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
<multiply4x4>:
  endbr64
  sub    $0x58,%rsp
  movups (%rdi),%xmm0
  movups 0x10(%rsi),%xmm5
  mov    $0x40,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movups (%rsi),%xmm4
  movups 0x20(%rsi),%xmm3
  movaps %xmm0,%xmm1
  movaps %xmm0,%xmm6
  movups 0x30(%rsi),%xmm2
  mov    %rsp,%rsi
  shufps $0x0,%xmm0,%xmm1
  shufps $0x55,%xmm0,%xmm6
  mulps  %xmm4,%xmm1
  mulps  %xmm5,%xmm6
  addps  %xmm6,%xmm1
  movaps %xmm0,%xmm6
  shufps $0xaa,%xmm0,%xmm6
  mulps  %xmm3,%xmm6
  shufps $0xff,%xmm0,%xmm0
  mulps  %xmm2,%xmm0
  addps  %xmm6,%xmm1
  addps  %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  movups 0x10(%rdi),%xmm0
  movaps %xmm0,%xmm1
  movaps %xmm0,%xmm6
  shufps $0x0,%xmm0,%xmm1
  shufps $0x55,%xmm0,%xmm6
  mulps  %xmm4,%xmm1
  mulps  %xmm5,%xmm6
  addps  %xmm6,%xmm1
  movaps %xmm0,%xmm6
  shufps $0xaa,%xmm0,%xmm6
  mulps  %xmm3,%xmm6
  shufps $0xff,%xmm0,%xmm0
  mulps  %xmm2,%xmm0
  addps  %xmm6,%xmm1
  addps  %xmm1,%xmm0
  movaps %xmm0,0x10(%rsp)
  movups 0x20(%rdi),%xmm0
  movaps %xmm0,%xmm1
  movaps %xmm0,%xmm6
  shufps $0x55,%xmm0,%xmm1
  shufps $0x0,%xmm0,%xmm6
  mulps  %xmm5,%xmm1
  mulps  %xmm4,%xmm6
  addps  %xmm6,%xmm1
  movaps %xmm0,%xmm6
  shufps $0xaa,%xmm0,%xmm6
  mulps  %xmm3,%xmm6
  shufps $0xff,%xmm0,%xmm0
  mulps  %xmm2,%xmm0
  addps  %xmm6,%xmm1
  addps  %xmm1,%xmm0
  movaps %xmm0,0x20(%rsp)
  movups 0x30(%rdi),%xmm0
  movaps %xmm0,%xmm1
  shufps $0x55,%xmm0,%xmm1
  mulps  %xmm5,%xmm1
  movaps %xmm0,%xmm5
  shufps $0x0,%xmm0,%xmm5
  mulps  %xmm5,%xmm4
  addps  %xmm4,%xmm1
  movaps %xmm0,%xmm4
  shufps $0xaa,%xmm0,%xmm4
  mulps  %xmm4,%xmm3
  shufps $0xff,%xmm0,%xmm0
  mulps  %xmm2,%xmm0
  addps  %xmm3,%xmm1
  addps  %xmm1,%xmm0
  movaps %xmm0,0x30(%rsp)
  call   <memcpy@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x58,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void multiply4x4(float *m1, const float *m2) {
  float temp[16];
  int x, y;
  for (x = 0; x < 4; x++) {
    for (y = 0; y < 4; y++) {
      temp[y + (x * 4)] = (m1[x * 4] * m2[y]) + (m1[(x * 4) + 1] * m2[y + 4]) + (m1[(x * 4) + 2] * m2[y + 8]) + (m1[(x * 4) + 3] * m2[y + 12]);
    }
  }
  memcpy(m1, temp, sizeof(float) << 4);
}
```
user: What is the c source code of the assembly code below:

```asm
<WriteDma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  mov    $0x5a,%esi
  mov    $0xb,%edi
  call   <outb@plt>
  mov    $0x2,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void WriteDma(void) {
  outb(0x0a, 0x06);
  outb(0x0b, 0x5A);
  outb(0x0a, 0x02);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<WriteDma>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  mov    $0x5a,%esi
  mov    $0xb,%edi
  call   <outb@plt>
  mov    $0x2,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void WriteDma(void) {
  outb(0x0a, 0x06);
  outb(0x0b, 0x5A);
  outb(0x0a, 0x02);
}
```
user: Translate the below assembly code into C programming language:

```asm
<WriteDma>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  mov    $0x5a,%esi
  mov    $0xb,%edi
  call   <outb@plt>
  mov    $0x2,%esi
  mov    $0xa,%edi
  add    $0x8,%rsp
  jmp    1050 <outb@plt>
```

assistant: ```c
void WriteDma(void) {
  outb(0x0a, 0x06);
  outb(0x0b, 0x5A);
  outb(0x0a, 0x02);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<WriteDma>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  mov    $0xa,%edi
  call   <outb@plt>
  mov    $0x5a,%esi
  mov    $0xb,%edi
  call   <outb@plt>
  mov    $0x2,%esi
  mov    $0xa,%edi
  add    $0x8,%rsp
  jmp    1050 <outb@plt>
```

assistant: ```c
void WriteDma(void) {
  outb(0x0a, 0x06);
  outb(0x0b, 0x5A);
  outb(0x0a, 0x02);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<HotPlugService>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <DisableInterrupts@plt>
  mov    $0x0,%eax
  call   <EnableTMDS@plt>
  mov    $0x0,%eax
  call   <TxPowerStateD0@plt>
  mov    $0x0,%eax
  call   <EnableInterrupts@plt>
  mov    $0x0,%eax
  call   <CheckTxFifoStable@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void HotPlugService(void) {
  DisableInterrupts();
  EnableTMDS();
  TxPowerStateD0();
  EnableInterrupts();
  CheckTxFifoStable();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<HotPlugService>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <DisableInterrupts@plt>
  mov    $0x0,%eax
  call   <EnableTMDS@plt>
  mov    $0x0,%eax
  call   <TxPowerStateD0@plt>
  mov    $0x0,%eax
  call   <EnableInterrupts@plt>
  mov    $0x0,%eax
  call   <CheckTxFifoStable@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void HotPlugService(void) {
  DisableInterrupts();
  EnableTMDS();
  TxPowerStateD0();
  EnableInterrupts();
  CheckTxFifoStable();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<HotPlugService>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DisableInterrupts@plt>
  xor    %eax,%eax
  call   <EnableTMDS@plt>
  xor    %eax,%eax
  call   <TxPowerStateD0@plt>
  xor    %eax,%eax
  call   <EnableInterrupts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <CheckTxFifoStable@plt>
```

assistant: ```c
void HotPlugService(void) {
  DisableInterrupts();
  EnableTMDS();
  TxPowerStateD0();
  EnableInterrupts();
  CheckTxFifoStable();
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HotPlugService>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <DisableInterrupts@plt>
  xor    %eax,%eax
  call   <EnableTMDS@plt>
  xor    %eax,%eax
  call   <TxPowerStateD0@plt>
  xor    %eax,%eax
  call   <EnableInterrupts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10c0 <CheckTxFifoStable@plt>
```

assistant: ```c
void HotPlugService(void) {
  DisableInterrupts();
  EnableTMDS();
  TxPowerStateD0();
  EnableInterrupts();
  CheckTxFifoStable();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L16
L15:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t len = strlen(s);
  char *new = malloc(len + 1);
  if (!new)
    return 0;
  memcpy(new, s, len);
  new[len] = '\0';
  return new;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  mov    %rax,%rbp
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%rbx,%rbp,1)
L8:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t len = strlen(s);
  char *new = malloc(len + 1);
  if (!new)
    return 0;
  memcpy(new, s, len);
  new[len] = '\0';
  return new;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L2:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t len = strlen(s);
  char *new = malloc(len + 1);
  if (!new)
    return 0;
  memcpy(new, s, len);
  new[len] = '\0';
  return new;
}
```
user: What is the c source code of the assembly code below?

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L5
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L5:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  size_t len = strlen(s);
  char *new = malloc(len + 1);
  if (!new)
    return 0;
  memcpy(new, s, len);
  new[len] = '\0';
  return new;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x34(%rbp)
  jne    L16
  movl   $0x1,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x7,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  movl   $0x7,-0x28(%rbp)
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x24(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%ecx
  lea    -0x20(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <search@plt>
  test   %rax,%rax
  jle    L40
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L33
L40:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L33:
  mov    $0x0,%eax
  jmp    L39
L16:
  nop
  nop
L39:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D26
data_type: string
name: disassemble_data
arguments: data_label: D2
data_type: string
name: disassemble_data
arguments: data_label: D1
data_type: string

tool: D26: This is our input key: %d


tool: D2: Key found


tool: D1: Sorry! Key not found, try again..


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2)
    return;
  int A[] = {1, 5, 3, 7, 0, 2};
  int n = sizeof(A) / sizeof(A[0]) + 1;
  int key = (int)atoi(argv[1]);
  printf("This is our input key: %d\n");
  getchar();
  search(A, n, key) > 0 ? printf("Key found\n") : printf("Sorry! Key not found, try again..\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L26
L2:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L45
  add    $0x20,%rsp
  pop    %rbx
  ret
L26:
  movl   $0x1,(%rsp)
  movl   $0x5,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x7,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x2,0x14(%rsp)
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  mov    %rax,%rbx
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %rsp,%rdi
  mov    %ebx,%edx
  mov    $0x7,%esi
  call   <search@plt>
  test   %rax,%rax
  jle    L29
  lea    D21(%rip),%rdi
  call   <printf@plt>
L41:
  mov    $0x0,%eax
  jmp    L2
L29:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L41
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D3
type_infered: string
name: retrieve_data
arguments: label: D21
type_infered: string
name: retrieve_data
arguments: label: D13
type_infered: string

tool: D3: This is our input key: %d


tool: D21: Key found


tool: D13: Sorry! Key not found, try again..


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2)
    return;
  int A[] = {1, 5, 3, 7, 0, 2};
  int n = sizeof(A) / sizeof(A[0]) + 1;
  int key = (int)atoi(argv[1]);
  printf("This is our input key: %d\n");
  getchar();
  search(A, n, key) > 0 ? printf("Key found\n") : printf("Sorry! Key not found, try again..\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L2
L48:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x20,%rsp
  pop    %rbx
  ret
L2:
  movabs $0x500000001,%rax
  mov    0x8(%rsi),%rdi
  mov    %rax,(%rsp)
  movabs $0x700000003,%rax
  mov    %rax,0x8(%rsp)
  mov    $0x1,%eax
  shl    $0x21,%rax
  mov    %rax,0x10(%rsp)
  call   <atoi@plt>
  lea    D29(%rip),%rdi
  mov    %rax,%rbx
  call   <printf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rsp,%rdi
  mov    %ebx,%edx
  mov    $0x7,%esi
  call   <search@plt>
  test   %rax,%rax
  jle    L13
  lea    D22(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L48
L13:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L48
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type: string
name: parse_data
arguments: data_label: D22
type: string
name: parse_data
arguments: data_label: D0
type: string

tool: D29: This is our input key: %d


tool: D22: Key found


tool: D0: Sorry! Key not found, try again..


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2)
    return;
  int A[] = {1, 5, 3, 7, 0, 2};
  int n = sizeof(A) / sizeof(A[0]) + 1;
  int key = (int)atoi(argv[1]);
  printf("This is our input key: %d\n");
  getchar();
  search(A, n, key) > 0 ? printf("Key found\n") : printf("Sorry! Key not found, try again..\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L31
L48:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x20,%rsp
  pop    %rbx
  ret
L31:
  mov    0x8(%rsi),%rdi
  movdqa D9(%rip),%xmm0
  mov    D12(%rip),%rax
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  call   <atoi@plt>
  lea    D35(%rip),%rdi
  mov    %rax,%rbx
  call   <printf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rsp,%rdi
  mov    %ebx,%edx
  mov    $0x7,%esi
  call   <search@plt>
  test   %rax,%rax
  jle    L24
  lea    D28(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L48
L24:
  lea    D11(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  jmp    L48
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
type: byte[16]
name: parse_data
arguments: tag: D12
type: byte[8]
name: parse_data
arguments: tag: D35
type: string
name: parse_data
arguments: tag: D28
type: string
name: parse_data
arguments: tag: D11
type: string

tool: {"D9": [0x01,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D12": [0x00,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: D35: This is our input key: %d


tool: D28: Key found


tool: D11: Sorry! Key not found, try again..


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2)
    return;
  int A[] = {1, 5, 3, 7, 0, 2};
  int n = sizeof(A) / sizeof(A[0]) + 1;
  int key = (int)atoi(argv[1]);
  printf("This is our input key: %d\n");
  getchar();
  search(A, n, key) > 0 ? printf("Key found\n") : printf("Sorry! Key not found, try again..\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<StrConcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <StrLength@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <StrLength@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cmp    $0xfe,%eax
  jle    L0
  mov    $0x0,%eax
  jmp    L20
L0:
  movl   $0x0,-0xc(%rbp)
  jmp    L21
L32:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L21:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L32
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x1,%eax
L20:
  leave
  ret
```

assistant: ```c
int StrConcat(char *s1, char *s2) {
  int len1 = StrLength(s1);
  int len2 = StrLength(s2);
  if (len1 + len2 > 255 - 1)
    return 0;
  int i = 0;
  while (s2[i] != '\0') {
    s1[len1 + i] = s2[i];
    ++i;
  }
  s1[len1 + i] = '\0';
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<StrConcat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  call   <StrLength@plt>
  mov    %eax,%ebp
  mov    %rbx,%rdi
  call   <StrLength@plt>
  add    %ebp,%eax
  mov    $0x0,%edx
  cmp    $0xfe,%eax
  jg     L7
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L12
  mov    $0x1,%eax
  movslq %ebp,%rcx
  add    %r12,%rcx
L10:
  mov    %dl,-0x1(%rcx,%rax,1)
  mov    %eax,%esi
  add    $0x1,%rax
  movzbl -0x1(%rbx,%rax,1),%edx
  test   %dl,%dl
  jne    L10
L3:
  add    %esi,%ebp
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
  mov    $0x1,%edx
L7:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    $0x0,%esi
  jmp    L3
```

assistant: ```c
int StrConcat(char *s1, char *s2) {
  int len1 = StrLength(s1);
  int len2 = StrLength(s2);
  if (len1 + len2 > 255 - 1)
    return 0;
  int i = 0;
  while (s2[i] != '\0') {
    s1[len1 + i] = s2[i];
    ++i;
  }
  s1[len1 + i] = '\0';
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<StrConcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <StrLength@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <StrLength@plt>
  xor    %r8d,%r8d
  add    %ebp,%eax
  cmp    $0xfe,%eax
  jg     L25
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L27
  movslq %ebp,%rax
  mov    $0x1,%edx
  add    %r12,%rax
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %cl,-0x1(%rax,%rdx,1)
  mov    %edx,%esi
  add    $0x1,%rdx
  movzbl -0x1(%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L1
  add    %esi,%ebp
L27:
  movslq %ebp,%rbp
  mov    $0x1,%r8d
  movb   $0x0,(%r12,%rbp,1)
L25:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int StrConcat(char *s1, char *s2) {
  int len1 = StrLength(s1);
  int len2 = StrLength(s2);
  if (len1 + len2 > 255 - 1)
    return 0;
  int i = 0;
  while (s2[i] != '\0') {
    s1[len1 + i] = s2[i];
    ++i;
  }
  s1[len1 + i] = '\0';
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<StrConcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <StrLength@plt>
  mov    %rbx,%rdi
  mov    %eax,%ebp
  call   <StrLength@plt>
  xor    %r8d,%r8d
  add    %ebp,%eax
  cmp    $0xfe,%eax
  jg     L26
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L16
  movslq %ebp,%rax
  mov    $0x1,%edx
  add    %r12,%rax
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %cl,-0x1(%rax,%rdx,1)
  mov    %edx,%esi
  add    $0x1,%rdx
  movzbl -0x1(%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L9
  add    %esi,%ebp
L16:
  movslq %ebp,%rbp
  mov    $0x1,%r8d
  movb   $0x0,(%r12,%rbp,1)
L26:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int StrConcat(char *s1, char *s2) {
  int len1 = StrLength(s1);
  int len2 = StrLength(s2);
  if (len1 + len2 > 255 - 1)
    return 0;
  int i = 0;
  while (s2[i] != '\0') {
    s1[len1 + i] = s2[i];
    ++i;
  }
  s1[len1 + i] = '\0';
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pi_digits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L15:
  cmp    %rdx,%rsp
  je     L116
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L116:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L20
  orq    $0x0,-0x8(%rsp,%rax,1)
L20:
  mov    %rsp,%rbx
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  js     L109
  mov    %rbx,%rax
  lea    0x1(%r12),%edx
  lea    (%rbx,%rdx,4),%rdx
L98:
  movl   $0x7d0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L98
L109:
  test   %r12d,%r12d
  jle    L49
  movslq %r12d,%r8
  lea    -0x1(%r12,%r12,1),%edi
  mov    $0x0,%r9d
  jmp    L27
L6:
  imul   %ecx,%eax
  imul   $0x2710,(%rbx,%rcx,4),%edx
  add    %edx,%eax
  cltd
  idiv   %esi
  mov    %edx,(%rbx,%rcx,4)
  sub    $0x1,%rcx
  sub    $0x2,%esi
  test   %ecx,%ecx
  jg     L6
L16:
  sub    $0xe,%r12d
  sub    $0xe,%r8
  sub    $0x1c,%edi
  test   %r12d,%r12d
  jle    L49
L27:
  mov    %edi,%esi
  mov    %r8,%rcx
  mov    %r9d,%eax
  test   %r12d,%r12d
  jg     L6
  jmp    L16
L49:
  mov    $0x0,%eax
  call   <clock@plt>
  sub    %r13,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  divss  D17(%rip),%xmm0
  mulss  D15(%rip),%xmm0
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D17
type_guessed: byte[4]
name: retrieve_data
arguments: data_label: D15
type_guessed: f32

tool: {"D17": [0x00,0x24,0x74,0x49]}

tool: D15: 1000

assistant: ```c
float pi_digits(int digits) {
  int carry = 0;
  int arr[digits + 1];
  int i, j;
  clock_t t1, t2;
  t1 = clock();
  for (i = 0; i <= digits; ++i)
    arr[i] = 2000;
  for (i = digits; i > 0; i -= 14) {
    int sum = 0;
    for (j = i; j > 0; --j) {
      sum = sum * j + 10000 * arr[j];
      arr[j] = sum % (j * 2 - 1);
      sum /= j * 2 - 1;
    }
    carry = sum % 10000;
  }
  t2 = clock();
  float diff = ((float)(t2 - t1) / 1000000.0F) * 1000;
  return diff;
}
```