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
<atomicio>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L26
L38:
  mov    -0x40(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rsi
  mov    -0x2c(%rbp),%eax
  mov    -0x28(%rbp),%rcx
  mov    %eax,%edi
  call   *%rcx
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L8
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  jne    L42
  mov    D34(%rip),%rax
  mov    (%rax),%edx
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L26
  mov    D34(%rip),%rax
  mov    (%rax),%edx
  mov    D37(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L8
  jmp    L26
L8:
  cmpq   $0x0,-0x18(%rbp)
  je     L7
  mov    -0x18(%rbp),%rax
  jmp    L46
L7:
  mov    -0x8(%rbp),%rax
  jmp    L46
L42:
  mov    -0x8(%rbp),%rax
  add    %rax,-0x18(%rbp)
L26:
  mov    -0x40(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  ja     L38
  mov    -0x18(%rbp),%rax
L46:
  leave
  ret
```

assistant: ```c
ssize_t atomicio(ssize_t (*f)(int, void *, size_t), int fd, void *_s, size_t n) {
  char *s = _s;
  ssize_t res, pos = 0;
  while (n > pos) {
    res = (f)(fd, s + pos, n - pos);
    switch (res) {
    case -1:
      if (errno == EINTR || errno == EAGAIN)
        continue;
    case 0:
      if (pos != 0)
        return (pos);
      return (res);
    default:
      pos += res;
    }
  }
  return (pos);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<atomicio>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rcx,%rbp
  test   %rcx,%rcx
  je     L11
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %rdx,%r14
  mov    $0x0,%ebx
L57:
  mov    %rbp,%rdx
  sub    %rbx,%rdx
  lea    (%r14,%rbx,1),%rsi
  mov    %r13d,%edi
  call   *%r12
  test   %rax,%rax
  je     L26
  cmp    $0xffffffffffffffff,%rax
  jne    L69
  mov    D22(%rip),%rdx
  mov    (%rdx),%edx
  mov    D1(%rip),%rcx
  cmp    (%rcx),%edx
  je     L52
  mov    D4(%rip),%rcx
  cmp    (%rcx),%edx
  je     L52
L26:
  test   %rbx,%rbx
  cmovne %rbx,%rax
L45:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L69:
  add    %rax,%rbx
L52:
  cmp    %rbp,%rbx
  jb     L57
  mov    %rbx,%rax
  jmp    L45
L11:
  mov    %rcx,%rax
  jmp    L45
```

assistant: ```c
ssize_t atomicio(ssize_t (*f)(int, void *, size_t), int fd, void *_s, size_t n) {
  char *s = _s;
  ssize_t res, pos = 0;
  while (n > pos) {
    res = (f)(fd, s + pos, n - pos);
    switch (res) {
    case -1:
      if (errno == EINTR || errno == EAGAIN)
        continue;
    case 0:
      if (pos != 0)
        return (pos);
      return (res);
    default:
      pos += res;
    }
  }
  return (pos);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<atomicio>:
  endbr64
  test   %rcx,%rcx
  je     L41
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  xor    %ebx,%ebx
L5:
  mov    %rbp,%rdx
  lea    0x0(%r13,%rbx,1),%rsi
  mov    %r14d,%edi
  sub    %rbx,%rdx
  call   *%r12
  test   %rax,%rax
  je     L57
  cmp    $0xffffffffffffffff,%rax
  jne    L43
  mov    D27(%rip),%rdx
  mov    D19(%rip),%rcx
  mov    (%rdx),%edx
  cmp    (%rcx),%edx
  je     L33
  mov    D8(%rip),%rcx
  cmp    (%rcx),%edx
  je     L33
L57:
  test   %rbx,%rbx
  cmovne %rbx,%rax
L30:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  add    %rax,%rbx
L33:
  cmp    %rbx,%rbp
  ja     L5
  mov    %rbx,%rax
  jmp    L30
  nopl   (%rax)
L41:
  xor    %eax,%eax
  ret
```

assistant: ```c
ssize_t atomicio(ssize_t (*f)(int, void *, size_t), int fd, void *_s, size_t n) {
  char *s = _s;
  ssize_t res, pos = 0;
  while (n > pos) {
    res = (f)(fd, s + pos, n - pos);
    switch (res) {
    case -1:
      if (errno == EINTR || errno == EAGAIN)
        continue;
    case 0:
      if (pos != 0)
        return (pos);
      return (res);
    default:
      pos += res;
    }
  }
  return (pos);
}
```
user: What is the c source code of the assembly code below?

```asm
<atomicio>:
  endbr64
  test   %rcx,%rcx
  je     L21
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  xor    %ebx,%ebx
L51:
  mov    %rbp,%rdx
  lea    0x0(%r13,%rbx,1),%rsi
  mov    %r14d,%edi
  sub    %rbx,%rdx
  call   *%r12
  test   %rax,%rax
  je     L1
  cmp    $0xffffffffffffffff,%rax
  jne    L30
  mov    D2(%rip),%rdx
  mov    D6(%rip),%rcx
  mov    (%rdx),%edx
  cmp    (%rcx),%edx
  je     L23
  mov    D21(%rip),%rcx
  cmp    (%rcx),%edx
  je     L23
L1:
  test   %rbx,%rbx
  cmovne %rbx,%rax
L55:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  add    %rax,%rbx
L23:
  cmp    %rbx,%rbp
  ja     L51
  mov    %rbx,%rax
  jmp    L55
  nopl   (%rax)
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
ssize_t atomicio(ssize_t (*f)(int, void *, size_t), int fd, void *_s, size_t n) {
  char *s = _s;
  ssize_t res, pos = 0;
  while (n > pos) {
    res = (f)(fd, s + pos, n - pos);
    switch (res) {
    case -1:
      if (errno == EINTR || errno == EAGAIN)
        continue;
    case 0:
      if (pos != 0)
        return (pos);
      return (res);
    default:
      pos += res;
    }
  }
  return (pos);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L10
L5:
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0xc(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <pow@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L10:
  cmpl   $0x19,-0xc(%rbp)
  jle    L5
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_guessed: string
name: unpack_data
arguments: data_tag: D26
type_guessed: string
name: unpack_data
arguments: data_tag: D15
type_guessed: string

tool: D2: %i	

tool: D26: %f	

tool: D15: %f	


assistant: ```c
int main(int argc, char **argv) {
  int x;
  double y;
  for (x = 1; x <= 25; x++) {
    y = x;
    printf("%i	", x);
    y = sqrt(x);
    printf("%f	", y);
    y = pow(x, (1 / 3.0));
    printf("%f	\n", y);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%ebx
  lea    D16(%rip),%r13
  lea    D18(%rip),%r12
  lea    D27(%rip),%rbp
L3:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <pow@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x1a,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
type: string
name: retrieve_data
arguments: label: D18
type: string
name: retrieve_data
arguments: label: D27
type: string

tool: D16: %i	

tool: D18: %f	

tool: D27: %f	


assistant: ```c
int main(int argc, char **argv) {
  int x;
  double y;
  for (x = 1; x <= 25; x++) {
    y = x;
    printf("%i	", x);
    y = sqrt(x);
    printf("%f	", y);
    y = pow(x, (1 / 3.0));
    printf("%f	\n", y);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  lea    D11(%rip),%r13
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L6:
  pxor   %xmm0,%xmm0
  mov    %r13,%rdi
  cvtsi2sd %ebx,%xmm0
  call   <printf@plt>
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  xor    %esi,%esi
  add    $0x1,%ebx
  call   <pow@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x1a,%ebx
  jne    L6
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D11
type_guessed: string
name: disassemble_data
arguments: tag: D5
type_guessed: string
name: disassemble_data
arguments: tag: D8
type_guessed: string

tool: D11: %i	

tool: D5: %f	

tool: D8: %f	


assistant: ```c
int main(int argc, char **argv) {
  int x;
  double y;
  for (x = 1; x <= 25; x++) {
    y = x;
    printf("%i	", x);
    y = sqrt(x);
    printf("%f	", y);
    y = pow(x, (1 / 3.0));
    printf("%f	\n", y);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D3(%rip),%r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  lea    D25(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  pxor   %xmm0,%xmm0
  mov    %r13,%rdi
  cvtsi2sd %ebx,%xmm0
  call   <printf@plt>
  mov    %ebx,%edi
  call   <sqrt@plt>
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  xor    %esi,%esi
  add    $0x1,%ebx
  call   <pow@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x1a,%ebx
  jne    L1
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type: string
name: retrieve_data
arguments: tag: D10
type: string
name: retrieve_data
arguments: tag: D25
type: string

tool: D3: %i	

tool: D10: %f	

tool: D25: %f	


assistant: ```c
int main(int argc, char **argv) {
  int x;
  double y;
  for (x = 1; x <= 25; x++) {
    y = x;
    printf("%i	", x);
    y = sqrt(x);
    printf("%f	", y);
    y = pow(x, (1 / 3.0));
    printf("%f	\n", y);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<dcc_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x4,-0x4(%rbp)
  jmp    L18
L27:
  nop
L15:
  mov    $0xe000edf8,%eax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L15
  mov    -0x14(%rbp),%eax
  shl    $0x8,%eax
  or     $0x1,%eax
  mov    %eax,%edx
  mov    $0xe000edf8,%eax
  mov    %dx,(%rax)
  sarl   $0x8,-0x14(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L27
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dcc_write(uint32_t dcc_data) {
  int len = 4;
  while (len--) {
    while ((*((volatile unsigned short *)0xE000EDF8)) & 1)
      ;
    (*((volatile unsigned short *)0xE000EDF8)) = (unsigned short)(((dcc_data & 0xff) << 8) | 1);
    dcc_data >>= 8;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<dcc_write>:
  endbr64
  mov    $0x4,%ecx
  mov    $0xe000edf8,%edx
L15:
  movzwl (%rdx),%eax
  test   $0x1,%al
  jne    L15
  mov    %edi,%eax
  shl    $0x8,%eax
  or     $0x1,%eax
  mov    %ax,(%rdx)
  sar    $0x8,%edi
  sub    $0x1,%ecx
  jne    L15
  ret
```

assistant: ```c
void dcc_write(uint32_t dcc_data) {
  int len = 4;
  while (len--) {
    while ((*((volatile unsigned short *)0xE000EDF8)) & 1)
      ;
    (*((volatile unsigned short *)0xE000EDF8)) = (unsigned short)(((dcc_data & 0xff) << 8) | 1);
    dcc_data >>= 8;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dcc_write>:
  endbr64
  mov    $0x4,%ecx
  mov    $0xe000edf8,%edx
  xchg   %ax,%ax
L17:
  movzwl (%rdx),%eax
  test   $0x1,%al
  jne    L17
  mov    %edi,%eax
  sar    $0x8,%edi
  shl    $0x8,%eax
  or     $0x1,%eax
  mov    %ax,(%rdx)
  sub    $0x1,%ecx
  jne    L17
  ret
```

assistant: ```c
void dcc_write(uint32_t dcc_data) {
  int len = 4;
  while (len--) {
    while ((*((volatile unsigned short *)0xE000EDF8)) & 1)
      ;
    (*((volatile unsigned short *)0xE000EDF8)) = (unsigned short)(((dcc_data & 0xff) << 8) | 1);
    dcc_data >>= 8;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dcc_write>:
  endbr64
  mov    $0x4,%ecx
  mov    $0xe000edf8,%edx
  xchg   %ax,%ax
L4:
  movzwl (%rdx),%eax
  test   $0x1,%al
  jne    L4
  mov    %edi,%eax
  sar    $0x8,%edi
  shl    $0x8,%eax
  or     $0x1,%eax
  mov    %ax,(%rdx)
  sub    $0x1,%ecx
  jne    L4
  ret
```

assistant: ```c
void dcc_write(uint32_t dcc_data) {
  int len = 4;
  while (len--) {
    while ((*((volatile unsigned short *)0xE000EDF8)) & 1)
      ;
    (*((volatile unsigned short *)0xE000EDF8)) = (unsigned short)(((dcc_data & 0xff) << 8) | 1);
    dcc_data >>= 8;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Round>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    0x10(%rbp),%rax
  add    %rax,%rdx
  mov    0x18(%rbp),%rax
  add    %rax,%rdx
  mov    0x20(%rbp),%eax
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <rol@plt>
  mov    -0x28(%rbp),%rdx
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <rol@plt>
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void Round(uint32_t *a, uint32_t b, uint32_t *c, uint32_t d, uint32_t e, uint32_t f, uint32_t x, uint32_t k, int r) {
  *a = rol(*a + f + x + k, r) + e;
  *c = rol(*c, 10);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Round>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rdx,%rbx
  mov    %r8,%r12
  mov    %r9,%rdi
  add    0x0(%rbp),%rdi
  add    0x20(%rsp),%rdi
  add    0x28(%rsp),%rdi
  mov    0x30(%rsp),%esi
  call   <rol@plt>
  add    %r12,%rax
  mov    %rax,0x0(%rbp)
  mov    $0xa,%esi
  mov    (%rbx),%rdi
  call   <rol@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Round(uint32_t *a, uint32_t b, uint32_t *c, uint32_t d, uint32_t e, uint32_t f, uint32_t x, uint32_t k, int r) {
  *a = rol(*a + f + x + k, r) + e;
  *c = rol(*c, 10);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Round>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  mov    %rdx,%rbx
  mov    0x30(%rsp),%esi
  add    %r9,%rdi
  add    0x20(%rsp),%rdi
  add    0x28(%rsp),%rdi
  call   <rol@plt>
  mov    $0xa,%esi
  add    %rbp,%rax
  mov    %rax,(%r12)
  mov    (%rbx),%rdi
  call   <rol@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Round(uint32_t *a, uint32_t b, uint32_t *c, uint32_t d, uint32_t e, uint32_t f, uint32_t x, uint32_t k, int r) {
  *a = rol(*a + f + x + k, r) + e;
  *c = rol(*c, 10);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Round>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  mov    %rdx,%rbx
  mov    0x30(%rsp),%esi
  add    %r9,%rdi
  add    0x20(%rsp),%rdi
  add    0x28(%rsp),%rdi
  call   <rol@plt>
  mov    $0xa,%esi
  add    %rbp,%rax
  mov    %rax,(%r12)
  mov    (%rbx),%rdi
  call   <rol@plt>
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Round(uint32_t *a, uint32_t b, uint32_t *c, uint32_t d, uint32_t e, uint32_t f, uint32_t x, uint32_t k, int r) {
  *a = rol(*a + f + x + k, r) + e;
  *c = rol(*c, 10);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<t17>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2008(%rsp)
  xor    %eax,%eax
  mov    $0x6,%ebx
  lea    -0x4(%rsp),%r12
  jmp    L26
L14:
  mov    $0x0,%r13d
  lea    0x1000(%rsp),%r14
L56:
  mov    %r13d,%eax
  xor    $0x1f,%eax
  shl    $0x3,%eax
  cmp    (%r14,%r13,4),%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x1,%r13
  cmp    $0x400,%r13
  jne    L56
  sub    $0x1,%ebp
  je     L74
L13:
  mov    $0x0,%eax
L33:
  mov    %eax,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x400,%rax
  jne    L33
  mov    %rsp,%r8
  lea    0x107c(%rsp),%rdi
  lea    0xffc(%rsp),%rsi
  mov    %r12,%r10
  lea    0x1ffc(%rsp),%r9
L18:
  mov    %rdi,%rax
  mov    %r8,%rcx
L93:
  mov    (%rcx),%edx
  add    %edx,%edx
  mov    %edx,(%rax)
  add    $0x4,%rcx
  sub    $0x4,%rax
  cmp    %rsi,%rax
  jne    L93
  lea    0x4(%rsi),%rcx
  lea    0x80(%r10),%r11
  mov    %r11,%rax
L54:
  mov    (%rcx),%edx
  add    %edx,%edx
  mov    %edx,(%rax)
  add    $0x4,%rcx
  sub    $0x4,%rax
  cmp    %r10,%rax
  jne    L54
  mov    %rdi,%rax
  mov    %r8,%rcx
L9:
  mov    (%rcx),%edx
  add    %edx,%edx
  mov    %edx,(%rax)
  add    $0x4,%rcx
  sub    $0x4,%rax
  cmp    %rsi,%rax
  jne    L9
  sub    $0xffffffffffffff80,%r8
  sub    $0xffffffffffffff80,%rdi
  sub    $0xffffffffffffff80,%rsi
  cmp    %r9,%rsi
  je     L14
  mov    %r11,%r10
  jmp    L18
L74:
  sub    $0x1,%ebx
  je     L47
L26:
  mov    $0x6,%ebp
  jmp    L13
L47:
  mov    0x2008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  add    $0x2010,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void t17() {
  int arr_a[32 * 32], arr_b[32 * 32], i;
  int num_workers, num_gangs;
  for (num_workers = 1; num_workers <= 32; num_workers <<= 1)
    for (num_gangs = 1; num_gangs <= 32; num_gangs <<= 1) {
      for (i = 0; i < 32 * 32; i++)
        arr_a[i] = i;
#pragma acc parallel copyin(arr_a) copyout(arr_b) num_gangs(num_gangs) num_workers(num_workers) vector_length(32)
      {
        int j;
#pragma acc loop gang
        for (j = 0; j < 32; j++) {
          int k;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_b[j * 32 + (31 - k)] = arr_a[j * 32 + k] * 2;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_a[j * 32 + (31 - k)] = arr_b[j * 32 + k] * 2;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_b[j * 32 + (31 - k)] = arr_a[j * 32 + k] * 2;
        }
      }
      for (i = 0; i < 32 * 32; i++)
        assert(arr_b[i] == (i ^ 31) * 8);
    }
}
```
user: What is the c source code of the assembly code below:

```asm
<t17>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x38,%rsp
  mov    $0x6,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x2028(%rsp)
  xor    %eax,%eax
  lea    0x109c(%rsp),%rax
  lea    0x20(%rsp),%rbx
  mov    %rax,0x8(%rsp)
  lea    0x101c(%rsp),%rax
  lea    0x201c(%rsp),%r12
  mov    %rax,0x18(%rsp)
  lea    0x1c(%rsp),%rax
  lea    0x1020(%rsp),%rbp
  mov    %rax,0x10(%rsp)
L24:
  mov    $0x6,%r14d
L36:
  xor    %eax,%eax
  nopl   0x0(%rax)
L11:
  mov    %eax,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x400,%rax
  jne    L11
  mov    0x8(%rsp),%rdi
  mov    0x18(%rsp),%rsi
  mov    %rbx,%r8
  mov    0x10(%rsp),%r9
  nopl   0x0(%rax)
L90:
  mov    %rdi,%rax
  mov    %r8,%rcx
  cs nopw 0x0(%rax,%rax,1)
L33:
  mov    (%rcx),%edx
  sub    $0x4,%rax
  add    $0x4,%rcx
  add    %edx,%edx
  mov    %edx,0x4(%rax)
  cmp    %rsi,%rax
  jne    L33
  lea    0x80(%r9),%r10
  lea    0x4(%rsi),%rcx
  mov    %r10,%rax
  nopw   0x0(%rax,%rax,1)
L45:
  mov    (%rcx),%edx
  sub    $0x4,%rax
  add    $0x4,%rcx
  add    %edx,%edx
  mov    %edx,0x4(%rax)
  cmp    %r9,%rax
  jne    L45
  mov    %rdi,%rax
  mov    %r8,%rcx
  nopw   0x0(%rax,%rax,1)
L79:
  mov    (%rcx),%edx
  sub    $0x4,%rax
  add    $0x4,%rcx
  add    %edx,%edx
  mov    %edx,0x4(%rax)
  cmp    %rsi,%rax
  jne    L79
  sub    $0xffffffffffffff80,%rsi
  sub    $0xffffffffffffff80,%r8
  sub    $0xffffffffffffff80,%rdi
  cmp    %r12,%rsi
  je     L29
  mov    %r10,%r9
  jmp    L90
L29:
  xor    %r15d,%r15d
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r15d,%eax
  xor    %edi,%edi
  xor    $0x1f,%eax
  shl    $0x3,%eax
  cmp    0x0(%rbp,%r15,4),%eax
  sete   %dil
  add    $0x1,%r15
  call   <assert@plt>
  cmp    $0x400,%r15
  jne    L13
  sub    $0x1,%r14d
  jne    L36
  sub    $0x1,%r13d
  jne    L24
  mov    0x2028(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x2038,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void t17() {
  int arr_a[32 * 32], arr_b[32 * 32], i;
  int num_workers, num_gangs;
  for (num_workers = 1; num_workers <= 32; num_workers <<= 1)
    for (num_gangs = 1; num_gangs <= 32; num_gangs <<= 1) {
      for (i = 0; i < 32 * 32; i++)
        arr_a[i] = i;
#pragma acc parallel copyin(arr_a) copyout(arr_b) num_gangs(num_gangs) num_workers(num_workers) vector_length(32)
      {
        int j;
#pragma acc loop gang
        for (j = 0; j < 32; j++) {
          int k;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_b[j * 32 + (31 - k)] = arr_a[j * 32 + k] * 2;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_a[j * 32 + (31 - k)] = arr_b[j * 32 + k] * 2;
#pragma acc loop worker
          for (k = 0; k < 32; k++)
            arr_b[j * 32 + (31 - k)] = arr_a[j * 32 + k] * 2;
        }
      }
      for (i = 0; i < 32 * 32; i++)
        assert(arr_b[i] == (i ^ 31) * 8);
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ne_path_has_trailing_slash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L26
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L26
  mov    $0x1,%eax
  jmp    L12
L26:
  mov    $0x0,%eax
L12:
  leave
  ret
```

assistant: ```c
int ne_path_has_trailing_slash(const char *uri) {
  size_t len = strlen(uri);
  return ((len > 0) && (uri[len - 1] == '/'));
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ne_path_has_trailing_slash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    $0x0,%edx
  test   %rax,%rax
  je     L4
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  sete   %dl
  movzbl %dl,%edx
L4:
  mov    %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ne_path_has_trailing_slash(const char *uri) {
  size_t len = strlen(uri);
  return ((len > 0) && (uri[len - 1] == '/'));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ne_path_has_trailing_slash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L2
  xor    %r8d,%r8d
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  sete   %r8b
L2:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ne_path_has_trailing_slash(const char *uri) {
  size_t len = strlen(uri);
  return ((len > 0) && (uri[len - 1] == '/'));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ne_path_has_trailing_slash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L0
  xor    %r8d,%r8d
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  sete   %r8b
L0:
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ne_path_has_trailing_slash(const char *uri) {
  size_t len = strlen(uri);
  return ((len > 0) && (uri[len - 1] == '/'));
}
```