ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_wcstombs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  jmp    L2
L15:
  movl   $0xffffffff,-0x20(%rbp)
  lea    -0xc(%rbp),%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <ft_bzero@plt>
  mov    -0x30(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    (%rax),%rax
  mov    %eax,%edx
  lea    -0xc(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <ft_wctomb@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0xffffffff,-0x1c(%rbp)
  jne    L33
  mov    $0xffffffffffffffff,%rax
  jmp    L29
L33:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x38(%rbp)
  jb     L21
  jmp    L92
L41:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  movzbl -0xc(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
L92:
  addl   $0x1,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L41
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
L2:
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L15
  jmp    L58
L21:
  nop
L58:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jae    L5
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
L5:
  mov    -0x18(%rbp),%rax
L29:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L34
  call   <__stack_chk_fail@plt>
L34:
  leave
  ret
```

assistant: ```c
size_t ft_wcstombs(char *s, const wchar_t *wcs, size_t n) {
  size_t converted_char;
  int current_char;
  int i;
  char temp[4];
  converted_char = 0;
  while (*wcs) {
    i = -1;
    ft_bzero(temp, 4);
    if ((current_char = ft_wctomb(temp, *wcs++)) == -1)
      return ((size_t)-1);
    if ((converted_char + current_char) <= n)
      while (++i < current_char)
        s[i] = temp[i];
    else
      break;
    s += current_char;
    converted_char += current_char;
  }
  if (converted_char < n)
    *s = '\0';
  return (converted_char);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_wcstombs>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmpq   $0x0,(%rsi)
  je     L87
  mov    %rsi,%r12
  mov    $0x0,%r14d
  lea    0x4(%rsp),%rbp
  jmp    L68
L0:
  add    %rsi,%rbx
  cmpq   $0x0,(%r12)
  je     L106
  mov    %rdi,%r14
L68:
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <ft_bzero@plt>
  add    $0x8,%r12
  mov    -0x8(%r12),%esi
  mov    %rbp,%rdi
  call   <ft_wctomb@plt>
  cmp    $0xffffffff,%eax
  je     L66
  movslq %eax,%rsi
  lea    (%rsi,%r14,1),%rdi
  cmp    %r13,%rdi
  ja     L94
  test   %eax,%eax
  jle    L0
  mov    $0x0,%edx
L76:
  movzbl (%rdx,%rbp,1),%ecx
  mov    %cl,(%rbx,%rdx,1)
  add    $0x1,%rdx
  cmp    %edx,%eax
  jg     L76
  jmp    L0
L106:
  mov    %rdi,%r14
L94:
  cmp    %r14,%r13
  jbe    L63
  movb   $0x0,(%rbx)
L63:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  mov    %r14,%rax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L87:
  mov    $0x0,%r14d
  jmp    L94
L66:
  mov    $0xffffffffffffffff,%r14
  jmp    L63
L49:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t ft_wcstombs(char *s, const wchar_t *wcs, size_t n) {
  size_t converted_char;
  int current_char;
  int i;
  char temp[4];
  converted_char = 0;
  while (*wcs) {
    i = -1;
    ft_bzero(temp, 4);
    if ((current_char = ft_wctomb(temp, *wcs++)) == -1)
      return ((size_t)-1);
    if ((converted_char + current_char) <= n)
      while (++i < current_char)
        s[i] = temp[i];
    else
      break;
    s += current_char;
    converted_char += current_char;
  }
  if (converted_char < n)
    *s = '\0';
  return (converted_char);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_wcstombs>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmpq   $0x0,(%rsi)
  je     L7
  mov    %rsi,%rbx
  xor    %r13d,%r13d
  lea    0x14(%rsp),%r12
  jmp    L28
  nopl   0x0(%rax)
L25:
  movslq %eax,%r14
  lea    (%r14,%r13,1),%r15
  cmp    0x8(%rsp),%r15
  ja     L29
  test   %eax,%eax
  jle    L72
  mov    %r14,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
L72:
  add    %r14,%rbp
  cmpq   $0x0,(%rbx)
  mov    %r15,%r13
  je     L29
L28:
  mov    $0x4,%esi
  mov    %r12,%rdi
  add    $0x8,%rbx
  call   <ft_bzero@plt>
  mov    -0x8(%rbx),%esi
  mov    %r12,%rdi
  call   <ft_wctomb@plt>
  cmp    $0xffffffff,%eax
  jne    L25
  mov    $0xffffffffffffffff,%r13
L34:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L78
  add    $0x28,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  xor    %r13d,%r13d
  nopl   0x0(%rax)
L29:
  cmp    %r13,0x8(%rsp)
  jbe    L34
  movb   $0x0,0x0(%rbp)
  jmp    L34
L78:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t ft_wcstombs(char *s, const wchar_t *wcs, size_t n) {
  size_t converted_char;
  int current_char;
  int i;
  char temp[4];
  converted_char = 0;
  while (*wcs) {
    i = -1;
    ft_bzero(temp, 4);
    if ((current_char = ft_wctomb(temp, *wcs++)) == -1)
      return ((size_t)-1);
    if ((converted_char + current_char) <= n)
      while (++i < current_char)
        s[i] = temp[i];
    else
      break;
    s += current_char;
    converted_char += current_char;
  }
  if (converted_char < n)
    *s = '\0';
  return (converted_char);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_wcstombs>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdx,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmpq   $0x0,(%rsi)
  je     L9
  mov    %rsi,%rbx
  xor    %r12d,%r12d
  lea    0x14(%rsp),%rbp
  jmp    L101
  nopl   0x0(%rax)
L33:
  mov    %r15,%rdi
  mov    %r14,%rdx
  mov    %rbp,%rsi
  add    %r14,%r15
  call   <memcpy@plt>
  cmpq   $0x0,(%rbx)
  je     L47
L76:
  mov    %r13,%r12
L101:
  mov    $0x4,%esi
  mov    %rbp,%rdi
  add    $0x8,%rbx
  call   <ft_bzero@plt>
  mov    -0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <ft_wctomb@plt>
  cmp    $0xffffffff,%eax
  je     L91
  movslq %eax,%r14
  lea    (%r14,%r12,1),%r13
  cmp    0x8(%rsp),%r13
  ja     L1
  test   %eax,%eax
  jg     L33
  add    %r14,%r15
  cmpq   $0x0,(%rbx)
  jne    L76
L47:
  cmp    %r13,0x8(%rsp)
  ja     L52
  nopl   (%rax)
L93:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x28,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L52:
  movb   $0x0,(%r15)
  jmp    L93
L91:
  mov    $0xffffffffffffffff,%r13
  jmp    L93
L1:
  mov    %r12,%r13
  cmp    %r13,0x8(%rsp)
  jbe    L93
  jmp    L52
L9:
  xor    %r13d,%r13d
  jmp    L47
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t ft_wcstombs(char *s, const wchar_t *wcs, size_t n) {
  size_t converted_char;
  int current_char;
  int i;
  char temp[4];
  converted_char = 0;
  while (*wcs) {
    i = -1;
    ft_bzero(temp, 4);
    if ((current_char = ft_wctomb(temp, *wcs++)) == -1)
      return ((size_t)-1);
    if ((converted_char + current_char) <= n)
      while (++i < current_char)
        s[i] = temp[i];
    else
      break;
    s += current_char;
    converted_char += current_char;
  }
  if (converted_char < n)
    *s = '\0';
  return (converted_char);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: string

tool: D6: The size of the array is: %d


assistant: ```c
int main() {
  int array[10];
  int size = sizeof(array) / sizeof(array[0]);
  printf("The size of the array is: %d\n", size);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type: string

tool: D3: The size of the array is: %d


assistant: ```c
int main() {
  int array[10];
  int size = sizeof(array) / sizeof(array[0]);
  printf("The size of the array is: %d\n", size);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D9
type: string

tool: D9: The size of the array is: %d


assistant: ```c
int main() {
  int array[10];
  int size = sizeof(array) / sizeof(array[0]);
  printf("The size of the array is: %d\n", size);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string

tool: D5: The size of the array is: %d


assistant: ```c
int main() {
  int array[10];
  int size = sizeof(array) / sizeof(array[0]);
  printf("The size of the array is: %d\n", size);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<multiply398>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L17
L2:
  movl   $0x0,-0x3c(%rbp)
  jmp    L23
L21:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L23:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L21
  addl   $0x1,-0x40(%rbp)
L17:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L2
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply398(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<multiply398>:
  endbr64
  test   %edi,%edi
  jle    L6
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L9:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L27:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L27
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  ret
```

assistant: ```c
void multiply398(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<multiply398>:
  endbr64
  test   %edi,%edi
  jle    L23
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L15
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L22
  pop    %rbx
  ret
L23:
  ret
```

assistant: ```c
void multiply398(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<multiply398>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L26
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L19:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L59
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L59
  nopl   0x0(%rax)
L71:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L71
  cmp    -0x24(%rsp),%r11d
  je     L8
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L8
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L8
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L8:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L19
L26:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L59:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L99:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L99
  mov    -0x18(%rsp),%r8
  jmp    L8
```

assistant: ```c
void multiply398(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L0:
  cmpl   $0x8,-0x4(%rbp)
  setle  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <__VERIFIER_assert@plt>
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x9,-0x4(%rbp)
  jle    L0
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    __VERIFIER_assert(i < 9);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%ebx
L2:
  cmp    $0x8,%ebx
  setle  %dil
  movzbl %dil,%edi
  call   <__VERIFIER_assert@plt>
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L2
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    __VERIFIER_assert(i < 9);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %edi,%edi
  cmp    $0x8,%ebx
  setle  %dil
  add    $0x1,%ebx
  call   <__VERIFIER_assert@plt>
  cmp    $0xa,%ebx
  jne    L5
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    __VERIFIER_assert(i < 9);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %edi,%edi
  cmp    $0x8,%ebx
  setle  %dil
  add    $0x1,%ebx
  call   <__VERIFIER_assert@plt>
  cmp    $0xa,%ebx
  jne    L4
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    __VERIFIER_assert(i < 9);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```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    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int *ret;
  ret = (int *)&ret + 2;
  (*ret) = (int)argv[1];
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int *ret;
  ret = (int *)&ret + 2;
  (*ret) = (int)argv[1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int *ret;
  ret = (int *)&ret + 2;
  (*ret) = (int)argv[1];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int *ret;
  ret = (int *)&ret + 2;
  (*ret) = (int)argv[1];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_delta_transfer_function_with_base>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9d,-0x28(%rbp)
  movsd  %xmm0,-0x40(%rbp)
  mov    -0x40(%rbp),%rdi
  mov    -0x28(%rbp),%r9d
  mov    -0x38(%rbp),%r8
  mov    -0x30(%rbp),%rcx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  movq   %rdi,%xmm0
  mov    %rax,%rdi
  call   <get_delta_transfer_function@plt>
  movl   $0x0,-0xc(%rbp)
  mov    -0x38(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L24:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L8:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L24
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L36:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L33:
  mov    -0xc(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L36
  nop
  nop
  leave
  ret
```

assistant: ```c
void get_delta_transfer_function_with_base(double b[], double b_out[], int b_size, double a[], double a_out[], int a_size, double delta) {
  get_delta_transfer_function(b, b_out, b_size, a, a_out, a_size, delta);
  int i = 0;
  double base = a_out[0];
  for (i = 0; i < b_size; i++) {
    b_out[i] = b_out[i] / base;
  }
  for (i = 0; i < a_size; i++) {
    a_out[i] = a_out[i] / base;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<get_delta_transfer_function_with_base>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r13
  mov    %edx,%r12d
  mov    %r8,%rbx
  mov    %r9d,%ebp
  call   <get_delta_transfer_function@plt>
  movsd  (%rbx),%xmm1
  test   %r12d,%r12d
  jle    L21
  mov    %r13,%rax
  lea    -0x1(%r12),%edx
  lea    0x8(%r13,%rdx,8),%rdx
L25:
  movsd  (%rax),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L25
L21:
  test   %ebp,%ebp
  jle    L14
  mov    %rbx,%rax
  lea    -0x1(%rbp),%edx
  lea    0x8(%rbx,%rdx,8),%rdx
L12:
  movsd  (%rax),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L12
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void get_delta_transfer_function_with_base(double b[], double b_out[], int b_size, double a[], double a_out[], int a_size, double delta) {
  get_delta_transfer_function(b, b_out, b_size, a, a_out, a_size, delta);
  int i = 0;
  double base = a_out[0];
  for (i = 0; i < b_size; i++) {
    b_out[i] = b_out[i] / base;
  }
  for (i = 0; i < a_size; i++) {
    a_out[i] = a_out[i] / base;
  }
}
```