ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<mmul>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  cmp    %edx,D37(%rip)
  jge    L0
  mov    D32(%rip),%rdi
  test   %rdi,%rdi
  je     L2
  call   <MemFree@plt>
L2:
  mov    %r12d,D23(%rip)
  lea    0x0(,%r12,8),%edi
  call   <MemAlloc@plt>
  mov    %rax,D23(%rip)
L0:
  test   %r12d,%r12d
  jle    L44
  mov    D23(%rip),%r8
  mov    %r8,%rdi
  movslq %r12d,%r9
  shl    $0x3,%r9
  mov    %rbp,%rcx
  lea    -0x1(%r12),%eax
  lea    0x8(%r8,%rax,8),%r10
  mov    %r12d,%esi
L13:
  mov    %rdi,%rdx
  movq   $0x0,(%rdi)
  mov    $0x0,%eax
L35:
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  (%rbx,%rax,8),%xmm0
  addsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdx)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L35
  add    $0x8,%rdi
  add    %r9,%rcx
  cmp    %r10,%rdi
  jne    L13
  mov    $0x0,%eax
L25:
  movsd  (%r8,%rax,8),%xmm0
  movsd  %xmm0,(%rbx,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L25
L44:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void mmul(double *a, double *b, int n) {
  int i, j;
  static double *r = 0;
  static int rsize = 0;
  if (rsize < n) {
    if (r)
      MemFree(r);
    rsize = n;
    r = (double *)MemAlloc(sizeof(double) * rsize);
  }
  for (i = 0; i < n; i++) {
    r[i] = 0.0;
    for (j = 0; j < n; j++)
      r[i] += a[i * n + j] * b[j];
  }
  for (i = 0; i < n; i++)
    b[i] = r[i];
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mmul>:
  endbr64
  cmp    %edx,D17(%rip)
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edx,%ebx
  jge    L48
  mov    D19(%rip),%rdi
  test   %rdi,%rdi
  je     L36
  call   <MemFree@plt>
L36:
  lea    0x0(,%rbx,8),%edi
  mov    %ebx,D36(%rip)
  call   <MemAlloc@plt>
  mov    %rax,D36(%rip)
L48:
  test   %ebx,%ebx
  jle    L42
  mov    D36(%rip),%rdi
  movslq %ebx,%r8
  lea    -0x1(%rbx),%eax
  mov    %r12,%rsi
  shl    $0x3,%r8
  pxor   %xmm2,%xmm2
  mov    %ebx,%edx
  mov    %rdi,%rcx
  lea    0x8(%rdi,%rax,8),%r9
  nopl   0x0(%rax,%rax,1)
L38:
  movq   $0x0,(%rcx)
  xor    %eax,%eax
  movapd %xmm2,%xmm1
  nopl   (%rax)
L46:
  movsd  (%rsi,%rax,8),%xmm0
  mulsd  0x0(%rbp,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rcx)
  cmp    %rax,%rdx
  jne    L46
  add    $0x8,%rcx
  add    %r8,%rsi
  cmp    %r9,%rcx
  jne    L38
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L31:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,0x0(%rbp,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L31
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void mmul(double *a, double *b, int n) {
  int i, j;
  static double *r = 0;
  static int rsize = 0;
  if (rsize < n) {
    if (r)
      MemFree(r);
    rsize = n;
    r = (double *)MemAlloc(sizeof(double) * rsize);
  }
  for (i = 0; i < n; i++) {
    r[i] = 0.0;
    for (j = 0; j < n; j++)
      r[i] += a[i * n + j] * b[j];
  }
  for (i = 0; i < n; i++)
    b[i] = r[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mmul>:
  endbr64
  cmp    %edx,D33(%rip)
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rsi,%rbx
  jge    L17
  mov    D37(%rip),%rdi
  test   %rdi,%rdi
  je     L42
  call   <MemFree@plt>
L42:
  lea    0x0(,%rbp,8),%edi
  mov    %ebp,D14(%rip)
  call   <MemAlloc@plt>
  mov    %rax,D14(%rip)
L17:
  test   %ebp,%ebp
  jle    L71
  mov    D14(%rip),%rdi
  lea    -0x1(%rbp),%eax
  movslq %ebp,%r8
  mov    %r12,%rcx
  shl    $0x3,%r8
  mov    %rax,%r11
  pxor   %xmm2,%xmm2
  mov    %ebp,%esi
  lea    0x8(%rdi),%r10
  mov    %rdi,%rdx
  lea    (%r10,%rax,8),%r9
  nopw   0x0(%rax,%rax,1)
L99:
  movq   $0x0,(%rdx)
  xor    %eax,%eax
  movapd %xmm2,%xmm1
  nopl   (%rax)
L40:
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  (%rbx,%rax,8),%xmm0
  add    $0x1,%rax
  addsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %rax,%rsi
  jne    L40
  add    $0x8,%rdx
  add    %r8,%rcx
  cmp    %r9,%rdx
  jne    L99
  cmp    %r10,%rbx
  je     L45
  cmp    $0x2,%r11d
  jbe    L45
  mov    %ebp,%edx
  xor    %eax,%eax
  shr    %edx
  shl    $0x4,%rdx
  nopl   0x0(%rax)
L13:
  movupd (%rdi,%rax,1),%xmm3
  movups %xmm3,(%rbx,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L13
  mov    %ebp,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ebp
  je     L71
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rbx,%rax,8)
L71:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L45:
  xor    %eax,%eax
  nopl   (%rax)
L67:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rbx,%rax,8)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L67
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void mmul(double *a, double *b, int n) {
  int i, j;
  static double *r = 0;
  static int rsize = 0;
  if (rsize < n) {
    if (r)
      MemFree(r);
    rsize = n;
    r = (double *)MemAlloc(sizeof(double) * rsize);
  }
  for (i = 0; i < n; i++) {
    r[i] = 0.0;
    for (j = 0; j < n; j++)
      r[i] += a[i * n + j] * b[j];
  }
  for (i = 0; i < n; i++)
    b[i] = r[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copystr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L11
  cmpq   $0x0,-0x20(%rbp)
  je     L11
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <reclaim@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L57
L17:
  addl   $0x1,-0x4(%rbp)
L57:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  jmp    L14
L55:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x4(%rbp)
L14:
  cmpl   $0x0,-0x4(%rbp)
  jns    L55
  jmp    L65
L11:
  nop
L65:
  leave
  ret
```

assistant: ```c
void copystr(char **target_string, char *source_string) {
  int i;
  if (target_string == NULL || source_string == NULL)
    return;
  reclaim(*target_string);
  i = 0;
  while (source_string[i] != '\0')
    i++;
  *target_string = (char *)calloc(i + 1, sizeof(char));
  while (i >= 0) {
    *(*target_string + i) = source_string[i];
    i--;
  }
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<copystr>:
  endbr64
  test   %rdi,%rdi
  je     L30
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  test   %rsi,%rsi
  je     L2
  mov    (%rdi),%rdi
  call   <reclaim@plt>
  cmpb   $0x0,(%r12)
  je     L0
  lea    0x1(%r12),%rax
  mov    $0x0,%ebx
L13:
  mov    %ebx,%edi
  add    $0x1,%ebx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L13
  add    $0x2,%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  test   %ebx,%ebx
  js     L2
L67:
  movslq %ebx,%rax
L6:
  movzbl (%r12,%rax,1),%ecx
  mov    0x0(%rbp),%rdx
  mov    %cl,(%rdx,%rax,1)
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L6
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,0x0(%rbp)
  mov    $0x0,%ebx
  jmp    L67
L30:
  ret
```

assistant: ```c
void copystr(char **target_string, char *source_string) {
  int i;
  if (target_string == NULL || source_string == NULL)
    return;
  reclaim(*target_string);
  i = 0;
  while (source_string[i] != '\0')
    i++;
  *target_string = (char *)calloc(i + 1, sizeof(char));
  while (i >= 0) {
    *(*target_string + i) = source_string[i];
    i--;
  }
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<copystr>:
  endbr64
  test   %rdi,%rdi
  je     L4
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  test   %rsi,%rsi
  je     L41
  mov    %rdi,%r12
  mov    (%rdi),%rdi
  call   <reclaim@plt>
  cmpb   $0x0,0x0(%rbp)
  je     L47
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbp,%rax,1)
  jne    L2
  lea    0x1(%rbx),%edi
L20:
  mov    $0x1,%esi
  call   <calloc@plt>
  movslq %ebx,%rdx
  mov    %rax,(%r12)
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L44:
  mov    (%r12),%rax
L50:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  sub    $0x1,%rdx
  cmp    $0xffffffff,%edx
  jne    L44
L41:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  ret
  nopl   0x0(%rax)
L47:
  mov    $0x1,%edi
  xor    %ebx,%ebx
  jmp    L20
```

assistant: ```c
void copystr(char **target_string, char *source_string) {
  int i;
  if (target_string == NULL || source_string == NULL)
    return;
  reclaim(*target_string);
  i = 0;
  while (source_string[i] != '\0')
    i++;
  *target_string = (char *)calloc(i + 1, sizeof(char));
  while (i >= 0) {
    *(*target_string + i) = source_string[i];
    i--;
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copystr>:
  endbr64
  test   %rdi,%rdi
  je     L23
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  test   %rsi,%rsi
  je     L28
  mov    %rdi,%r12
  mov    (%rdi),%rdi
  call   <reclaim@plt>
  cmpb   $0x0,0x0(%rbp)
  je     L39
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L27:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rbp,%rax,1)
  jne    L27
  lea    0x1(%rbx),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  movslq %ebx,%rdx
  mov    %rax,(%r12)
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  lea    -0x1(%rbx),%eax
  cltq
  xchg   %ax,%ax
L57:
  movzbl 0x0(%rbp,%rax,1),%ecx
  mov    (%r12),%rdx
  mov    %cl,(%rdx,%rax,1)
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L57
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  ret
  nopl   0x0(%rax)
L39:
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,(%r12)
  movzbl 0x0(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L28
```

assistant: ```c
void copystr(char **target_string, char *source_string) {
  int i;
  if (target_string == NULL || source_string == NULL)
    return;
  reclaim(*target_string);
  i = 0;
  while (source_string[i] != '\0')
    i++;
  *target_string = (char *)calloc(i + 1, sizeof(char));
  while (i >= 0) {
    *(*target_string + i) = source_string[i];
    i--;
  }
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<op_0xe7>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x8,%edx
  mov    $0x5,%esi
  mov    $0x20,%edi
  call   <RST@plt>
  mov    $0xb,%edi
  call   <T_WAIT_UNTIL@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void op_0xe7(void) {
  RST(0x20, 5, 8);
  T_WAIT_UNTIL(11);
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<op_0xe7>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edx
  mov    $0x5,%esi
  mov    $0x20,%edi
  call   <RST@plt>
  mov    $0xb,%edi
  call   <T_WAIT_UNTIL@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void op_0xe7(void) {
  RST(0x20, 5, 8);
  T_WAIT_UNTIL(11);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<op_0xe7>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x20,%edi
  mov    $0x8,%edx
  mov    $0x5,%esi
  call   <RST@plt>
  mov    $0xb,%edi
  add    $0x8,%rsp
  jmp    1070 <T_WAIT_UNTIL@plt>
```

assistant: ```c
void op_0xe7(void) {
  RST(0x20, 5, 8);
  T_WAIT_UNTIL(11);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<op_0xe7>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x20,%edi
  mov    $0x8,%edx
  mov    $0x5,%esi
  call   <RST@plt>
  mov    $0xb,%edi
  add    $0x8,%rsp
  jmp    1070 <T_WAIT_UNTIL@plt>
```

assistant: ```c
void op_0xe7(void) {
  RST(0x20, 5, 8);
  T_WAIT_UNTIL(11);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<empty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x64,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D7
type: string

tool: 

assistant: ```c
void empty(char **name) {
  *name = (char *)malloc(100 * sizeof(char));
  sprintf(*name, "");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x64,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%rbx)
  lea    D2(%rip),%rsi
  call   <sprintf@plt>
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D2
type_infered: string

tool: 

assistant: ```c
void empty(char **name) {
  *name = (char *)malloc(100 * sizeof(char));
  sprintf(*name, "");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x64,%edi
  call   <malloc@plt>
  lea    D5(%rip),%rsi
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  pop    %rbx
  jmp    1070 <sprintf@plt>
```

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

tool: 

assistant: ```c
void empty(char **name) {
  *name = (char *)malloc(100 * sizeof(char));
  sprintf(*name, "");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<empty>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x64,%edi
  call   <malloc@plt>
  lea    D0(%rip),%rsi
  mov    %rax,(%rbx)
  mov    %rax,%rdi
  pop    %rbx
  jmp    1070 <sprintf@plt>
```

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

tool: 

assistant: ```c
void empty(char **name) {
  *name = (char *)malloc(100 * sizeof(char));
  sprintf(*name, "");
}
```
user: Please show me the C implementation that matches the assembly code listed 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)
  jmp    L9
L48:
  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)
  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    L52
  mov    -0x4(%rbp),%eax
  jmp    L32
L52:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jle    L53
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L9
L53:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L9:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L48
  mov    $0xffffffff,%eax
L32:
  pop    %rbp
  ret
```

assistant: ```c
int binarySearch(int *A, int n, int key) {
  int p = 0, q, r = (n - 1);
  while (p < r) {
    q = (p + r) / 2;
    if (key == *(A + q)) {
      return q;
    } else if (key > *(A + q)) {
      p = q + 1;
    } else {
      r = q - 1;
    }
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<binarySearch>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L39
  mov    $0x0,%r8d
  jmp    L58
L20:
  lea    -0x1(%rax),%esi
L11:
  cmp    %esi,%r8d
  jge    L67
L58:
  lea    (%r8,%rsi,1),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  mov    %eax,%r9d
  movslq %eax,%rcx
  mov    (%rdi,%rcx,4),%ecx
  cmp    %edx,%ecx
  je     L61
  jge    L20
  lea    0x1(%rax),%r8d
  jmp    L11
L67:
  mov    $0xffffffff,%r9d
L61:
  mov    %r9d,%eax
  ret
L39:
  mov    $0xffffffff,%r9d
  jmp    L61
```

assistant: ```c
int binarySearch(int *A, int n, int key) {
  int p = 0, q, r = (n - 1);
  while (p < r) {
    q = (p + r) / 2;
    if (key == *(A + q)) {
      return q;
    } else if (key > *(A + q)) {
      p = q + 1;
    } else {
      r = q - 1;
    }
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<binarySearch>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L38
  xor    %ecx,%ecx
  jmp    L22
  nop
L16:
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  jle    L38
L22:
  lea    (%rsi,%rcx,1),%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  je     L15
  jl     L16
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jg     L22
L38:
  mov    $0xffffffff,%eax
L15:
  ret
```

assistant: ```c
int binarySearch(int *A, int n, int key) {
  int p = 0, q, r = (n - 1);
  while (p < r) {
    q = (p + r) / 2;
    if (key == *(A + q)) {
      return q;
    } else if (key > *(A + q)) {
      p = q + 1;
    } else {
      r = q - 1;
    }
  }
  return -1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<binarySearch>:
  endbr64
  sub    $0x1,%esi
  test   %esi,%esi
  jle    L22
  xor    %ecx,%ecx
  nopl   (%rax)
L53:
  lea    (%rsi,%rcx,1),%eax
  sar    %eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  je     L32
  jge    L9
  lea    0x1(%rax),%ecx
  cmp    %esi,%ecx
  jl     L53
L22:
  mov    $0xffffffff,%eax
L32:
  ret
  nopl   (%rax)
L9:
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jg     L53
  mov    $0xffffffff,%eax
  jmp    L32
```

assistant: ```c
int binarySearch(int *A, int n, int key) {
  int p = 0, q, r = (n - 1);
  while (p < r) {
    q = (p + r) / 2;
    if (key == *(A + q)) {
      return q;
    } else if (key > *(A + q)) {
      p = q + 1;
    } else {
      r = q - 1;
    }
  }
  return -1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<timestamp_worker>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <test__date_to_timestamp@plt>
  mov    -0x8(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
void *timestamp_worker(void *data) {
  int *ret = (int *)data;
  *ret = test__date_to_timestamp();
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<timestamp_worker>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x0,%eax
  call   <test__date_to_timestamp@plt>
  mov    %eax,(%rbx)
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
void *timestamp_worker(void *data) {
  int *ret = (int *)data;
  *ret = test__date_to_timestamp();
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<timestamp_worker>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %rdi,%rbx
  call   <test__date_to_timestamp@plt>
  mov    %eax,(%rbx)
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
void *timestamp_worker(void *data) {
  int *ret = (int *)data;
  *ret = test__date_to_timestamp();
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<timestamp_worker>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %rdi,%rbx
  call   <test__date_to_timestamp@plt>
  mov    %eax,(%rbx)
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
void *timestamp_worker(void *data) {
  int *ret = (int *)data;
  *ret = test__date_to_timestamp();
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_revstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L37
  mov    $0x0,%eax
  jmp    L12
L37:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <my_put_nbr@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L31
L13:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
L31:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x8(%rbp)
  jl     L13
  mov    -0x18(%rbp),%rax
L12:
  leave
  ret
```

assistant: ```c
char *my_revstr(char *str) {
  int len;
  int lp;
  char cp;
  if (str == NULL)
    return (NULL);
  len = my_strlen(str);
  my_put_nbr(len);
  lp = 0;
  while (lp < (len / 2)) {
    cp = str[lp];
    str[lp] = str[len - lp - 1];
    str[len - lp - 1] = cp;
    ++lp;
  }
  return (str);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<my_revstr>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L3
  call   <my_strlen@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <my_put_nbr@plt>
  mov    %ebp,%edi
  shr    $0x1f,%edi
  add    %ebp,%edi
  sar    %edi
  cmp    $0x1,%ebp
  jle    L3
  movslq %ebp,%rbp
  lea    -0x1(%rbx,%rbp,1),%rdx
  mov    $0x0,%eax
L2:
  movzbl (%rbx,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,(%rbx,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%edi
  jg     L2
L3:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *my_revstr(char *str) {
  int len;
  int lp;
  char cp;
  if (str == NULL)
    return (NULL);
  len = my_strlen(str);
  my_put_nbr(len);
  lp = 0;
  while (lp < (len / 2)) {
    cp = str[lp];
    str[lp] = str[len - lp - 1];
    str[len - lp - 1] = cp;
    ++lp;
  }
  return (str);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_revstr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L10
  call   <my_strlen@plt>
  movslq %eax,%rbx
  mov    %ebx,%edi
  call   <my_put_nbr@plt>
  mov    %ebx,%edi
  shr    $0x1f,%edi
  add    %ebx,%edi
  sar    %edi
  cmp    $0x1,%ebx
  jle    L10
  lea    -0x1(%r12,%rbx,1),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L5:
  movzbl (%rdx),%esi
  movzbl (%r12,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r12,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L5
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *my_revstr(char *str) {
  int len;
  int lp;
  char cp;
  if (str == NULL)
    return (NULL);
  len = my_strlen(str);
  my_put_nbr(len);
  lp = 0;
  while (lp < (len / 2)) {
    cp = str[lp];
    str[lp] = str[len - lp - 1];
    str[len - lp - 1] = cp;
    ++lp;
  }
  return (str);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_revstr>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L7
  call   <my_strlen@plt>
  movslq %eax,%rbx
  mov    %ebx,%edi
  call   <my_put_nbr@plt>
  mov    %ebx,%edi
  shr    $0x1f,%edi
  add    %ebx,%edi
  sar    %edi
  cmp    $0x1,%ebx
  jle    L7
  lea    -0x1(%r12,%rbx,1),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L6:
  movzbl (%rdx),%esi
  movzbl (%r12,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%r12,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%edi
  jg     L6
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *my_revstr(char *str) {
  int len;
  int lp;
  char cp;
  if (str == NULL)
    return (NULL);
  len = my_strlen(str);
  my_put_nbr(len);
  lp = 0;
  while (lp < (len / 2)) {
    cp = str[lp];
    str[lp] = str[len - lp - 1];
    str[len - lp - 1] = cp;
    ++lp;
  }
  return (str);
}
```