub    %r13,%rdi
  xor    %r12d,%r12d
  mov    $0x100,%ebp
  mov    $0x20,%ebx
  add    %rdx,%r9
  add    %rdx,%r8
  nopl   0x0(%rax,%rax,1)
L2:
  lea    (%r11,%rbp,1),%rsi
  lea    (%r11,%r12,1),%rcx
  add    $0x20,%ebx
  lea    -0x78(%rdi),%rax
  add    %rdx,%rsi
  add    %rdx,%rcx
  nopl   (%rax)
L28:
  movss  0x4(%rdx,%rax,1),%xmm0
  addss  0x8(%rdx,%rax,1),%xmm0
  addss  (%rdx,%rax,1),%xmm0
  addss  0x4(%rsi,%rax,1),%xmm0
  addss  0x4(%rcx,%rax,1),%xmm0
  addss  0x4(%r9,%rax,1),%xmm0
  addss  0x4(%r8,%rax,1),%xmm0
  movss  %xmm0,0x4(%r10,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rdi
  jne    L28
  sub    $0xffffffffffffff80,%rbp
  sub    $0xffffffffffffff80,%r12
  add    $0xffffffffffffff80,%r11
  sub    $0xffffffffffffff80,%rdi
  cmp    $0x3e0,%ebx
  jne    L2
  addq   $0x400,-0x8(%rsp)
  mov    -0x8(%rsp),%rax
  add    $0x1000,%r14
  sub    $0x1000,%r13
  add    $0x1000,%r15
  cmp    $0x7c00,%rax
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void kernel(float *g1, float *g2) {
  int x, y, z;
  int halo_width = 1;
  for (z = halo_width; z < 32 - halo_width; ++z) {
    for (y = halo_width; y < 32 - halo_width; ++y) {
      for (x = halo_width; x < 32 - halo_width; ++x) {
        float v = g1[((x) + (y)*32 + (z)*32 * 32)] + g1[((x + 1) + (y)*32 + (z)*32 * 32)] + g1[((x - 1) + (y)*32 + (z)*32 * 32)] + g1[((x) + (y + 1) * 32 + (z)*32 * 32)] + g1[((x) + (y - 1) * 32 + (z)*32 * 32)] + g1[((x) + (y)*32 + (z - 1) * 32 * 32)] + g1[((x) + (y)*32 + (z + 1) * 32 * 32)];
        g2[((x) + (y)*32 + (z)*32 * 32)] = v;
      }
    }
  }
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L31
L20:
  subq   $0x1,-0x8(%rbp)
L31:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jbe    L94
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L20
L94:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,(%rax)
  jmp    L9
L71:
  subq   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jbe    L103
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L71
  jmp    L103
L60:
  addq   $0x1,-0x8(%rbp)
L103:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L66
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L60
L66:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L104
  mov    -0x18(%rbp),%rax
  jmp    L46
L104:
  mov    -0x8(%rbp),%rax
L46:
  leave
  ret
```

assistant: ```c
char *basename(char *path) {
  char *startp = path + strlen(path) - 1;
  while (startp > path && *startp == '/')
    startp--;
  startp[1] = '\0';
  while (startp > path && *startp != '/')
    startp--;
  while (*startp && *startp == '/')
    startp++;
  if (!*startp)
    return path;
  return startp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<basename>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jae    L40
L48:
  cmpb   $0x2f,(%rax)
  jne    L5
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jne    L48
  mov    %rbx,%rax
L40:
  movb   $0x0,0x1(%rax)
  jmp    L53
L5:
  movb   $0x0,0x1(%rax)
  cmp    %rax,%rbx
  jae    L53
L28:
  cmpb   $0x2f,(%rax)
  je     L55
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jne    L28
  mov    %rbx,%rax
L53:
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  jne    L15
L55:
  add    $0x1,%rax
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  je     L55
L15:
  test   %dl,%dl
  cmove  %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *basename(char *path) {
  char *startp = path + strlen(path) - 1;
  while (startp > path && *startp == '/')
    startp--;
  startp[1] = '\0';
  while (startp > path && *startp != '/')
    startp--;
  while (*startp && *startp == '/')
    startp++;
  if (!*startp)
    return path;
  return startp;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<basename>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmp    %rax,%rbx
  jb     L36
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L48:
  sub    $0x1,%rax
  cmp    %rax,%rbx
  je     L10
L36:
  cmpb   $0x2f,(%rax)
  je     L48
  movb   $0x0,0x1(%rax)
  cmp    %rax,%rbx
  jae    L40
  nopw   0x0(%rax,%rax,1)
L23:
  sub    $0x1,%rax
  movzbl (%rax),%edx
  cmp    %rax,%rbx
  je     L40
  cmp    $0x2f,%dl
  jne    L23
  nopl   0x0(%rax)
L51:
  add    $0x1,%rax
L40:
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  je     L51
  test   %dl,%dl
  cmove  %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L10:
  movb   $0x0,0x1(%rax)
  jmp    L40
```

assistant: ```c
char *basename(char *path) {
  char *startp = path + strlen(path) - 1;
  while (startp > path && *startp == '/')
    startp--;
  startp[1] = '\0';
  while (startp > path && *startp != '/')
    startp--;
  while (*startp && *startp == '/')
    startp++;
  if (!*startp)
    return path;
  return startp;
}
```