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

```asm
<test2>:
  endbr64
  push   %r14
  mov    $0x2328,%edi
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <heap_create@plt>
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%r12
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%r14
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%rbp
  call   <ant_alloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <heap_dump@plt>
  mov    %r14,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  mov    %r13,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  xor    %eax,%eax
  call   <heap_dump@plt>
  mov    %r12,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  mov    %rbp,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  xor    %eax,%eax
  call   <heap_dump@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <heap_destroy@plt>
```

assistant: ```c
void test2() {
  size_t heap_sz = 9000;
  size_t request_sz = 1000;
  heap_create(heap_sz);
  void *p1 = ant_alloc(request_sz);
  void *p2 = ant_alloc(request_sz);
  void *p3 = ant_alloc(request_sz);
  void *p4 = ant_alloc(request_sz);
  heap_dump();
  ant_free(p2, request_sz);
  ant_free(p4, request_sz);
  heap_dump();
  ant_free(p1, request_sz);
  ant_free(p3, request_sz);
  heap_dump();
  heap_destroy();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test2>:
  endbr64
  push   %r14
  mov    $0x2328,%edi
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  call   <heap_create@plt>
  mov    $0x3e8,%edi
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%r12
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%r14
  call   <ant_alloc@plt>
  mov    $0x3e8,%edi
  mov    %rax,%rbp
  call   <ant_alloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <heap_dump@plt>
  mov    %r14,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  mov    %r13,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  xor    %eax,%eax
  call   <heap_dump@plt>
  mov    %r12,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  mov    %rbp,%rdi
  mov    $0x3e8,%esi
  call   <ant_free@plt>
  xor    %eax,%eax
  call   <heap_dump@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1090 <heap_destroy@plt>
```

assistant: ```c
void test2() {
  size_t heap_sz = 9000;
  size_t request_sz = 1000;
  heap_create(heap_sz);
  void *p1 = ant_alloc(request_sz);
  void *p2 = ant_alloc(request_sz);
  void *p3 = ant_alloc(request_sz);
  void *p4 = ant_alloc(request_sz);
  heap_dump();
  ant_free(p2, request_sz);
  ant_free(p4, request_sz);
  heap_dump();
  ant_free(p1, request_sz);
  ant_free(p3, request_sz);
  heap_dump();
  heap_destroy();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x14(%rbp)
  jmp    L33
L55:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  jmp    L29
L48:
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x18(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L5
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  mov    -0x18(%rbp),%edx
  mov    %edx,-0x10(%rbp,%rax,4)
L5:
  addl   $0x1,-0x18(%rbp)
L29:
  cmpl   $0x1,-0x1c(%rbp)
  jg     L16
  mov    -0x1c(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  cmp    %eax,%edx
  jle    L48
L16:
  mov    -0x20(%rbp),%eax
  mov    -0xc(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %eax,%edi
  mov    -0xc(%rbp),%esi
  mov    -0x20(%rbp),%eax
  mov    -0x10(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %eax,%r8d
  mov    -0x10(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x14(%rbp),%eax
  sub    $0x8,%rsp
  push   %rdi
  mov    %esi,%r9d
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  add    $0x10,%rsp
  addl   $0x1,-0x14(%rbp)
L33:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L55
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
type: string
name: read_data
arguments: data_label: D29
type: string

tool: D16: %d


tool: D29: Case #%d: %d = %d * %d = %d * %d


assistant: ```c
int main() {
  int test, k, j, i, x;
  int v[2];
  scanf("%d\n", &test);
  for (x = 1; x <= test; ++x) {
    scanf("%d\n", &k);
    j = 0;
    for (i = 2; j < 2 && (j * j) <= k; ++i)
      if (k % i == 0)
        v[j++] = i;
    printf("Case #%d: %d = %d * %d = %d * %d\n", x, k, v[0], k / v[0], v[1], k / v[1]);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L37
  mov    $0x1,%ebx
  lea    0xc(%rsp),%r14
  lea    D25(%rip),%r13
  mov    $0x0,%r12d
  lea    D0(%rip),%rbp
  jmp    L78
L60:
  add    $0x1,%esi
  cmp    $0x1,%ecx
  jg     L38
L27:
  mov    %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  jg     L38
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L60
  movslq %ecx,%rax
  mov    %esi,0x10(%rsp,%rax,4)
  lea    0x1(%rcx),%ecx
  jmp    L60
L38:
  mov    0x14(%rsp),%r9d
  mov    0x10(%rsp),%ecx
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r8d
  sub    $0x8,%rsp
  mov    %edi,%eax
  cltd
  idiv   %r9d
  push   %rax
  mov    %edi,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  add    $0x10,%rsp
  cmp    %ebx,0x8(%rsp)
  jl     L37
L78:
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  mov    $0x2,%esi
  mov    %r12d,%ecx
  jmp    L27
L37:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L79
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L79:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
type_guessed: string
name: retrieve_data
arguments: data_tag: D0
type_guessed: string

tool: D25: %d


tool: D0: Case #%d: %d = %d * %d = %d * %d


assistant: ```c
int main() {
  int test, k, j, i, x;
  int v[2];
  scanf("%d\n", &test);
  for (x = 1; x <= test; ++x) {
    scanf("%d\n", &k);
    j = 0;
    for (i = 2; j < 2 && (j * j) <= k; ++i)
      if (k % i == 0)
        v[j++] = i;
    printf("Case #%d: %d = %d * %d = %d * %d\n", x, k, v[0], k / v[0], v[1], k / v[1]);
  }
  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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%ecx
  test   %ecx,%ecx
  jle    L49
  mov    $0x1,%r13d
  lea    0xc(%rsp),%r12
  lea    D8(%rip),%rbp
L39:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  mov    $0x2,%ecx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L9:
  cmp    %esi,%r10d
  jl     L38
L58:
  mov    %r10d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L50
  movslq %esi,%rax
  lea    0x1(%rsi),%edx
  mov    %ecx,0x10(%rsp,%rax,4)
  add    $0x1,%ecx
  cmp    $0x1,%esi
  je     L38
  mov    %edx,%esi
  cmp    %esi,%r10d
  jge    L58
L38:
  mov    0x10(%rsp),%ecx
  mov    %r10d,%eax
  mov    0x14(%rsp),%r9d
  sub    $0x8,%rsp
  cltd
  mov    %r13d,%esi
  mov    %rbp,%rdi
  add    $0x1,%r13d
  idiv   %ecx
  mov    %eax,%r8d
  mov    %r10d,%eax
  cltd
  idiv   %r9d
  mov    %r10d,%edx
  push   %rax
  call   <printf@plt>
  pop    %rax
  pop    %rdx
  cmp    %r13d,0x8(%rsp)
  jge    L39
L49:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  add    $0x1,%ecx
  jmp    L9
L60:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type: string
name: extract_symbolic_data
arguments: data_tag: D8
type: string

tool: D4: %d


tool: D8: Case #%d: %d = %d * %d = %d * %d


assistant: ```c
int main() {
  int test, k, j, i, x;
  int v[2];
  scanf("%d\n", &test);
  for (x = 1; x <= test; ++x) {
    scanf("%d\n", &k);
    j = 0;
    for (i = 2; j < 2 && (j * j) <= k; ++i)
      if (k % i == 0)
        v[j++] = i;
    printf("Case #%d: %d = %d * %d = %d * %d\n", x, k, v[0], k / v[0], v[1], k / v[1]);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D16(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%ecx
  test   %ecx,%ecx
  jle    L70
  mov    $0x1,%r13d
  lea    0xc(%rsp),%r12
  lea    D4(%rip),%rbp
L5:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r10d
  mov    $0x2,%ecx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L15:
  cmp    %esi,%r10d
  jl     L1
L79:
  mov    %r10d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L36
  movslq %esi,%rax
  mov    %ecx,0x10(%rsp,%rax,4)
  add    $0x1,%ecx
  cmp    $0x1,%esi
  je     L1
  mov    $0x1,%esi
  cmp    %esi,%r10d
  jge    L79
L1:
  mov    0x10(%rsp),%ecx
  mov    %r10d,%eax
  mov    0x14(%rsp),%r9d
  sub    $0x8,%rsp
  cltd
  mov    %r13d,%esi
  mov    %rbp,%rdi
  add    $0x1,%r13d
  idiv   %ecx
  mov    %eax,%r8d
  mov    %r10d,%eax
  cltd
  idiv   %r9d
  mov    %r10d,%edx
  push   %rax
  call   <printf@plt>
  pop    %rax
  pop    %rdx
  cmp    %r13d,0x8(%rsp)
  jge    L5
L70:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  add    $0x1,%ecx
  jmp    L15
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D16
type_infered: string
name: unpack_data
arguments: data_tag: D4
type_infered: string

tool: D16: %d


tool: D4: Case #%d: %d = %d * %d = %d * %d


assistant: ```c
int main() {
  int test, k, j, i, x;
  int v[2];
  scanf("%d\n", &test);
  for (x = 1; x <= test; ++x) {
    scanf("%d\n", &k);
    j = 0;
    for (i = 2; j < 2 && (j * j) <= k; ++i)
      if (k % i == 0)
        v[j++] = i;
    printf("Case #%d: %d = %d * %d = %d * %d\n", x, k, v[0], k / v[0], v[1], k / v[1]);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void *f() {
  void *i = &i;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<f>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  mov    %rsp,%rax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *f() {
  void *i = &i;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
void *f() {
  void *i = &i;
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
void *f() {
  void *i = &i;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<create_allocate_tiff_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
create_allocate_tiff_file(file_name, image_header) char file_name[];
struct tiff_header_struct *image_header;
{
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<create_allocate_tiff_file>:
  endbr64
  ret
```

assistant: ```c
create_allocate_tiff_file(file_name, image_header) char file_name[];
struct tiff_header_struct *image_header;
{
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<create_allocate_tiff_file>:
  endbr64
  ret
```

assistant: ```c
create_allocate_tiff_file(file_name, image_header) char file_name[];
struct tiff_header_struct *image_header;
{
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<create_allocate_tiff_file>:
  endbr64
  ret
```

assistant: ```c
create_allocate_tiff_file(file_name, image_header) char file_name[];
struct tiff_header_struct *image_header;
{
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<int_cmp_desc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int int_cmp_desc(const void *b, const void *a) {
  const int *ia = (const int *)a;
  const int *ib = (const int *)b;
  return *ia - *ib;
}
```
user: Translate the below assembly code into C programming language:

```asm
<int_cmp_desc>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  ret
```

assistant: ```c
int int_cmp_desc(const void *b, const void *a) {
  const int *ia = (const int *)a;
  const int *ib = (const int *)b;
  return *ia - *ib;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<int_cmp_desc>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  ret
```

assistant: ```c
int int_cmp_desc(const void *b, const void *a) {
  const int *ia = (const int *)a;
  const int *ib = (const int *)b;
  return *ia - *ib;
}
```
user: What is the c source code of the assembly code below?

```asm
<int_cmp_desc>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  ret
```

assistant: ```c
int int_cmp_desc(const void *b, const void *a) {
  const int *ia = (const int *)a;
  const int *ib = (const int *)b;
  return *ia - *ib;
}
```
user: What is the c source code of the assembly code below:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L9
L10:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %rax,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L9:
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jl     L10
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long sum(long n, long *a) {
  long i, s;
  s = 0;
  for (i = 0; i < n; i = i + 1) {
    s = s + a[i];
  }
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  jle    L15
  mov    %rsi,%rax
  lea    (%rsi,%rdi,8),%rcx
  mov    $0x0,%edx
L5:
  add    (%rax),%rdx
  add    $0x8,%rax
  cmp    %rcx,%rax
  jne    L5
L21:
  mov    %rdx,%rax
  ret
L15:
  mov    $0x0,%edx
  jmp    L21
```

assistant: ```c
long sum(long n, long *a) {
  long i, s;
  s = 0;
  for (i = 0; i < n; i = i + 1) {
    s = s + a[i];
  }
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  jle    L2
  lea    (%rsi,%rdi,8),%rdx
  xor    %eax,%eax
  nop
L8:
  add    (%rsi),%rax
  add    $0x8,%rsi
  cmp    %rdx,%rsi
  jne    L8
  ret
  nopl   (%rax)
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
long sum(long n, long *a) {
  long i, s;
  s = 0;
  for (i = 0; i < n; i = i + 1) {
    s = s + a[i];
  }
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  jle    L5
  cmp    $0x1,%rdi
  je     L31
  mov    %rdi,%rdx
  mov    %rsi,%rax
  pxor   %xmm0,%xmm0
  shr    %rdx
  shl    $0x4,%rdx
  add    %rsi,%rdx
  nopl   0x0(%rax,%rax,1)
L17:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddq  %xmm2,%xmm0
  cmp    %rax,%rdx
  jne    L17
  movdqa %xmm0,%xmm1
  mov    %rdi,%rdx
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffe,%rdx
  and    $0x1,%edi
  paddq  %xmm1,%xmm0
  movq   %xmm0,%rax
  je     L28
L3:
  add    (%rsi,%rdx,8),%rax
  ret
  nopl   0x0(%rax)
L28:
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  ret
L31:
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L3
```

assistant: ```c
long sum(long n, long *a) {
  long i, s;
  s = 0;
  for (i = 0; i < n; i = i + 1) {
    s = s + a[i];
  }
  return s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<subs_si_test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L11
  mov    -0x14(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  jmp    L4
L11:
  mov    -0x4(%rbp),%eax
L4:
  pop    %rbp
  ret
```

assistant: ```c
int subs_si_test1(int a, int b, int c) {
  int d = a - c;
  if (d == 0)
    return a + c;
  else
    return d;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<subs_si_test1>:
  endbr64
  mov    %edi,%eax
  lea    (%rdi,%rdx,1),%ecx
  sub    %edx,%eax
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int subs_si_test1(int a, int b, int c) {
  int d = a - c;
  if (d == 0)
    return a + c;
  else
    return d;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<subs_si_test1>:
  endbr64
  mov    %edi,%eax
  lea    (%rdi,%rdx,1),%ecx
  sub    %edx,%eax
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int subs_si_test1(int a, int b, int c) {
  int d = a - c;
  if (d == 0)
    return a + c;
  else
    return d;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<subs_si_test1>:
  endbr64
  mov    %edi,%eax
  lea    (%rdi,%rdx,1),%ecx
  sub    %edx,%eax
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int subs_si_test1(int a, int b, int c) {
  int d = a - c;
  if (d == 0)
    return a + c;
  else
    return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<minimum_distance>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L33:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L21
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L21
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L21:
  addl   $0x1,-0xc(%rbp)
L0:
  cmpl   $0x7,-0xc(%rbp)
  jle    L33
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int minimum_distance(const int min_distances[], const bool shortest_paths[]) {
  int i;
  int minimum = INT_MAX, index;
  for (i = 0; i < 8; ++i) {
    if (!shortest_paths[i] && min_distances[i] <= minimum) {
      minimum = min_distances[i];
      index = i;
    }
  }
  return index;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<minimum_distance>:
  endbr64
  mov    D8(%rip),%rax
  mov    (%rax),%r8d
  mov    $0x0,%eax
  jmp    L11
L20:
  add    $0x1,%rax
  cmp    $0x8,%rax
  je     L42
L11:
  cmpl   $0x0,(%rsi,%rax,4)
  jne    L20
  mov    (%rdi,%rax,4),%edx
  cmp    %r8d,%edx
  jg     L20
  mov    %eax,%r9d
  mov    %edx,%r8d
  jmp    L20
L42:
  mov    %r9d,%eax
  ret
```

assistant: ```c
int minimum_distance(const int min_distances[], const bool shortest_paths[]) {
  int i;
  int minimum = INT_MAX, index;
  for (i = 0; i < 8; ++i) {
    if (!shortest_paths[i] && min_distances[i] <= minimum) {
      minimum = min_distances[i];
      index = i;
    }
  }
  return index;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<minimum_distance>:
  endbr64
  mov    D9(%rip),%rax
  xor    %r8d,%r8d
  mov    (%rax),%ecx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L0:
  mov    (%rsi,%rax,4),%edx
  test   %edx,%edx
  jne    L11
  mov    (%rdi,%rax,4),%edx
  cmp    %ecx,%edx
  jg     L11
  mov    %eax,%r8d
  mov    %edx,%ecx
L11:
  add    $0x1,%rax
  cmp    $0x8,%rax
  jne    L0
  mov    %r8d,%eax
  ret
```

assistant: ```c
int minimum_distance(const int min_distances[], const bool shortest_paths[]) {
  int i;
  int minimum = INT_MAX, index;
  for (i = 0; i < 8; ++i) {
    if (!shortest_paths[i] && min_distances[i] <= minimum) {
      minimum = min_distances[i];
      index = i;
    }
  }
  return index;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<minimum_distance>:
  endbr64
  mov    D3(%rip),%rax
  mov    (%rsi),%ecx
  mov    (%rax),%eax
  test   %ecx,%ecx
  jne    L11
  mov    (%rdi),%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
L11:
  mov    0x4(%rsi),%r8d
  test   %r8d,%r8d
  jne    L195
  mov    0x4(%rdi),%edx
  cmp    %edx,%eax
  jge    L33
L17:
  mov    0x8(%rsi),%edx
  test   %edx,%edx
  jne    L12
  mov    0x8(%rdi),%edx
  cmp    %edx,%eax
  jge    L65
L12:
  mov    0xc(%rsi),%r11d
  test   %r11d,%r11d
  jne    L23
  mov    0xc(%rdi),%edx
  cmp    %edx,%eax
  jge    L206
L23:
  mov    0x10(%rsi),%r10d
  test   %r10d,%r10d
  jne    L26
  mov    0x10(%rdi),%edx
  cmp    %edx,%eax
  jge    L187
L26:
  mov    0x14(%rsi),%r9d
  test   %r9d,%r9d
  jne    L166
  mov    0x14(%rdi),%edx
  cmp    %edx,%eax
  jge    L141
L166:
  mov    0x18(%rsi),%ecx
  test   %ecx,%ecx
  jne    L172
  mov    0x18(%rdi),%edx
  cmp    %edx,%eax
  jge    L89
L172:
  mov    0x1c(%rsi),%edx
  test   %edx,%edx
  jne    L169
  cmp    %eax,0x1c(%rdi)
  mov    $0x7,%eax
  cmovle %eax,%r8d
L169:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L195:
  xor    %r8d,%r8d
  jmp    L17
  nopl   (%rax)
L89:
  mov    %edx,%eax
  mov    $0x6,%r8d
  jmp    L172
  nopw   0x0(%rax,%rax,1)
L141:
  mov    %edx,%eax
  mov    $0x5,%r8d
  jmp    L166
  nopw   0x0(%rax,%rax,1)
L187:
  mov    %edx,%eax
  mov    $0x4,%r8d
  jmp    L26
  nopw   0x0(%rax,%rax,1)
L206:
  mov    %edx,%eax
  mov    $0x3,%r8d
  jmp    L23
  nopl   (%rax)
L65:
  mov    %edx,%eax
  mov    $0x2,%r8d
  jmp    L12
  nopl   (%rax)
L33:
  mov    %edx,%eax
  mov    $0x1,%r8d
  jmp    L17
```

assistant: ```c
int minimum_distance(const int min_distances[], const bool shortest_paths[]) {
  int i;
  int minimum = INT_MAX, index;
  for (i = 0; i < 8; ++i) {
    if (!shortest_paths[i] && min_distances[i] <= minimum) {
      minimum = min_distances[i];
      index = i;
    }
  }
  return index;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <one_three@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
type_infered: string
name: disassemble_data
arguments: data_tag: D7
type_infered: string

tool: D12: starting now:


tool: D7: done!


assistant: ```c
int main(void) {
  printf("starting now:\n");
  one_three();
  printf("done!\n");
  return 0;
}
```