ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x7e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x7d8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x3f0(%rsp),%r13
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L43:
  cmp    %ebx,%ebp
  jl     L12
L34:
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <m_getline@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L43
  test   %ebp,%ebp
  jne    L55
L2:
  mov    0x7d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x7e8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %ebx,%ebp
  call   <m_copy@plt>
  jmp    L34
L55:
  lea    0x3f0(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  jmp    L2
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: %s

assistant: ```c
main() {
  int len;
  int max;
  char line[1000];
  char longest[1000];
  max = 0;
  while ((len = m_getline(line, 1000)) > 0)
    if (len > max) {
      max = len;
      m_copy(longest, line);
    }
  if (max > 0)
    printf("%s", longest);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x180,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x174(%rbp)
  jmp    L18
L1:
  lea    -0x170(%rbp),%rax
  mov    $0x24,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x140(%rbp),%rdx
  lea    -0x170(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <reconstructInitBoard@plt>
  mov    -0x174(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x170(%rbp),%rax
  mov    %rax,%rdi
  call   <printBoard@plt>
  addl   $0x1,-0x174(%rbp)
L18:
  lea    -0x140(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

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

tool: D2: Case #%d:


assistant: ```c
int main(void) {
  int gameBoard[3][3], caseId;
  char line[300 + 2];
  caseId = 1;
  while (gets(line) != NULL) {
    memset(gameBoard, 0, 3 * 3 * sizeof(int));
    reconstructInitBoard(gameBoard, line);
    printf("Case #%d:\n", caseId);
    printBoard(gameBoard);
    caseId++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x170,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  mov    $0x1,%ebp
  lea    D3(%rip),%r12
  jmp    L27
L23:
  mov    %rsp,%rbx
  mov    $0x24,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  lea    0x30(%rsp),%rsi
  mov    %rbx,%rdi
  call   <reconstructInitBoard@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <printBoard@plt>
  add    $0x1,%ebp
L27:
  lea    0x30(%rsp),%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L23
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x170,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D3
type_guessed: string

tool: D3: Case #%d:


assistant: ```c
int main(void) {
  int gameBoard[3][3], caseId;
  char line[300 + 2];
  caseId = 1;
  while (gets(line) != NULL) {
    memset(gameBoard, 0, 3 * 3 * sizeof(int));
    reconstructInitBoard(gameBoard, line);
    printf("Case #%d:\n", caseId);
    printBoard(gameBoard);
    caseId++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x178,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r12
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %rsp,%rbp
  mov    $0x24,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <reconstructInitBoard@plt>
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printBoard@plt>
L16:
  mov    %r12,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L28
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x178,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D9
data_type: string

tool: D9: Case #%d:


assistant: ```c
int main(void) {
  int gameBoard[3][3], caseId;
  char line[300 + 2];
  caseId = 1;
  while (gets(line) != NULL) {
    memset(gameBoard, 0, 3 * 3 * sizeof(int));
    reconstructInitBoard(gameBoard, line);
    printf("Case #%d:\n", caseId);
    printBoard(gameBoard);
    caseId++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x178,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r12
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %rsp,%rbp
  mov    $0x24,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <reconstructInitBoard@plt>
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <printBoard@plt>
L11:
  mov    %r12,%rdi
  call   <gets@plt>
  test   %rax,%rax
  jne    L29
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x178,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D0
datatype: string

tool: D0: Case #%d:


assistant: ```c
int main(void) {
  int gameBoard[3][3], caseId;
  char line[300 + 2];
  caseId = 1;
  while (gets(line) != NULL) {
    memset(gameBoard, 0, 3 * 3 * sizeof(int));
    reconstructInitBoard(gameBoard, line);
    printf("Case #%d:\n", caseId);
    printBoard(gameBoard);
    caseId++;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xd,-0xc(%rbp)
  movl   $0x42,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <multiply@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: string

tool: D5: 
 Our result is [%u]


assistant: ```c
int main(void) {
  int x = 13;
  int y = 66;
  int result = 0;
  result = multiply(x, y);
  printf("\n Our result is [%u]\n", result);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x42,%esi
  mov    $0xd,%edi
  call   <multiply@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: string

tool: D0: 
 Our result is [%u]


assistant: ```c
int main(void) {
  int x = 13;
  int y = 66;
  int result = 0;
  result = multiply(x, y);
  printf("\n Our result is [%u]\n", result);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x42,%esi
  mov    $0xd,%edi
  call   <multiply@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type: string

tool: D8: 
 Our result is [%u]


assistant: ```c
int main(void) {
  int x = 13;
  int y = 66;
  int result = 0;
  result = multiply(x, y);
  printf("\n Our result is [%u]\n", result);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x42,%esi
  mov    $0xd,%edi
  call   <multiply@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
data_type: string

tool: D6: 
 Our result is [%u]


assistant: ```c
int main(void) {
  int x = 13;
  int y = 66;
  int result = 0;
  result = multiply(x, y);
  printf("\n Our result is [%u]\n", result);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sig_permanent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movq   $0x0,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sigaction@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int sig_permanent(int sig, void (*handler)(int)) {
  struct sigaction sa;
  memset(&sa, 0, sizeof sa);
  sa.sa_flags = 0;
  sa.sa_handler = handler;
  return sigaction(sig, &sa, NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sig_permanent>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  movq   $0x0,0x8(%rsp)
  mov    %rbp,(%rsp)
  mov    $0x0,%edx
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <sigaction@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sig_permanent(int sig, void (*handler)(int)) {
  struct sigaction sa;
  memset(&sa, 0, sizeof sa);
  sa.sa_flags = 0;
  sa.sa_handler = handler;
  return sigaction(sig, &sa, NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sig_permanent>:
  endbr64
  push   %r12
  mov    $0x10,%edx
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %edx,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  movq   $0x0,0x8(%rsp)
  mov    %rbx,(%rsp)
  call   <sigaction@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sig_permanent(int sig, void (*handler)(int)) {
  struct sigaction sa;
  memset(&sa, 0, sizeof sa);
  sa.sa_flags = 0;
  sa.sa_handler = handler;
  return sigaction(sig, &sa, NULL);
}
```
user: What is the c source code of the assembly code below:

```asm
<sig_permanent>:
  endbr64
  push   %r12
  mov    $0x10,%edx
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  xor    %esi,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %edx,%edx
  mov    %r12,%rsi
  mov    %ebp,%edi
  movq   $0x0,0x8(%rsp)
  mov    %rbx,(%rsp)
  call   <sigaction@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int sig_permanent(int sig, void (*handler)(int)) {
  struct sigaction sa;
  memset(&sa, 0, sizeof sa);
  sa.sa_flags = 0;
  sa.sa_handler = handler;
  return sigaction(sig, &sa, NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<idx_add2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x38(%rsp),%r13d
  mov    0x40(%rsp),%r15d
  test   %r9d,%r9d
  jle    L45
  mov    %rsi,%r11
  mov    %rdx,%rbx
  mov    %rcx,%rsi
  mov    %r9d,%r12d
  movslq %r13d,%r14
  shl    $0x2,%r14
  lea    -0x1(%r13),%ebp
  lea    0x4(,%rbp,4),%r10
  mov    $0x0,%ecx
  mov    $0x0,%r9d
  not    %rbp
  shl    $0x2,%rbp
  jmp    L41
L42:
  mov    %ecx,%edx
  add    (%r11,%rax,1),%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  movsd  (%r8,%rax,2),%xmm1
  mulsd  (%rsi,%rax,2),%xmm1
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  mov    %ecx,%edx
  add    (%rbx,%rax,1),%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,2),%xmm1
  mulsd  (%r8,%rax,2),%xmm1
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  add    $0x4,%rax
  cmp    %r10,%rax
  jne    L42
L11:
  add    %r15d,%ecx
  add    $0x1,%r9d
  add    %r14,%r10
  cmp    %r9d,%r12d
  je     L45
L41:
  lea    0x0(%rbp,%r10,1),%rax
  test   %r13d,%r13d
  jg     L42
  jmp    L11
L45:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void idx_add2(double output[], int idx1[], int idx2[], double data[], double mult[], double ds, int ix, int jx1, int jx2) {
  int i, j, lim1 = 0, lim2 = 0;
  for (i = 0; i < ix; i++) {
    for (j = 0; j < jx1; j++) {
      output[lim2 + idx1[lim1 + j]] = output[lim2 + idx1[lim1 + j]] + mult[lim1 + j] * data[lim1 + j];
      output[lim2 + idx2[lim1 + j]] = output[lim2 + idx2[lim1 + j]] + mult[lim1 + j] * (ds - data[lim1 + j]);
    }
    lim1 = lim1 + jx1;
    lim2 = lim2 + jx2;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<idx_add2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x38(%rsp),%r13d
  mov    0x40(%rsp),%r15d
  test   %r9d,%r9d
  jle    L34
  lea    -0x1(%r13),%ebp
  movslq %r13d,%r14
  mov    %rsi,%r11
  mov    %r9d,%r12d
  add    $0x1,%rbp
  mov    %rcx,%rsi
  mov    %rdx,%rbx
  shl    $0x2,%r14
  lea    0x0(,%rbp,4),%r10
  neg    %rbp
  xor    %ecx,%ecx
  xor    %r9d,%r9d
  shl    $0x2,%rbp
L18:
  lea    0x0(%rbp,%r10,1),%rax
  test   %r13d,%r13d
  jle    L11
  nopw   0x0(%rax,%rax,1)
L14:
  mov    (%r11,%rax,1),%edx
  movsd  (%r8,%rax,2),%xmm1
  mulsd  (%rsi,%rax,2),%xmm1
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  mov    (%rbx,%rax,1),%edx
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,2),%xmm1
  mulsd  (%r8,%rax,2),%xmm1
  add    $0x4,%rax
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %r10,%rax
  jne    L14
L11:
  add    $0x1,%r9d
  add    %r15d,%ecx
  add    %r14,%r10
  cmp    %r9d,%r12d
  jne    L18
L34:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void idx_add2(double output[], int idx1[], int idx2[], double data[], double mult[], double ds, int ix, int jx1, int jx2) {
  int i, j, lim1 = 0, lim2 = 0;
  for (i = 0; i < ix; i++) {
    for (j = 0; j < jx1; j++) {
      output[lim2 + idx1[lim1 + j]] = output[lim2 + idx1[lim1 + j]] + mult[lim1 + j] * data[lim1 + j];
      output[lim2 + idx2[lim1 + j]] = output[lim2 + idx2[lim1 + j]] + mult[lim1 + j] * (ds - data[lim1 + j]);
    }
    lim1 = lim1 + jx1;
    lim2 = lim2 + jx2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<idx_add2>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x30(%rsp),%eax
  mov    0x38(%rsp),%r14d
  test   %r9d,%r9d
  jle    L2
  test   %eax,%eax
  jle    L2
  mov    %eax,%ebp
  movslq %eax,%r13
  mov    %rsi,%r11
  mov    %r9d,%r12d
  mov    %rcx,%rsi
  lea    0x0(,%rbp,4),%r10
  neg    %rbp
  xor    %ecx,%ecx
  shl    $0x2,%r13
  mov    %rdx,%rbx
  xor    %r9d,%r9d
  shl    $0x2,%rbp
  nop
L28:
  lea    0x0(%rbp,%r10,1),%rax
  nopl   (%rax)
L20:
  mov    (%r11,%rax,1),%edx
  movsd  (%r8,%rax,2),%xmm1
  mulsd  (%rsi,%rax,2),%xmm1
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  mov    (%rbx,%rax,1),%edx
  movapd %xmm0,%xmm1
  subsd  (%rsi,%rax,2),%xmm1
  mulsd  (%r8,%rax,2),%xmm1
  add    $0x4,%rax
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,8),%rdx
  addsd  (%rdx),%xmm1
  movsd  %xmm1,(%rdx)
  cmp    %r10,%rax
  jne    L20
  add    $0x1,%r9d
  add    %r14d,%ecx
  add    %r13,%r10
  cmp    %r9d,%r12d
  jne    L28
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void idx_add2(double output[], int idx1[], int idx2[], double data[], double mult[], double ds, int ix, int jx1, int jx2) {
  int i, j, lim1 = 0, lim2 = 0;
  for (i = 0; i < ix; i++) {
    for (j = 0; j < jx1; j++) {
      output[lim2 + idx1[lim1 + j]] = output[lim2 + idx1[lim1 + j]] + mult[lim1 + j] * data[lim1 + j];
      output[lim2 + idx2[lim1 + j]] = output[lim2 + idx2[lim1 + j]] + mult[lim1 + j] * (ds - data[lim1 + j]);
    }
    lim1 = lim1 + jx1;
    lim2 = lim2 + jx2;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```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>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L12
  mov    $0x0,%eax
  jmp    L4
L12:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L4:
  leave
  ret
```

assistant: ```c
char *strdup(const char *s) {
  char *ret;
  ret = malloc(strlen(s) + 1);
  if (ret == NULL) {
    return NULL;
  }
  strcpy(ret, s);
  return ret;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L8:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strdup(const char *s) {
  char *ret;
  ret = malloc(strlen(s) + 1);
  if (ret == NULL) {
    return NULL;
  }
  strcpy(ret, s);
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  char *ret;
  ret = malloc(strlen(s) + 1);
  if (ret == NULL) {
    return NULL;
  }
  strcpy(ret, s);
  return ret;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(const char *s) {
  char *ret;
  ret = malloc(strlen(s) + 1);
  if (ret == NULL) {
    return NULL;
  }
  strcpy(ret, s);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kr1_hash_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L2
L5:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L5
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int kr1_hash_str(const char *str, size_t str_len) {
  unsigned int hash;
  size_t i;
  hash = 0;
  for (i = 0; i < str_len; ++i) {
    hash += (unsigned int)str[i];
  }
  return hash;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<kr1_hash_str>:
  endbr64
  test   %rsi,%rsi
  je     L19
  mov    %rdi,%rax
  add    %rsi,%rdi
  mov    $0x0,%edx
L17:
  movsbl (%rax),%ecx
  add    %ecx,%edx
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L17
L11:
  mov    %edx,%eax
  ret
L19:
  mov    $0x0,%edx
  jmp    L11
```

assistant: ```c
unsigned int kr1_hash_str(const char *str, size_t str_len) {
  unsigned int hash;
  size_t i;
  hash = 0;
  for (i = 0; i < str_len; ++i) {
    hash += (unsigned int)str[i];
  }
  return hash;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kr1_hash_str>:
  endbr64
  test   %rsi,%rsi
  je     L4
  add    %rdi,%rsi
  xor    %eax,%eax
  xchg   %ax,%ax
L5:
  movsbl (%rdi),%edx
  add    $0x1,%rdi
  add    %edx,%eax
  cmp    %rsi,%rdi
  jne    L5
  ret
  nop
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int kr1_hash_str(const char *str, size_t str_len) {
  unsigned int hash;
  size_t i;
  hash = 0;
  for (i = 0; i < str_len; ++i) {
    hash += (unsigned int)str[i];
  }
  return hash;
}
```