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
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L9
  lea    0xc(%rsp),%rax
  mov    %rax,0x10(%rsp)
L9:
  mov    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

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

tool: D7: %u %u

assistant: ```c
int main() {
  int *p, a;
  if (a)
    p = &a;
  printf("%u %u", p, &p);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L13
  lea    0xc(%rsp),%rax
  mov    %rax,0x10(%rsp)
L13:
  mov    0x10(%rsp),%rsi
  xor    %eax,%eax
  lea    0x10(%rsp),%rdx
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_infered: string

tool: D2: %u %u

assistant: ```c
int main() {
  int *p, a;
  if (a)
    p = &a;
  printf("%u %u", p, &p);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movb   $0x0,-0x1(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void foo(void) {
  enum foo_tag { e_foo };
  volatile unsigned char a = e_foo;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<foo>:
  endbr64
  movb   $0x0,-0x1(%rsp)
  ret
```

assistant: ```c
void foo(void) {
  enum foo_tag { e_foo };
  volatile unsigned char a = e_foo;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  movb   $0x0,-0x1(%rsp)
  ret
```

assistant: ```c
void foo(void) {
  enum foo_tag { e_foo };
  volatile unsigned char a = e_foo;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atoll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L57
L38:
  addq   $0x1,-0x18(%rbp)
L57:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L38
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x2b,%eax
  je     L53
  cmp    $0x2d,%eax
  jne    L12
  movl   $0x1,-0xc(%rbp)
L53:
  addq   $0x1,-0x18(%rbp)
  jmp    L12
L26:
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  add    %rcx,%rax
  mov    %rax,-0x8(%rbp)
L12:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L26
  cmpl   $0x0,-0xc(%rbp)
  je     L48
  mov    -0x8(%rbp),%rax
  neg    %rax
  jmp    L65
L48:
  mov    -0x8(%rbp),%rax
L65:
  leave
  ret
```

assistant: ```c
long long atoll(const char *str) {
  long long val = 0;
  bool neg = false;
  while (isspace(*str)) {
    str++;
  }
  switch (*str) {
  case '-':
    neg = true;
  case '+':
    str++;
  }
  while (isdigit(*str)) {
    val = (10 * val) + (*str++ - '0');
  }
  return neg ? -val : val;
}
```
user: What is the c source code of the assembly code below:

```asm
<atoll>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  jmp    L47
L69:
  add    $0x1,%rbp
L47:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L69
  movzbl 0x0(%rbp),%eax
  cmp    $0x2b,%al
  je     L14
  mov    $0x0,%r12d
  cmp    $0x2d,%al
  jne    L18
  mov    $0x1,%r12d
L57:
  add    $0x1,%rbp
  jmp    L18
L14:
  mov    $0x0,%r12d
  jmp    L57
L56:
  lea    (%rbx,%rbx,4),%rdx
  add    $0x1,%rbp
  movsbl -0x1(%rbp),%eax
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%rbx
L18:
  movsbl 0x0(%rbp),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L56
  mov    %rbx,%rax
  neg    %rax
  test   %r12d,%r12d
  cmovne %rax,%rbx
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long long atoll(const char *str) {
  long long val = 0;
  bool neg = false;
  while (isspace(*str)) {
    str++;
  }
  switch (*str) {
  case '-':
    neg = true;
  case '+':
    str++;
  }
  while (isdigit(*str)) {
    val = (10 * val) + (*str++ - '0');
  }
  return neg ? -val : val;
}
```
user: Translate the below assembly code into C programming language:

```asm
<atoll>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L5
  nopl   (%rax)
L15:
  add    $0x1,%rbx
L5:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L15
  movsbl (%rbx),%edi
  xor    %ebp,%ebp
  cmp    $0x2b,%dil
  je     L31
  cmp    $0x2d,%dil
  jne    L42
  mov    $0x1,%ebp
L31:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  jmp    L42
  nopl   0x0(%rax)
L44:
  movsbl (%rbx),%eax
  lea    (%r12,%r12,4),%rdx
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%r12
L42:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L44
  mov    %r12,%rax
  pop    %rbx
  neg    %rax
  test   %ebp,%ebp
  pop    %rbp
  cmovne %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long long atoll(const char *str) {
  long long val = 0;
  bool neg = false;
  while (isspace(*str)) {
    str++;
  }
  switch (*str) {
  case '-':
    neg = true;
  case '+':
    str++;
  }
  while (isdigit(*str)) {
    val = (10 * val) + (*str++ - '0');
  }
  return neg ? -val : val;
}
```
user: What is the c source code of the assembly code below?

```asm
<atoll>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L55
  nopl   (%rax)
L45:
  add    $0x1,%rbx
L55:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L45
  movzbl (%rbx),%eax
  xor    %ebp,%ebp
  cmp    $0x2b,%al
  je     L15
  cmp    $0x2d,%al
  jne    L27
  mov    $0x1,%ebp
L15:
  add    $0x1,%rbx
  jmp    L27
  nopl   0x0(%rax)
L34:
  movsbl (%rbx),%eax
  lea    (%r12,%r12,4),%rdx
  add    $0x1,%rbx
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%r12
L27:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L34
  mov    %r12,%rax
  pop    %rbx
  neg    %rax
  test   %ebp,%ebp
  pop    %rbp
  cmovne %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long long atoll(const char *str) {
  long long val = 0;
  bool neg = false;
  while (isspace(*str)) {
    str++;
  }
  switch (*str) {
  case '-':
    neg = true;
  case '+':
    str++;
  }
  while (isdigit(*str)) {
    val = (10 * val) + (*str++ - '0');
  }
  return neg ? -val : val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sieve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movl   $0x0,(%rax)
  movl   $0x2,-0x8(%rbp)
  jmp    L19
L35:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L35
  movl   $0x2,-0x8(%rbp)
  jmp    L3
L10:
  mov    -0x8(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L54
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
L54:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  addl   $0x1,-0x8(%rbp)
L3:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,-0x8(%rbp)
  jl     L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void sieve(int *arr, int n) {
  int i, j;
  arr[0] = 0;
  arr[1] = 0;
  for (i = 2; i < n; i++)
    arr[i] = 1;
  for (i = 2; i < sqrt(n); i++) {
    for (j = i * i; j < n; j += i)
      arr[j] = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sieve>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  movl   $0x0,(%rdi)
  movl   $0x0,0x4(%rdi)
  cmp    $0x2,%esi
  jle    L62
  lea    0x8(%rdi),%rax
  lea    -0x3(%rsi),%edx
  lea    0xc(%rdi,%rdx,4),%rdx
L28:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L28
L62:
  mov    $0x2,%ebx
  jmp    L53
L24:
  add    $0x1,%rbx
L53:
  mov    %ebp,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jle    L25
  mov    %ebx,%eax
  imul   %ebx,%eax
  cmp    %eax,%ebp
  jle    L24
  cltq
L33:
  movl   $0x0,(%r12,%rax,4)
  add    %rbx,%rax
  cmp    %eax,%ebp
  jg     L33
  jmp    L24
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void sieve(int *arr, int n) {
  int i, j;
  arr[0] = 0;
  arr[1] = 0;
  for (i = 2; i < n; i++)
    arr[i] = 1;
  for (i = 2; i < sqrt(n); i++) {
    for (j = i * i; j < n; j += i)
      arr[j] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sieve>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  movq   $0x0,(%rdi)
  cmp    $0x2,%esi
  jle    L43
  lea    -0x3(%rsi),%edx
  lea    0x8(%rdi),%rax
  lea    0xc(%rdi,%rdx,4),%rdx
  nopl   (%rax)
L1:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L1
L43:
  mov    %ebp,%edi
  mov    $0x2,%ebx
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jle    L40
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %ebx,%eax
  imul   %ebx,%eax
  cmp    %eax,%ebp
  jle    L53
  cltq
  nopl   0x0(%rax,%rax,1)
L3:
  movl   $0x0,(%r12,%rax,4)
  add    %rbx,%rax
  cmp    %eax,%ebp
  jg     L3
L53:
  mov    %ebp,%edi
  add    $0x1,%rbx
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jg     L48
L40:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void sieve(int *arr, int n) {
  int i, j;
  arr[0] = 0;
  arr[1] = 0;
  for (i = 2; i < n; i++)
    arr[i] = 1;
  for (i = 2; i < sqrt(n); i++) {
    for (j = i * i; j < n; j += i)
      arr[j] = 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sieve>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  movq   $0x0,(%rdi)
  cmp    $0x2,%esi
  jle    L103
  lea    -0x3(%rsi),%eax
  lea    -0x2(%rsi),%ecx
  cmp    $0x2,%eax
  jbe    L0
  mov    %ecx,%edx
  movdqa D6(%rip),%xmm0
  lea    0x8(%rdi),%rax
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x18(%rdi,%rdx,1),%rdx
  nopl   (%rax)
L69:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L69
  mov    %ecx,%edx
  and    $0xfffffffc,%edx
  lea    0x2(%rdx),%eax
  cmp    %edx,%ecx
  je     L103
L92:
  movslq %eax,%rdx
  movl   $0x1,(%r12,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%rax),%edx
  cmp    %edx,%ebp
  jle    L103
  movl   $0x1,0x4(%r12,%rcx,1)
  add    $0x2,%eax
  cmp    %eax,%ebp
  jle    L103
  movl   $0x1,0x8(%r12,%rcx,1)
L103:
  mov    %ebp,%edi
  mov    $0x2,%ebx
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jle    L66
  nopl   0x0(%rax,%rax,1)
L70:
  mov    %ebx,%eax
  imul   %ebx,%eax
  cmp    %eax,%ebp
  jle    L54
  cltq
  nopl   0x0(%rax,%rax,1)
L46:
  movl   $0x0,(%r12,%rax,4)
  add    %rbx,%rax
  cmp    %eax,%ebp
  jg     L46
L54:
  mov    %ebp,%edi
  add    $0x1,%rbx
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jg     L70
L66:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    $0x2,%eax
  jmp    L92
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type: byte[16]

tool: {"D6": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
void sieve(int *arr, int n) {
  int i, j;
  arr[0] = 0;
  arr[1] = 0;
  for (i = 2; i < n; i++)
    arr[i] = 1;
  for (i = 2; i < sqrt(n); i++) {
    for (j = i * i; j < n; j += i)
      arr[j] = 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xf,%edi
  call   <fixed_test@plt>
  mov    $0x5,%edi
  call   <fixed_test@plt>
  mov    $0x11,%edi
  call   <fixed_test@plt>
  mov    $0x6,%edi
  call   <fixed_test@plt>
  mov    $0x1f,%edi
  call   <fixed_test@plt>
  mov    $0x60,%edi
  call   <fixed_test@plt>
  mov    $0x0,%eax
  call   <random_test@plt>
  mov    $0x0,%eax
  call   <null_test@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  fixed_test(15);
  fixed_test(5);
  fixed_test(17);
  fixed_test(6);
  fixed_test(31);
  fixed_test(96);
  random_test();
  null_test();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf,%edi
  call   <fixed_test@plt>
  mov    $0x5,%edi
  call   <fixed_test@plt>
  mov    $0x11,%edi
  call   <fixed_test@plt>
  mov    $0x6,%edi
  call   <fixed_test@plt>
  mov    $0x1f,%edi
  call   <fixed_test@plt>
  mov    $0x60,%edi
  call   <fixed_test@plt>
  mov    $0x0,%eax
  call   <random_test@plt>
  mov    $0x0,%eax
  call   <null_test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  fixed_test(15);
  fixed_test(5);
  fixed_test(17);
  fixed_test(6);
  fixed_test(31);
  fixed_test(96);
  random_test();
  null_test();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xf,%edi
  call   <fixed_test@plt>
  mov    $0x5,%edi
  call   <fixed_test@plt>
  mov    $0x11,%edi
  call   <fixed_test@plt>
  mov    $0x6,%edi
  call   <fixed_test@plt>
  mov    $0x1f,%edi
  call   <fixed_test@plt>
  mov    $0x60,%edi
  call   <fixed_test@plt>
  xor    %eax,%eax
  call   <random_test@plt>
  xor    %eax,%eax
  call   <null_test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  fixed_test(15);
  fixed_test(5);
  fixed_test(17);
  fixed_test(6);
  fixed_test(31);
  fixed_test(96);
  random_test();
  null_test();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<vbench_median>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  sub    %eax,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  add    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int vbench_median(int a, int b, int c) {
  int t = (a - b) & ((a - b) >> 31);
  a -= t;
  b += t;
  b -= (b - c) & ((b - c) >> 31);
  b += (a - b) & ((a - b) >> 31);
  return b;
}
```
user: What is the c source code of the assembly code below?

```asm
<vbench_median>:
  endbr64
  mov    %edx,%r8d
  mov    %edi,%edx
  sub    %esi,%edx
  mov    %edx,%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  add    %eax,%esi
  mov    %esi,%ecx
  sub    %r8d,%ecx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  and    %ecx,%edx
  sub    %edx,%esi
  sub    %eax,%edi
  sub    %esi,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  and    %edi,%eax
  add    %esi,%eax
  ret
```

assistant: ```c
int vbench_median(int a, int b, int c) {
  int t = (a - b) & ((a - b) >> 31);
  a -= t;
  b += t;
  b -= (b - c) & ((b - c) >> 31);
  b += (a - b) & ((a - b) >> 31);
  return b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<vbench_median>:
  endbr64
  mov    %edx,%r8d
  mov    %edi,%edx
  sub    %esi,%edx
  mov    %edx,%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  add    %eax,%esi
  sub    %eax,%edi
  mov    %esi,%ecx
  sub    %r8d,%ecx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  and    %ecx,%edx
  sub    %edx,%esi
  sub    %esi,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  and    %edi,%eax
  add    %esi,%eax
  ret
```

assistant: ```c
int vbench_median(int a, int b, int c) {
  int t = (a - b) & ((a - b) >> 31);
  a -= t;
  b += t;
  b -= (b - c) & ((b - c) >> 31);
  b += (a - b) & ((a - b) >> 31);
  return b;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<vbench_median>:
  endbr64
  mov    %edx,%r8d
  mov    %edi,%edx
  sub    %esi,%edx
  mov    %edx,%eax
  sar    $0x1f,%eax
  and    %edx,%eax
  add    %eax,%esi
  sub    %eax,%edi
  mov    %esi,%ecx
  sub    %r8d,%ecx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  and    %ecx,%edx
  sub    %edx,%esi
  sub    %esi,%edi
  mov    %edi,%eax
  sar    $0x1f,%eax
  and    %edi,%eax
  add    %esi,%eax
  ret
```

assistant: ```c
int vbench_median(int a, int b, int c) {
  int t = (a - b) & ((a - b) >> 31);
  a -= t;
  b += t;
  b -= (b - c) & ((b - c) >> 31);
  b += (a - b) & ((a - b) >> 31);
  return b;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<printSomeCharAddresses>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x6e,0x6(%rsp)
  lea    0x6(%rsp),%rbx
  mov    %rbx,%rdx
  mov    $0x6e,%esi
  lea    D44(%rip),%rdi
  call   <printf@plt>
  movb   $0x6d,0x7(%rsp)
  lea    0x7(%rsp),%r12
  mov    %r12,%rdx
  mov    $0x6d,%esi
  lea    D56(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D85(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  sub    %ebx,%esi
  lea    D83(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D97(%rip),%rdi
  call   <puts@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  cmp    %rbx,%r12
  lea    D96(%rip),%rsi
  lea    D34(%rip),%rax
  cmovbe %rax,%rsi
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  sub    %ebp,%esi
  lea    D76(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D44
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D56
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D85
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D83
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D97
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D28
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D96
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D34
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D51
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D76
type_infered: string

tool: D44: Value of n: %c, Location of n: %p


tool: D56: Value of m: %c, Location of m: %p


tool: D85: Size of a char: %d


tool: D83: Difference between char pointers: %d



tool: D97: m declared after n

tool: D28: Where are they comparatively in memory?

tool: D96: Yes

tool: D34: No

tool: D51: Is mLoc > nLoc?: %s



tool: D76: Difference between char and ptr from main: %d



assistant: ```c
void printSomeCharAddresses(int *ptrFromMain) {
  char n = 'n';
  char *nLoc = &n;
  printf("Value of n: %c, Location of n: %p\n", n, nLoc);
  char m = 'm';
  char *mLoc = &m;
  printf("Value of m: %c, Location of m: %p\n", m, mLoc);
  printf("Size of a char: %d\n", (int)sizeof(char));
  printf("Difference between char pointers: %d\n\n", (int)mLoc - (int)nLoc);
  printf("m declared after n\n");
  printf("Where are they comparatively in memory?\n");
  printf("Is mLoc > nLoc?: %s\n\n", mLoc > nLoc ? "Yes" : "No");
  printf("Difference between char and ptr from main: %d\n\n", (int)nLoc - (int)ptrFromMain);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printSomeCharAddresses>:
  endbr64
  push   %r12
  mov    $0x6e,%esi
  push   %rbp
  mov    %rdi,%rbp
  lea    D8(%rip),%rdi
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbx
  lea    0x7(%rsp),%r12
  movb   $0x6e,0x6(%rsp)
  mov    %rbx,%rdx
  call   <printf@plt>
  mov    %r12,%rdx
  mov    $0x6d,%esi
  xor    %eax,%eax
  lea    D61(%rip),%rdi
  movb   $0x6d,0x7(%rsp)
  call   <printf@plt>
  mov    $0x1,%esi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  sub    %ebx,%esi
  call   <printf@plt>
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  cmp    %rbx,%r12
  lea    D63(%rip),%rax
  lea    D77(%rip),%rsi
  cmovbe %rax,%rsi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  xor    %eax,%eax
  lea    D91(%rip),%rdi
  sub    %ebp,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D8
type_guessed: string
name: parse_data
arguments: data_label: D61
type_guessed: string
name: parse_data
arguments: data_label: D56
type_guessed: string
name: parse_data
arguments: data_label: D10
type_guessed: string
name: parse_data
arguments: data_label: D22
type_guessed: string
name: parse_data
arguments: data_label: D12
type_guessed: string
name: parse_data
arguments: data_label: D63
type_guessed: string
name: parse_data
arguments: data_label: D77
type_guessed: string
name: parse_data
arguments: data_label: D40
type_guessed: string
name: parse_data
arguments: data_label: D91
type_guessed: string

tool: D8: Value of n: %c, Location of n: %p


tool: D61: Value of m: %c, Location of m: %p


tool: D56: Size of a char: %d


tool: D10: Difference between char pointers: %d



tool: D22: m declared after n

tool: D12: Where are they comparatively in memory?

tool: D63: No

tool: D77: Yes

tool: D40: Is mLoc > nLoc?: %s



tool: D91: Difference between char and ptr from main: %d



assistant: ```c
void printSomeCharAddresses(int *ptrFromMain) {
  char n = 'n';
  char *nLoc = &n;
  printf("Value of n: %c, Location of n: %p\n", n, nLoc);
  char m = 'm';
  char *mLoc = &m;
  printf("Value of m: %c, Location of m: %p\n", m, mLoc);
  printf("Size of a char: %d\n", (int)sizeof(char));
  printf("Difference between char pointers: %d\n\n", (int)mLoc - (int)nLoc);
  printf("m declared after n\n");
  printf("Where are they comparatively in memory?\n");
  printf("Is mLoc > nLoc?: %s\n\n", mLoc > nLoc ? "Yes" : "No");
  printf("Difference between char and ptr from main: %d\n\n", (int)nLoc - (int)ptrFromMain);
}
```