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
<isinteger>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L15
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L5
L15:
  addq   $0x1,-0x8(%rbp)
  jmp    L5
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  je     L23
  addq   $0x1,-0x8(%rbp)
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  jmp    L20
L23:
  nop
L20:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int isinteger(const char *s) {
  if (((unsigned char)*s) == '-' || ((unsigned char)*s) == '+')
    s++;
  while (*s) {
    if (!isdigit((unsigned char)*s))
      break;
    s++;
  }
  return (*s == '\0');
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isinteger>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%eax
  sub    $0x2b,%eax
  and    $0xfffffffd,%eax
  cmp    $0x1,%al
  adc    $0x0,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L5
L18:
  movzbl %dil,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  je     L5
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L18
L5:
  cmpb   $0x0,(%rbx)
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int isinteger(const char *s) {
  if (((unsigned char)*s) == '-' || ((unsigned char)*s) == '+')
    s++;
  while (*s) {
    if (!isdigit((unsigned char)*s))
      break;
    s++;
  }
  return (*s == '\0');
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<isinteger>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  lea    -0x2b(%rdi),%eax
  test   $0xfd,%al
  jne    L19
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  jmp    L19
  nopl   0x0(%rax)
L13:
  call   <isdigit@plt>
  test   %eax,%eax
  je     L0
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
L19:
  test   %dil,%dil
  jne    L13
  xor    %eax,%eax
  test   %dil,%dil
  pop    %rbx
  sete   %al
  ret
L0:
  movzbl (%rbx),%edi
  xor    %eax,%eax
  pop    %rbx
  test   %dil,%dil
  sete   %al
  ret
```

assistant: ```c
int isinteger(const char *s) {
  if (((unsigned char)*s) == '-' || ((unsigned char)*s) == '+')
    s++;
  while (*s) {
    if (!isdigit((unsigned char)*s))
      break;
    s++;
  }
  return (*s == '\0');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<isinteger>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  lea    -0x2b(%rdi),%eax
  test   $0xfd,%al
  jne    L8
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  jmp    L8
  nopl   0x0(%rax)
L0:
  call   <isdigit@plt>
  test   %eax,%eax
  je     L13
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
L8:
  test   %dil,%dil
  jne    L0
  xor    %eax,%eax
  test   %dil,%dil
  pop    %rbx
  sete   %al
  ret
L13:
  movzbl (%rbx),%edi
  xor    %eax,%eax
  pop    %rbx
  test   %dil,%dil
  sete   %al
  ret
```

assistant: ```c
int isinteger(const char *s) {
  if (((unsigned char)*s) == '-' || ((unsigned char)*s) == '+')
    s++;
  while (*s) {
    if (!isdigit((unsigned char)*s))
      break;
    s++;
  }
  return (*s == '\0');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x2065737265766572,%rax
  mov    %rax,-0x15(%rbp)
  movl   $0x73696874,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0x15(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: string

tool: D7: %s
%s


assistant: ```c
int main() {
  char str[] = "reverse this";
  char *reversed = reverse(str);
  printf("%s\n%s\n", str, reversed);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x2065737265766572,%rax
  mov    %rax,0xb(%rsp)
  movl   $0x73696874,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  lea    0xb(%rsp),%rbx
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rax,%rdx
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D4
type_infered: string

tool: D4: %s
%s


assistant: ```c
int main() {
  char str[] = "reverse this";
  char *reversed = reverse(str);
  printf("%s\n%s\n", str, reversed);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x2065737265766572,%rax
  lea    0xb(%rsp),%rbp
  movl   $0x73696874,0x13(%rsp)
  mov    %rbp,%rdi
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xb(%rsp)
  call   <reverse@plt>
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  mov    %rax,%rdx
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
datatype: string

tool: D4: %s
%s


assistant: ```c
int main() {
  char str[] = "reverse this";
  char *reversed = reverse(str);
  printf("%s\n%s\n", str, reversed);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x2065737265766572,%rax
  lea    0xb(%rsp),%rbp
  movl   $0x73696874,0x13(%rsp)
  mov    %rbp,%rdi
  movb   $0x0,0x17(%rsp)
  mov    %rax,0xb(%rsp)
  call   <reverse@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  mov    %rax,%rdx
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
data_type: string

tool: D2: %s
%s


assistant: ```c
int main() {
  char str[] = "reverse this";
  char *reversed = reverse(str);
  printf("%s\n%s\n", str, reversed);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <a@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  a();
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <a@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  a();
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <a@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  a();
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <a@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  a();
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int *arr, int first, int second) {
  int tmp = arr[first];
  arr[first] = arr[second];
  arr[second] = tmp;
}
```
user: What is the c source code of the assembly code below?

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int *arr, int first, int second) {
  int tmp = arr[first];
  arr[first] = arr[second];
  arr[second] = tmp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<swap>:
  endbr64
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
  ret
```

assistant: ```c
void swap(int *arr, int first, int second) {
  int tmp = arr[first];
  arr[first] = arr[second];
  arr[second] = tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<S>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%edx
  mov    -0x4(%rbp),%eax
  sar    $0x6,%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t S(uint8_t x, uint8_t y, uint8_t a) {
  uint8_t t = x + y + a;
  return (t << 2) | (t >> 6);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<S>:
  endbr64
  add    %esi,%edi
  add    %edx,%edi
  lea    0x0(,%rdi,4),%eax
  sar    $0x6,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
uint32_t S(uint8_t x, uint8_t y, uint8_t a) {
  uint8_t t = x + y + a;
  return (t << 2) | (t >> 6);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<S>:
  endbr64
  add    %esi,%edi
  add    %edx,%edi
  lea    0x0(,%rdi,4),%eax
  sar    $0x6,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
uint32_t S(uint8_t x, uint8_t y, uint8_t a) {
  uint8_t t = x + y + a;
  return (t << 2) | (t >> 6);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<S>:
  endbr64
  add    %esi,%edi
  add    %edx,%edi
  lea    0x0(,%rdi,4),%eax
  sar    $0x6,%edi
  or     %edi,%eax
  ret
```

assistant: ```c
uint32_t S(uint8_t x, uint8_t y, uint8_t a) {
  uint8_t t = x + y + a;
  return (t << 2) | (t >> 6);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<randomNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  -0x8(%rbp)
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int randomNumber(int min, int max) {
  max -= min;
  return (rand() % max + min);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<randomNumber>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  sub    %edi,%esi
  mov    %esi,%ebp
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %ebp
  lea    (%rdx,%rbx,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int randomNumber(int min, int max) {
  max -= min;
  return (rand() % max + min);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<randomNumber>:
  endbr64
  push   %rbp
  sub    %edi,%esi
  xor    %eax,%eax
  push   %rbx
  mov    %esi,%ebp
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <rand@plt>
  add    $0x8,%rsp
  cltd
  idiv   %ebp
  lea    (%rdx,%rbx,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int randomNumber(int min, int max) {
  max -= min;
  return (rand() % max + min);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<randomNumber>:
  endbr64
  push   %rbp
  sub    %edi,%esi
  xor    %eax,%eax
  push   %rbx
  mov    %esi,%ebp
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <rand@plt>
  add    $0x8,%rsp
  cltd
  idiv   %ebp
  lea    (%rdx,%rbx,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int randomNumber(int min, int max) {
  max -= min;
  return (rand() % max + min);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<invalid_work_group_size>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %edi,%r13d
  mov    %esi,%r14d
  mov    %edx,%ebx
  mov    %rcx,%rbp
  mov    %r8,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rcx
  mov    $0x0,%r8d
  mov    $0x8,%edx
  mov    D7(%rip),%rax
  mov    (%rax),%esi
  call   <clGetDeviceInfo@plt>
  lea    0x10(%rsp),%rcx
  mov    $0x0,%r8d
  mov    $0x18,%edx
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  mov    %r13d,%edi
  call   <clGetDeviceInfo@plt>
  mov    $0x0,%r9d
  lea    0x8(%rsp),%r8
  mov    $0x8,%ecx
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    %r13d,%esi
  mov    %r14d,%edi
  call   <clGetKernelWorkGroupInfo@plt>
  test   %ebx,%ebx
  jle    L30
  mov    $0x0,%eax
  mov    $0x1,%esi
  mov    $0x0,%edx
  mov    $0x1,%edi
  lea    0x10(%rsp),%r8
L14:
  mov    0x0(%rbp,%rax,8),%rcx
  cmp    (%r12,%rax,8),%rcx
  cmova  %edi,%edx
  cmp    (%r8,%rax,8),%rcx
  cmova  %edi,%edx
  imul   %rcx,%rsi
  add    $0x1,%rax
  cmp    %eax,%ebx
  jg     L14
L21:
  cmp    %rsi,(%rsp)
  mov    $0x1,%eax
  cmovb  %eax,%edx
  cmp    %rsi,0x8(%rsp)
  cmovb  %eax,%edx
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    %edx,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L30:
  mov    $0x1,%esi
  mov    $0x0,%edx
  jmp    L21
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int invalid_work_group_size(cl_device_id device, cl_kernel kernel, int dim, const size_t *local_work_size, const size_t *global_work_size) {
  size_t max_work_group_size;
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_GROUP_SIZE, sizeof(size_t), &max_work_group_size, NULL);
  size_t max_work_item_sizes[3];
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_ITEM_SIZES, sizeof(max_work_item_sizes), max_work_item_sizes, NULL);
  size_t kernel_max_work_group_size;
  clGetKernelWorkGroupInfo(kernel, device, CL_KERNEL_WORK_GROUP_SIZE, sizeof(kernel_max_work_group_size), &kernel_max_work_group_size, NULL);
  int invalid = 0;
  size_t total_size = 1;
  for (int i = 0; i < dim; i++) {
    if (local_work_size[i] > global_work_size[i])
      invalid = 1;
    if (local_work_size[i] > max_work_item_sizes[i])
      invalid = 1;
    total_size *= local_work_size[i];
  }
  if (total_size > max_work_group_size)
    invalid = 1;
  if (total_size > kernel_max_work_group_size)
    invalid = 1;
  return invalid;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<invalid_work_group_size>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %edi,%r14d
  push   %r13
  push   %r12
  mov    %r8,%r12
  xor    %r8d,%r8d
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %edx,%ebx
  mov    $0x8,%edx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D6(%rip),%rax
  mov    %rsp,%rcx
  lea    0x10(%rsp),%r13
  mov    (%rax),%esi
  call   <clGetDeviceInfo@plt>
  xor    %r8d,%r8d
  mov    %r13,%rcx
  mov    $0x18,%edx
  mov    D20(%rip),%rax
  mov    %r14d,%edi
  mov    (%rax),%esi
  call   <clGetDeviceInfo@plt>
  xor    %r9d,%r9d
  lea    0x8(%rsp),%r8
  mov    %r14d,%esi
  mov    D7(%rip),%rax
  mov    $0x8,%ecx
  mov    %r15d,%edi
  mov    (%rax),%edx
  call   <clGetKernelWorkGroupInfo@plt>
  test   %ebx,%ebx
  jle    L14
  xor    %eax,%eax
  mov    $0x1,%ecx
  xor    %r9d,%r9d
  mov    $0x1,%esi
L24:
  mov    0x0(%rbp,%rax,8),%rdx
  cmp    (%r12,%rax,8),%rdx
  cmova  %esi,%r9d
  cmp    0x0(%r13,%rax,8),%rdx
  cmova  %esi,%r9d
  imul   %rdx,%rcx
  add    $0x1,%rax
  cmp    %eax,%ebx
  jg     L24
L3:
  mov    $0x1,%eax
  cmp    %rcx,(%rsp)
  cmovb  %eax,%r9d
  cmp    %rcx,0x8(%rsp)
  cmovb  %eax,%r9d
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x38,%rsp
  mov    %r9d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0x1,%ecx
  xor    %r9d,%r9d
  jmp    L3
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int invalid_work_group_size(cl_device_id device, cl_kernel kernel, int dim, const size_t *local_work_size, const size_t *global_work_size) {
  size_t max_work_group_size;
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_GROUP_SIZE, sizeof(size_t), &max_work_group_size, NULL);
  size_t max_work_item_sizes[3];
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_ITEM_SIZES, sizeof(max_work_item_sizes), max_work_item_sizes, NULL);
  size_t kernel_max_work_group_size;
  clGetKernelWorkGroupInfo(kernel, device, CL_KERNEL_WORK_GROUP_SIZE, sizeof(kernel_max_work_group_size), &kernel_max_work_group_size, NULL);
  int invalid = 0;
  size_t total_size = 1;
  for (int i = 0; i < dim; i++) {
    if (local_work_size[i] > global_work_size[i])
      invalid = 1;
    if (local_work_size[i] > max_work_item_sizes[i])
      invalid = 1;
    total_size *= local_work_size[i];
  }
  if (total_size > max_work_group_size)
    invalid = 1;
  if (total_size > kernel_max_work_group_size)
    invalid = 1;
  return invalid;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<invalid_work_group_size>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %r8,%r12
  xor    %r8d,%r8d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  mov    $0x8,%edx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D13(%rip),%rax
  mov    %rsp,%rcx
  mov    (%rax),%esi
  call   <clGetDeviceInfo@plt>
  xor    %r8d,%r8d
  lea    0x10(%rsp),%rcx
  mov    %ebp,%edi
  mov    D10(%rip),%rax
  mov    $0x18,%edx
  mov    (%rax),%esi
  call   <clGetDeviceInfo@plt>
  xor    %r9d,%r9d
  lea    0x8(%rsp),%r8
  mov    %ebp,%esi
  mov    D8(%rip),%rax
  mov    $0x8,%ecx
  mov    %r14d,%edi
  mov    (%rax),%edx
  call   <clGetKernelWorkGroupInfo@plt>
  test   %ebx,%ebx
  jle    L45
  mov    0x0(%r13),%rdx
  xor    %eax,%eax
  cmp    (%r12),%rdx
  mov    $0x1,%ecx
  seta   %al
  cmp    0x10(%rsp),%rdx
  cmova  %ecx,%eax
  cmp    $0x1,%ebx
  je     L49
  mov    0x8(%r13),%rsi
  cmp    %rsi,0x8(%r12)
  cmovb  %ecx,%eax
  cmp    %rsi,0x18(%rsp)
  cmovb  %ecx,%eax
  imul   %rsi,%rdx
  cmp    $0x2,%ebx
  je     L49
  mov    0x10(%r13),%rsi
  cmp    0x10(%r12),%rsi
  cmova  %ecx,%eax
  cmp    0x20(%rsp),%rsi
  cmova  %ecx,%eax
  imul   %rsi,%rdx
L49:
  mov    $0x1,%ecx
  cmp    %rdx,(%rsp)
  cmovb  %ecx,%eax
  cmp    %rdx,0x8(%rsp)
  cmovb  %ecx,%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L33
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L45:
  mov    $0x1,%edx
  xor    %eax,%eax
  jmp    L49
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int invalid_work_group_size(cl_device_id device, cl_kernel kernel, int dim, const size_t *local_work_size, const size_t *global_work_size) {
  size_t max_work_group_size;
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_GROUP_SIZE, sizeof(size_t), &max_work_group_size, NULL);
  size_t max_work_item_sizes[3];
  clGetDeviceInfo(device, CL_DEVICE_MAX_WORK_ITEM_SIZES, sizeof(max_work_item_sizes), max_work_item_sizes, NULL);
  size_t kernel_max_work_group_size;
  clGetKernelWorkGroupInfo(kernel, device, CL_KERNEL_WORK_GROUP_SIZE, sizeof(kernel_max_work_group_size), &kernel_max_work_group_size, NULL);
  int invalid = 0;
  size_t total_size = 1;
  for (int i = 0; i < dim; i++) {
    if (local_work_size[i] > global_work_size[i])
      invalid = 1;
    if (local_work_size[i] > max_work_item_sizes[i])
      invalid = 1;
    total_size *= local_work_size[i];
  }
  if (total_size > max_work_group_size)
    invalid = 1;
  if (total_size > kernel_max_work_group_size)
    invalid = 1;
  return invalid;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L15
  mov    -0x8(%rbp),%eax
  jmp    L14
L15:
  mov    -0x4(%rbp),%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
int min(int a, int b) {
  if (a > b)
    return b;
  return a;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<min>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int min(int a, int b) {
  if (a > b)
    return b;
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<min>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int min(int a, int b) {
  if (a > b)
    return b;
  return a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<min>:
  endbr64
  cmp    %esi,%edi
  mov    %esi,%eax
  cmovle %edi,%eax
  ret
```

assistant: ```c
int min(int a, int b) {
  if (a > b)
    return b;
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L6
  mov    -0xc(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L19
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jle    L19
L6:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L21
L19:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L21:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D29
type_guessed: string
name: read_data
arguments: data_tag: D17
type_guessed: string
name: read_data
arguments: data_tag: D9
type_guessed: string
name: read_data
arguments: data_tag: D4
type_guessed: string

tool: D29: 
Please enter a year: 

tool: D17: %d

tool: D9: 
Your year, %d, is a leap year.


tool: D4: 
Your year, %d, is not a leap year.


assistant: ```c
int main() {
  int year;
  printf("\nPlease enter a year: ");
  scanf("%d", &year);
  if ((year % 400 == 0) || (year % 4 == 0 && year % 100 > 0)) {
    printf("\nYour year, %d, is a leap year.\n", year);
  } else
    printf("\nYour year, %d, is not a leap year.\n", year);
  return 0;
}
```