ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<vector_memalloc>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <malloc@plt>
  test   %rax,%rax
  je     L63
  test   %rbx,%rbx
  je     L63
  lea    -0x8(%rbx),%rcx
  mov    $0xfffffff7,%edx
  cmp    %rdx,%rcx
  ja     L149
  lea    -0x1(%rbx),%rdx
  cmp    $0xe,%rdx
  jbe    L73
  mov    %rbx,%rsi
  mov    %rax,%rcx
  pxor   %xmm0,%xmm0
  xor    %edx,%edx
  shr    $0x4,%rsi
  nop
L7:
  add    $0x1,%rdx
  movups %xmm0,(%rcx)
  add    $0x10,%rcx
  cmp    %rdx,%rsi
  ja     L7
  mov    %rbx,%rsi
  and    $0xfffffffffffffff0,%rsi
  lea    (%rax,%rsi,1),%rcx
  lea    0x1(%rsi),%edx
  cmp    %rsi,%rbx
  je     L63
  mov    %rbx,%rdi
  sub    %rsi,%rdi
  lea    -0x1(%rdi),%r8
  cmp    $0x6,%r8
  jbe    L67
L38:
  movq   $0x0,(%rax,%rsi,1)
  mov    %rdi,%rsi
  and    $0xfffffffffffffff8,%rsi
  add    %rsi,%rcx
  add    %esi,%edx
  cmp    %rdi,%rsi
  je     L63
L67:
  mov    %edx,%edi
  movb   $0x0,(%rcx)
  lea    0x1(%rdx),%esi
  cmp    %rdi,%rbx
  jbe    L63
  movb   $0x0,0x1(%rcx)
  lea    0x2(%rdx),%edi
  cmp    %rsi,%rbx
  jbe    L63
  movb   $0x0,0x2(%rcx)
  lea    0x3(%rdx),%esi
  cmp    %rdi,%rbx
  jbe    L63
  movb   $0x0,0x3(%rcx)
  lea    0x4(%rdx),%edi
  cmp    %rsi,%rbx
  jbe    L63
  movb   $0x0,0x4(%rcx)
  add    $0x5,%edx
  cmp    %rdi,%rbx
  jbe    L63
  movb   $0x0,0x5(%rcx)
  cmp    %rdx,%rbx
  jbe    L63
  movb   $0x0,0x6(%rcx)
L63:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L149:
  mov    $0x1,%edi
  mov    %rax,%rdx
  sub    %eax,%edi
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %rdx,%rsi
  lea    (%rdi,%rdx,1),%ecx
  add    $0x1,%rdx
  movb   $0x0,(%rsi)
  cmp    %rcx,%rbx
  ja     L10
  pop    %rbx
  ret
L73:
  mov    %rbx,%rdi
  mov    %rax,%rcx
  xor    %esi,%esi
  mov    $0x1,%edx
  jmp    L38
```

assistant: ```c
void *vector_memalloc(size_t size) {
  void *memory;
  unsigned int counter;
  char *explorer;
  counter = 0;
  memory = (void *)malloc(size);
  explorer = (char *)memory;
  if (memory != NULL) {
    while (counter++ < size)
      *explorer++ = 0;
  }
  return (memory);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_guessed: string

tool: D2: 
Usage: rm FILENAME ...



assistant: ```c
void usage(void) {
  printf("\nUsage: rm FILENAME ...\n\n");
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
datatype: string

tool: D7: 
Usage: rm FILENAME ...



assistant: ```c
void usage(void) {
  printf("\nUsage: rm FILENAME ...\n\n");
  exit(1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
data_type: string

tool: D0: 
Usage: rm FILENAME ...



assistant: ```c
void usage(void) {
  printf("\nUsage: rm FILENAME ...\n\n");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_guessed: string

tool: D3: 
Usage: rm FILENAME ...



assistant: ```c
void usage(void) {
  printf("\nUsage: rm FILENAME ...\n\n");
  exit(1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rot13>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%r9
  mov    %rdi,%rax
  lea    0x9(%rdi),%r8
  mov    $0x61,%r11d
  mov    $0x30,%r10d
  jmp    L74
L39:
  cmp    $0x9,%sil
  ja     L60
  movsbl %dl,%edx
  sub    $0x2b,%edx
  movslq %edx,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  lea    (%rcx,%rcx,4),%esi
  add    %esi,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %r10d,%edx
L11:
  add    %ecx,%edx
  mov    %dl,(%rdi)
L54:
  add    $0x1,%rax
  cmp    %r8,%rax
  je     L40
L74:
  mov    %rax,%rdi
  movzbl (%rax),%edx
  lea    -0x30(%rdx),%esi
  lea    -0x3a(%rdx),%ecx
  cmp    $0x6,%cl
  jbe    L54
  cmp    $0x4a,%sil
  ja     L54
  lea    -0x5b(%rdx),%ecx
  cmp    $0x5,%cl
  jbe    L54
  lea    -0x41(%rdx),%ecx
  cmp    $0x19,%cl
  ja     L39
  movb   $0x41,(%rax)
  jmp    L54
L60:
  movsbl %dl,%edx
  sub    $0x54,%edx
  movslq %edx,%rcx
  imul   $0x4ec4ec4f,%rcx,%rcx
  sar    $0x23,%rcx
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  imul   $0x1a,%ecx,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %r11d,%edx
  jmp    L11
L40:
  mov    %r9,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type: string

tool: D2: %s


assistant: ```c
void rot13(char arr[1000]) {
  int i, x = sizeof(arr);
  for (i = 0; i <= x; i++) {
    int number, n = 13;
    char ch = arr[i], ch1 = arr[i];
    if (((arr[i] <= '/') || (arr[i] >= ':') && (arr[i] <= 'A' - 1) || (arr[i] >= '[') && (arr[i] <= '`') || (arr[i] >= '{'))) {
      arr[i] = ch1;
    } else {
      if ((arr[i] >= 'A') && (arr[i] <= 'Z')) {
        arr[i] = 'A';
        number = ch - 'A';
        number = (number + n) % 26;
      } else {
        if ((arr[i] >= '0') && (arr[i] <= '9')) {
          number = ch - '0';
          arr[i] = '0';
          number = (number + 5) % 10;
        } else {
          number = ch - 'a';
          arr[i] = 'a';
          number = (number + n) % 26;
        }
        arr[i] = arr[i] + number;
      }
    }
  }
  printf("%s\n", arr);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rot13>:
  endbr64
  mov    %rdi,%r8
  mov    %rdi,%rdx
  lea    0x9(%rdi),%rdi
  mov    $0xcccccccd,%r9d
  jmp    L66
  cs nopw 0x0(%rax,%rax,1)
L28:
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  je     L14
L66:
  movzbl (%rdx),%eax
  lea    -0x3a(%rax),%ecx
  lea    -0x30(%rax),%esi
  cmp    $0x6,%cl
  jbe    L28
  cmp    $0x4a,%sil
  ja     L28
  lea    -0x5b(%rax),%ecx
  cmp    $0x5,%cl
  jbe    L28
  lea    -0x41(%rax),%ecx
  cmp    $0x19,%cl
  ja     L93
  movb   $0x41,(%rdx)
  add    $0x1,%rdx
  cmp    %rdx,%rdi
  jne    L66
L14:
  mov    %r8,%rsi
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L93:
  cmp    $0x9,%sil
  jbe    L77
  sub    $0x54,%eax
  movsbl %al,%ecx
  movsbq %al,%rax
  imul   $0x4ec4ec4f,%rax,%rax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sar    $0x23,%rax
  sub    %esi,%eax
  imul   $0x1a,%eax,%esi
  mov    %ecx,%eax
  mov    $0x61,%ecx
  sub    %esi,%eax
L0:
  add    %ecx,%eax
  mov    %al,(%rdx)
  jmp    L28
  nopw   0x0(%rax,%rax,1)
L77:
  sub    $0x2b,%eax
  movsbl %al,%eax
  mov    %rax,%rsi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  mov    %esi,%eax
  add    %ecx,%ecx
  sub    %ecx,%eax
  mov    $0x30,%ecx
  jmp    L0
```

tool_calls: name: read_data
arguments: label: D5
type_guessed: string

tool: D5: %s


assistant: ```c
void rot13(char arr[1000]) {
  int i, x = sizeof(arr);
  for (i = 0; i <= x; i++) {
    int number, n = 13;
    char ch = arr[i], ch1 = arr[i];
    if (((arr[i] <= '/') || (arr[i] >= ':') && (arr[i] <= 'A' - 1) || (arr[i] >= '[') && (arr[i] <= '`') || (arr[i] >= '{'))) {
      arr[i] = ch1;
    } else {
      if ((arr[i] >= 'A') && (arr[i] <= 'Z')) {
        arr[i] = 'A';
        number = ch - 'A';
        number = (number + n) % 26;
      } else {
        if ((arr[i] >= '0') && (arr[i] <= '9')) {
          number = ch - '0';
          arr[i] = '0';
          number = (number + 5) % 10;
        } else {
          number = ch - 'a';
          arr[i] = 'a';
          number = (number + n) % 26;
        }
        arr[i] = arr[i] + number;
      }
    }
  }
  printf("%s\n", arr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x10(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x10(%rbp),%eax
  imul   %eax,%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  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: read_data
arguments: data_label: D28
data_type: string
name: read_data
arguments: data_label: D29
data_type: string
name: read_data
arguments: data_label: D14
data_type: string

tool: D28: Enter an integer value: 

tool: D29: %d

tool: D14: The square of %d is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int num, sqr;
  printf("Enter an integer value: ");
  scanf("%d", &num);
  sqr = (num * num);
  printf("The square of %d is %d\n", num, (num * num));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    %esi,%edx
  imul   %esi,%edx
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D15
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D21
type_infered: string

tool: D6: Enter an integer value: 

tool: D15: %d

tool: D21: The square of %d is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int num, sqr;
  printf("Enter an integer value: ");
  scanf("%d", &num);
  sqr = (num * num);
  printf("The square of %d is %d\n", num, (num * num));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D18(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  mov    %esi,%edx
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D18
type_infered: string
name: read_data
arguments: data_label: D9
type_infered: string
name: read_data
arguments: data_label: D16
type_infered: string

tool: D18: Enter an integer value: 

tool: D9: %d

tool: D16: The square of %d is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int num, sqr;
  printf("Enter an integer value: ");
  scanf("%d", &num);
  sqr = (num * num);
  printf("The square of %d is %d\n", num, (num * num));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  mov    %esi,%edx
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D19
type: string
name: read_data
arguments: label: D17
type: string
name: read_data
arguments: label: D24
type: string

tool: D19: Enter an integer value: 

tool: D17: %d

tool: D24: The square of %d is %d


assistant: ```c
int main(int argc, char *argv[]) {
  int num, sqr;
  printf("Enter an integer value: ");
  scanf("%d", &num);
  sqr = (num * num);
  printf("The square of %d is %d\n", num, (num * num));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<binarySearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  jmp    L4
L45:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jge    L21
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L17
L21:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L17:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
L4:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jg     L18
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L45
L18:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L34
  mov    $0xffffffff,%eax
  jmp    L42
L34:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
L42:
  pop    %rbp
  ret
```

assistant: ```c
int binarySearch(int a[], int size, int item) {
  int beg, end, mid;
  beg = 0, end = size - 1;
  mid = (beg + end) / 2;
  while ((beg <= end) && a[mid] != item) {
    if (item < a[mid])
      end = mid - 1;
    else
      beg = mid + 1;
    mid = (beg + end) / 2;
  }
  if (beg > end)
    return -1;
  else
    return mid + 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<binarySearch>:
  endbr64
  mov    %rdi,%r8
  sub    $0x1,%esi
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  mov    $0x0,%edi
  test   %esi,%esi
  jns    L51
  mov    $0xffffffff,%eax
  ret
L17:
  lea    0x1(%rax),%edi
L53:
  lea    (%rdi,%rsi,1),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  cmp    %esi,%edi
  jg     L13
L51:
  movslq %eax,%rcx
  mov    (%r8,%rcx,4),%ecx
  cmp    %edx,%ecx
  je     L18
  jle    L17
  lea    -0x1(%rax),%esi
  jmp    L53
L18:
  cmp    %esi,%edi
  jg     L5
  add    $0x1,%eax
  ret
L13:
  mov    $0xffffffff,%eax
  ret
L5:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int binarySearch(int a[], int size, int item) {
  int beg, end, mid;
  beg = 0, end = size - 1;
  mid = (beg + end) / 2;
  while ((beg <= end) && a[mid] != item) {
    if (item < a[mid])
      end = mid - 1;
    else
      beg = mid + 1;
    mid = (beg + end) / 2;
  }
  if (beg > end)
    return -1;
  else
    return mid + 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<binarySearch>:
  endbr64
  sub    $0x1,%esi
  xor    %r8d,%r8d
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  test   %esi,%esi
  jns    L31
  jmp    L15
  nopl   0x0(%rax)
L28:
  lea    -0x1(%rax),%esi
L35:
  lea    (%r8,%rsi,1),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  cmp    %esi,%r8d
  jg     L15
L31:
  movslq %eax,%rcx
  cmp    %edx,(%rdi,%rcx,4)
  je     L41
  jg     L28
  lea    0x1(%rax),%r8d
  jmp    L35
  nopl   (%rax)
L41:
  add    $0x1,%eax
  ret
  nopl   0x0(%rax)
L15:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int binarySearch(int a[], int size, int item) {
  int beg, end, mid;
  beg = 0, end = size - 1;
  mid = (beg + end) / 2;
  while ((beg <= end) && a[mid] != item) {
    if (item < a[mid])
      end = mid - 1;
    else
      beg = mid + 1;
    mid = (beg + end) / 2;
  }
  if (beg > end)
    return -1;
  else
    return mid + 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<binarySearch>:
  endbr64
  sub    $0x1,%esi
  mov    %rdi,%r8
  xor    %ecx,%ecx
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  test   %esi,%esi
  js     L4
  nopl   0x0(%rax)
L1:
  movslq %eax,%rdi
  cmp    %edx,(%r8,%rdi,4)
  je     L9
L57:
  jle    L30
  lea    -0x1(%rax),%esi
  lea    (%rsi,%rcx,1),%edi
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  cmp    %ecx,%esi
  jge    L1
L4:
  mov    $0xffffffff,%eax
  ret
  nopl   0x0(%rax)
L30:
  lea    0x1(%rax),%ecx
  lea    (%rcx,%rsi,1),%edi
  mov    %edi,%eax
  shr    $0x1f,%eax
  add    %edi,%eax
  sar    %eax
  cmp    %esi,%ecx
  jg     L4
  movslq %eax,%rdi
  cmp    %edx,(%r8,%rdi,4)
  jne    L57
L9:
  add    $0x1,%eax
  ret
```

assistant: ```c
int binarySearch(int a[], int size, int item) {
  int beg, end, mid;
  beg = 0, end = size - 1;
  mid = (beg + end) / 2;
  while ((beg <= end) && a[mid] != item) {
    if (item < a[mid])
      end = mid - 1;
    else
      beg = mid + 1;
    mid = (beg + end) / 2;
  }
  if (beg > end)
    return -1;
  else
    return mid + 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0xa,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L9
L5:
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movl   $0x0,(%rax)
L9:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  jb     L5
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  const int SZ = 10;
  int *array = malloc(SZ * sizeof(int));
  for (int *p = array; p < &array[SZ];)
    *p++ = 0;
  free(array);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  lea    0x28(%rax),%rcx
  mov    %rax,%rdx
L5:
  add    $0x4,%rdx
  movl   $0x0,-0x4(%rdx)
  cmp    %rcx,%rdx
  jne    L5
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  const int SZ = 10;
  int *array = malloc(SZ * sizeof(int));
  for (int *p = array; p < &array[SZ];)
    *p++ = 0;
  free(array);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x28,%edi
  call   <malloc@plt>
  pxor   %xmm0,%xmm0
  movq   $0x0,0x20(%rax)
  mov    %rax,%rdi
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  const int SZ = 10;
  int *array = malloc(SZ * sizeof(int));
  for (int *p = array; p < &array[SZ];)
    *p++ = 0;
  free(array);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x28,%edi
  call   <malloc@plt>
  pxor   %xmm0,%xmm0
  movq   $0x0,0x20(%rax)
  mov    %rax,%rdi
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  call   <free@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  const int SZ = 10;
  int *array = malloc(SZ * sizeof(int));
  for (int *p = array; p < &array[SZ];)
    *p++ = 0;
  free(array);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<vec2_multiply_mat2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  movss  0x4(%rax),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x28(%rbp),%rax
  movss  (%rax),%xmm0
  movaps %xmm0,%xmm1
  mulss  -0x8(%rbp),%xmm1
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mulss  -0x4(%rbp),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  movaps %xmm0,%xmm1
  mulss  -0x8(%rbp),%xmm1
  mov    -0x28(%rbp),%rax
  add    $0xc,%rax
  movss  (%rax),%xmm0
  mulss  -0x4(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
float *vec2_multiply_mat2(float *result, float *a, float *m) {
  float x = a[0];
  float y = a[1];
  result[0] = m[0] * x + m[2] * y;
  result[1] = m[1] * x + m[3] * y;
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<vec2_multiply_mat2>:
  endbr64
  mov    %rdi,%rax
  movss  (%rsi),%xmm0
  movss  0x4(%rsi),%xmm1
  movaps %xmm0,%xmm2
  mulss  (%rdx),%xmm2
  movaps %xmm1,%xmm3
  mulss  0x8(%rdx),%xmm3
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rdx),%xmm0
  mulss  0xc(%rdx),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
float *vec2_multiply_mat2(float *result, float *a, float *m) {
  float x = a[0];
  float y = a[1];
  result[0] = m[0] * x + m[2] * y;
  result[1] = m[1] * x + m[3] * y;
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<vec2_multiply_mat2>:
  endbr64
  movss  (%rsi),%xmm0
  movss  0x4(%rsi),%xmm1
  mov    %rdi,%rax
  movss  (%rdx),%xmm2
  movss  0x8(%rdx),%xmm3
  mulss  %xmm0,%xmm2
  mulss  %xmm1,%xmm3
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rdx),%xmm0
  mulss  0xc(%rdx),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
float *vec2_multiply_mat2(float *result, float *a, float *m) {
  float x = a[0];
  float y = a[1];
  result[0] = m[0] * x + m[2] * y;
  result[1] = m[1] * x + m[3] * y;
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<vec2_multiply_mat2>:
  endbr64
  movss  (%rsi),%xmm0
  movss  0x4(%rsi),%xmm1
  mov    %rdi,%rax
  movss  (%rdx),%xmm2
  movss  0x8(%rdx),%xmm3
  mulss  %xmm0,%xmm2
  mulss  %xmm1,%xmm3
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rdx),%xmm0
  mulss  0xc(%rdx),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,0x4(%rdi)
  ret
```

assistant: ```c
float *vec2_multiply_mat2(float *result, float *a, float *m) {
  float x = a[0];
  float y = a[1];
  result[0] = m[0] * x + m[2] * y;
  result[1] = m[1] * x + m[3] * y;
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Ego>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <pthread_self@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int Ego() {
  int return_val;
  return_val = (int)pthread_self();
  return (return_val);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Ego>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <pthread_self@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int Ego() {
  int return_val;
  return_val = (int)pthread_self();
  return (return_val);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

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

assistant: ```c
int Ego() {
  int return_val;
  return_val = (int)pthread_self();
  return (return_val);
}
```
user: What is the c source code of the assembly code below?

```asm
<putc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  nop
L4:
  mov    $0xe2900010,%eax
  mov    (%rax),%rax
  and    $0x4,%eax
  test   %rax,%rax
  je     L4
  mov    $0xe2900020,%edx
  movsbq -0x4(%rbp),%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void putc(char c) {
  while (!((*((volatile unsigned long *)0xE2900010)) & (1 << 2)))
    ;
  (*((volatile unsigned long *)0xE2900020)) = c;
}
```