ylfeng/ReF-Decompile-dataset huggingface dataset:
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <execv@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string

tool: D9: helloworld

assistant: ```c
int main(int argc, char *argv[]) {
  (void)argc;
  (void)argv;
  execv("helloworld", NULL);
  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,%esi
  lea    D6(%rip),%rdi
  call   <execv@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_infered: string

tool: D6: helloworld

assistant: ```c
int main(int argc, char *argv[]) {
  (void)argc;
  (void)argv;
  execv("helloworld", NULL);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D5(%rip),%rdi
  call   <execv@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type_infered: string

tool: D5: helloworld

assistant: ```c
int main(int argc, char *argv[]) {
  (void)argc;
  (void)argv;
  execv("helloworld", NULL);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D6(%rip),%rdi
  call   <execv@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
datatype: string

tool: D6: helloworld

assistant: ```c
int main(int argc, char *argv[]) {
  (void)argc;
  (void)argv;
  execv("helloworld", NULL);
  return 0;
}
```
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
  sub    $0x10,%rsp
  movl   $0xa,-0x8(%rbp)
  movl   $0x8,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D49
datatype: string
name: parse_data
arguments: data_tag: D32
datatype: string
name: parse_data
arguments: data_tag: D39
datatype: string
name: parse_data
arguments: data_tag: D38
datatype: string
name: parse_data
arguments: data_tag: D33
datatype: string

tool: D49: Simples calculo com valores de x = %d e y = %d



tool: D32: %d / %d = %d


tool: D39: %d * %d = %d


tool: D38: %d + %d = %d


tool: D33: %d - %d = %d


assistant: ```c
int main() {
  int x = 10, y = 8;
  printf("Simples calculo com valores de x = %d e y = %d\n\n", x, y);
  printf("%d / %d = %d\n", x, y, x / y);
  printf("%d * %d = %d\n", x, y, x * y);
  printf("%d + %d = %d\n", x, y, x + y);
  printf("%d - %d = %d\n", x, y, x - y);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ecx
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x50,%ecx
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x12,%ecx
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2,%ecx
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string
name: read_data
arguments: data_tag: D9
type_infered: string
name: read_data
arguments: data_tag: D6
type_infered: string
name: read_data
arguments: data_tag: D17
type_infered: string
name: read_data
arguments: data_tag: D41
type_infered: string

tool: D4: Simples calculo com valores de x = %d e y = %d



tool: D9: %d / %d = %d


tool: D6: %d * %d = %d


tool: D17: %d + %d = %d


tool: D41: %d - %d = %d


assistant: ```c
int main() {
  int x = 10, y = 8;
  printf("Simples calculo com valores de x = %d e y = %d\n\n", x, y);
  printf("%d / %d = %d\n", x, y, x / y);
  printf("%d * %d = %d\n", x, y, x * y);
  printf("%d + %d = %d\n", x, y, x + y);
  printf("%d - %d = %d\n", x, y, x - y);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edx
  mov    $0xa,%esi
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x50,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D42(%rip),%rdi
  call   <printf@plt>
  mov    $0x12,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x2,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D44(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_guessed: string
name: parse_data
arguments: data_tag: D13
type_guessed: string
name: parse_data
arguments: data_tag: D42
type_guessed: string
name: parse_data
arguments: data_tag: D30
type_guessed: string
name: parse_data
arguments: data_tag: D44
type_guessed: string

tool: D8: Simples calculo com valores de x = %d e y = %d



tool: D13: %d / %d = %d


tool: D42: %d * %d = %d


tool: D30: %d + %d = %d


tool: D44: %d - %d = %d


assistant: ```c
int main() {
  int x = 10, y = 8;
  printf("Simples calculo com valores de x = %d e y = %d\n\n", x, y);
  printf("%d / %d = %d\n", x, y, x / y);
  printf("%d * %d = %d\n", x, y, x * y);
  printf("%d + %d = %d\n", x, y, x + y);
  printf("%d - %d = %d\n", x, y, x - y);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edx
  mov    $0xa,%esi
  xor    %eax,%eax
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D37(%rip),%rdi
  call   <printf@plt>
  mov    $0x50,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x12,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D42(%rip),%rdi
  call   <printf@plt>
  mov    $0x2,%ecx
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0xa,%esi
  lea    D41(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D38
type_guessed: string
name: read_data
arguments: data_tag: D37
type_guessed: string
name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D42
type_guessed: string
name: read_data
arguments: data_tag: D41
type_guessed: string

tool: D38: Simples calculo com valores de x = %d e y = %d



tool: D37: %d / %d = %d


tool: D14: %d * %d = %d


tool: D42: %d + %d = %d


tool: D41: %d - %d = %d


assistant: ```c
int main() {
  int x = 10, y = 8;
  printf("Simples calculo com valores de x = %d e y = %d\n\n", x, y);
  printf("%d / %d = %d\n", x, y, x / y);
  printf("%d * %d = %d\n", x, y, x * y);
  printf("%d + %d = %d\n", x, y, x + y);
  printf("%d - %d = %d\n", x, y, x - y);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Join>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   $0x2710,%eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int Join(int a, int b) {
  int ret = a * 10000 + b;
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Join>:
  endbr64
  imul   $0x2710,%edi,%edi
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int Join(int a, int b) {
  int ret = a * 10000 + b;
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Join>:
  endbr64
  imul   $0x2710,%edi,%edi
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int Join(int a, int b) {
  int ret = a * 10000 + b;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Join>:
  endbr64
  imul   $0x2710,%edi,%edi
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int Join(int a, int b) {
  int ret = a * 10000 + b;
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<BitReverse64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    $0x20,%eax
  mov    %eax,%ecx
  sarl   %cl,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  rol    $0x10,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  and    $0xff00ff,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x8,%eax
  and    $0xff00ff00,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x4,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  and    $0xf0f0f0f0,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    $0x2,%eax
  and    $0x33333333,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shl    $0x2,%eax
  and    $0xcccccccc,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sar    %eax
  and    $0x55555555,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  and    $0xaaaaaaaa,%eax
  or     %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t BitReverse64(uint64_t value) {
  value = (((value) >> (32)) & (0x00000000FFFFFFFFull)) | (((value) & (0x00000000FFFFFFFFull)) << (32));
  value = (((value) >> (16)) & (0x0000FFFF0000FFFFull)) | (((value) & (0x0000FFFF0000FFFFull)) << (16));
  value = (((value) >> (8)) & (0x00FF00FF00FF00FFull)) | (((value) & (0x00FF00FF00FF00FFull)) << (8));
  value = (((value) >> (4)) & (0x0F0F0F0F0F0F0F0Full)) | (((value) & (0x0F0F0F0F0F0F0F0Full)) << (4));
  value = (((value) >> (2)) & (0x3333333333333333ull)) | (((value) & (0x3333333333333333ull)) << (2));
  value = (((value) >> (1)) & (0x5555555555555555ull)) | (((value) & (0x5555555555555555ull)) << (1));
  return value;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<BitReverse64>:
  endbr64
  mov    $0x20,%ecx
  sar    %cl,%edi
  rol    $0x10,%edi
  mov    %edi,%edx
  sar    $0x8,%edx
  and    $0xff00ff,%edx
  shl    $0x8,%edi
  and    $0xff00ff00,%edi
  or     %edi,%edx
  mov    %edx,%eax
  sar    $0x4,%eax
  and    $0xf0f0f0f,%eax
  shl    $0x4,%edx
  and    $0xf0f0f0f0,%edx
  or     %edx,%eax
  mov    %eax,%edx
  sar    $0x2,%edx
  and    $0x33333333,%edx
  shl    $0x2,%eax
  and    $0xcccccccc,%eax
  or     %eax,%edx
  mov    %edx,%eax
  sar    %eax
  and    $0x55555555,%eax
  add    %edx,%edx
  and    $0xaaaaaaaa,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
uint64_t BitReverse64(uint64_t value) {
  value = (((value) >> (32)) & (0x00000000FFFFFFFFull)) | (((value) & (0x00000000FFFFFFFFull)) << (32));
  value = (((value) >> (16)) & (0x0000FFFF0000FFFFull)) | (((value) & (0x0000FFFF0000FFFFull)) << (16));
  value = (((value) >> (8)) & (0x00FF00FF00FF00FFull)) | (((value) & (0x00FF00FF00FF00FFull)) << (8));
  value = (((value) >> (4)) & (0x0F0F0F0F0F0F0F0Full)) | (((value) & (0x0F0F0F0F0F0F0F0Full)) << (4));
  value = (((value) >> (2)) & (0x3333333333333333ull)) | (((value) & (0x3333333333333333ull)) << (2));
  value = (((value) >> (1)) & (0x5555555555555555ull)) | (((value) & (0x5555555555555555ull)) << (1));
  return value;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<BitReverse64>:
  endbr64
  mov    $0x20,%ecx
  sar    %cl,%edi
  bswap  %edi
  mov    %edi,%eax
  shl    $0x4,%edi
  sar    $0x4,%eax
  and    $0xf0f0f0f0,%edi
  and    $0xf0f0f0f,%eax
  or     %edi,%eax
  mov    %eax,%edx
  shl    $0x2,%eax
  sar    $0x2,%edx
  and    $0xcccccccc,%eax
  and    $0x33333333,%edx
  or     %eax,%edx
  mov    %edx,%eax
  add    %edx,%edx
  sar    %eax
  and    $0xaaaaaaaa,%edx
  and    $0x55555555,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
uint64_t BitReverse64(uint64_t value) {
  value = (((value) >> (32)) & (0x00000000FFFFFFFFull)) | (((value) & (0x00000000FFFFFFFFull)) << (32));
  value = (((value) >> (16)) & (0x0000FFFF0000FFFFull)) | (((value) & (0x0000FFFF0000FFFFull)) << (16));
  value = (((value) >> (8)) & (0x00FF00FF00FF00FFull)) | (((value) & (0x00FF00FF00FF00FFull)) << (8));
  value = (((value) >> (4)) & (0x0F0F0F0F0F0F0F0Full)) | (((value) & (0x0F0F0F0F0F0F0F0Full)) << (4));
  value = (((value) >> (2)) & (0x3333333333333333ull)) | (((value) & (0x3333333333333333ull)) << (2));
  value = (((value) >> (1)) & (0x5555555555555555ull)) | (((value) & (0x5555555555555555ull)) << (1));
  return value;
}
```
user: What is the c source code of the assembly code below:

```asm
<BitReverse64>:
  endbr64
  mov    $0x20,%ecx
  sar    %cl,%edi
  bswap  %edi
  mov    %edi,%eax
  shl    $0x4,%edi
  sar    $0x4,%eax
  and    $0xf0f0f0f0,%edi
  and    $0xf0f0f0f,%eax
  or     %edi,%eax
  mov    %eax,%edx
  shl    $0x2,%eax
  sar    $0x2,%edx
  and    $0xcccccccc,%eax
  and    $0x33333333,%edx
  or     %eax,%edx
  mov    %edx,%eax
  add    %edx,%edx
  sar    %eax
  and    $0xaaaaaaaa,%edx
  and    $0x55555555,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
uint64_t BitReverse64(uint64_t value) {
  value = (((value) >> (32)) & (0x00000000FFFFFFFFull)) | (((value) & (0x00000000FFFFFFFFull)) << (32));
  value = (((value) >> (16)) & (0x0000FFFF0000FFFFull)) | (((value) & (0x0000FFFF0000FFFFull)) << (16));
  value = (((value) >> (8)) & (0x00FF00FF00FF00FFull)) | (((value) & (0x00FF00FF00FF00FFull)) << (8));
  value = (((value) >> (4)) & (0x0F0F0F0F0F0F0F0Full)) | (((value) & (0x0F0F0F0F0F0F0F0Full)) << (4));
  value = (((value) >> (2)) & (0x3333333333333333ull)) | (((value) & (0x3333333333333333ull)) << (2));
  value = (((value) >> (1)) & (0x5555555555555555ull)) | (((value) & (0x5555555555555555ull)) << (1));
  return value;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cfs_size_round0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L16
L15:
  mov    -0x4(%rbp),%eax
  add    $0x8,%eax
  and    $0xfffffff8,%eax
L16:
  pop    %rbp
  ret
```

assistant: ```c
int cfs_size_round0(int val) {
  if (!val)
    return 0;
  return (val + 1 + 7) & (~0x7);
}
```
user: What is the c source code of the assembly code below?

```asm
<cfs_size_round0>:
  endbr64
  lea    0x8(%rdi),%eax
  and    $0xfffffff8,%eax
  test   %edi,%edi
  cmove  %edi,%eax
  ret
```

assistant: ```c
int cfs_size_round0(int val) {
  if (!val)
    return 0;
  return (val + 1 + 7) & (~0x7);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cfs_size_round0>:
  endbr64
  lea    0x8(%rdi),%eax
  xor    %edx,%edx
  and    $0xfffffff8,%eax
  test   %edi,%edi
  cmove  %edx,%eax
  ret
```

assistant: ```c
int cfs_size_round0(int val) {
  if (!val)
    return 0;
  return (val + 1 + 7) & (~0x7);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cfs_size_round0>:
  endbr64
  lea    0x8(%rdi),%eax
  xor    %edx,%edx
  and    $0xfffffff8,%eax
  test   %edi,%edi
  cmove  %edx,%eax
  ret
```

assistant: ```c
int cfs_size_round0(int val) {
  if (!val)
    return 0;
  return (val + 1 + 7) & (~0x7);
}
```
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   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    0x24(%rsp),%r12
  mov    $0x0,%r15d
  lea    D4(%rip),%r14
L32:
  mov    %r15d,%r13d
  lea    -0x14(%r12),%rbx
  mov    $0x0,%ebp
L19:
  mov    %ebp,%edi
  call   <sqrt@plt>
  movss  %xmm0,(%rbx)
  cvtss2sd %xmm0,%xmm0
  mov    %r14,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    %r13d,%ebp
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r15d
  add    $0x14,%r12
  cmp    $0x5,%r15d
  jne    L32
  mov    0x8(%rsp),%rax
  lea    0x10(%rax),%rbp
  lea    0x60(%rax),%r13
  lea    D4(%rip),%r12
L20:
  lea    -0x10(%rbp),%rbx
L11:
  movss  (%rbx),%xmm0
  addss  0x14(%rbx),%xmm0
  addss  0x4(%rbx),%xmm0
  addss  0x18(%rbx),%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x14,%rbp
  cmp    %rbp,%r13
  jne    L20
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x2a,%eax
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D18
type: string
name: parse_data
arguments: data_tag: D4
type: string

tool: D18: Hi this is a program to test the performance characteristics of a couple specific things from lesson 8 of Halide's tutorial.

tool: D4: |%f|

assistant: ```c
int main() {
  printf("Hi this is a program to test the performance characteristics of a couple specific things from lesson 8 of Halide's tutorial.\n");
  float result[4][4];
  float result2[4][4];
  float producer_storage[5][5];
  int y, x;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      producer_storage[y][x] = sqrt(x * y);
      printf("|%f|", producer_storage[y][x]);
    }
    printf("\n");
  }
  for (y = 0; y < 4; y++) {
    for (x = 0; x < 4; x++) {
      result[y][x] = (producer_storage[y][x] + producer_storage[y + 1][x] + producer_storage[y][x + 1] + producer_storage[y + 1][x + 1]);
      printf("|%f|", result[y][x]);
    }
    printf("\n");
  }
  for (y = 0; y < 4; y++) {
    for (x = 0; x < 4; x++) {
      result2[x][y] = (producer_storage[y][x] + producer_storage[y + 1][x] + producer_storage[y][x + 1] + producer_storage[y + 1][x + 1]);
    }
  }
  return 42;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D3(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  call   <puts@plt>
  lea    0x24(%rsp),%rax
  mov    %rax,(%rsp)
  mov    %rax,%r13
L23:
  lea    -0x14(%r13),%rbx
  xor    %r15d,%r15d
L12:
  mov    %r15d,%edi
  add    $0x4,%rbx
  add    %r12d,%r15d
  call   <sqrt@plt>
  mov    %rbp,%rdi
  mov    $0x1,%eax
  movss  %xmm0,-0x4(%rbx)
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    %r13,%rbx
  jne    L12
  mov    $0xa,%edi
  add    $0x1,%r12d
  lea    0x14(%rbx),%r13
  call   <putchar@plt>
  cmp    $0x5,%r12d
  jne    L23
  mov    (%rsp),%r15
  lea    0x50(%r14),%rbx
L25:
  movss  (%r14),%xmm2
  movss  (%r15),%xmm1
  mov    $0x1,%r12d
L32:
  addss  %xmm1,%xmm2
  movss  (%r15,%r12,4),%xmm1
  mov    %rbp,%rdi
  mov    $0x1,%eax
  movss  %xmm1,(%rsp)
  movaps %xmm2,%xmm0
  movss  (%r14,%r12,4),%xmm2
  add    $0x1,%r12
  addss  %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  addss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0x5,%r12
  movss  (%rsp),%xmm1
  movss  0xc(%rsp),%xmm2
  jne    L32
  mov    $0xa,%edi
  add    $0x14,%r14
  add    $0x14,%r15
  call   <putchar@plt>
  cmp    %r14,%rbx
  jne    L25
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x88,%rsp
  mov    $0x2a,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string
name: extract_symbolic_data
arguments: data_label: D2
data_type: string

tool: D3: Hi this is a program to test the performance characteristics of a couple specific things from lesson 8 of Halide's tutorial.

tool: D2: |%f|

assistant: ```c
int main() {
  printf("Hi this is a program to test the performance characteristics of a couple specific things from lesson 8 of Halide's tutorial.\n");
  float result[4][4];
  float result2[4][4];
  float producer_storage[5][5];
  int y, x;
  for (y = 0; y < 5; y++) {
    for (x = 0; x < 5; x++) {
      producer_storage[y][x] = sqrt(x * y);
      printf("|%f|", producer_storage[y][x]);
    }
    printf("\n");
  }
  for (y = 0; y < 4; y++) {
    for (x = 0; x < 4; x++) {
      result[y][x] = (producer_storage[y][x] + producer_storage[y + 1][x] + producer_storage[y][x + 1] + producer_storage[y + 1][x + 1]);
      printf("|%f|", result[y][x]);
    }
    printf("\n");
  }
  for (y = 0; y < 4; y++) {
    for (x = 0; x < 4; x++) {
      result2[x][y] = (producer_storage[y][x] + producer_storage[y + 1][x] + producer_storage[y][x + 1] + producer_storage[y + 1][x + 1]);
    }
  }
  return 42;
}
```