ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  movsd  D79(%rip),%xmm1
  movsd  D85(%rip),%xmm0
  call   <__BUILTIN_DAED_DBETWEEN_WITH_ULP@plt>
  movsd  D73(%rip),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movsd  D1(%rip),%xmm0
  call   <__BUILTIN_DAED_DBETWEEN_WITH_ULP@plt>
  movsd  D86(%rip),%xmm1
  movsd  %xmm0,(%rsp)
  movsd  D77(%rip),%xmm0
  call   <__BUILTIN_DAED_DBETWEEN_WITH_ULP@plt>
  mulsd  D88(%rip),%xmm0
  movsd  0x8(%rsp),%xmm2
  addsd  D70(%rip),%xmm0
  addsd  %xmm0,%xmm2
  xorpd  D55(%rip),%xmm0
  mulsd  (%rsp),%xmm0
  movapd %xmm2,%xmm1
  mulsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  call   <DSENSITIVITY@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D79
data_type: f64
name: retrieve_data
arguments: data_tag: D85
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D73
data_type: f64
name: retrieve_data
arguments: data_tag: D1
data_type: f64
name: retrieve_data
arguments: data_tag: D86
data_type: f64
name: retrieve_data
arguments: data_tag: D77
data_type: byte[8]
name: retrieve_data
arguments: data_tag: D88
data_type: f64
name: retrieve_data
arguments: data_tag: D70
data_type: f64
name: retrieve_data
arguments: data_tag: D55
data_type: byte[16]

tool: D79: 125

tool: {"D85": [0x00,0x00,0x00,0x00,0x00,0x40,0x5f,0xc0]}

tool: D73: 25000

tool: D1: 15

tool: D86: 60

tool: {"D77": [0x00,0x00,0x00,0x00,0x00,0x00,0x44,0xc0]}

tool: D88: 0.6

tool: D70: 331.4

tool: {"D55": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  double u, v, T;
  double t1;
  double r;
  u = __BUILTIN_DAED_DBETWEEN_WITH_ULP(-125.0, 125.0);
  v = __BUILTIN_DAED_DBETWEEN_WITH_ULP(15.0, 25000.0);
  T = __BUILTIN_DAED_DBETWEEN_WITH_ULP(-40.0, 60.0);
  t1 = 331.4 + 0.6 * T;
  r = (-t1 * v) / ((t1 + u) * (t1 + u));
  DSENSITIVITY(r);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<malloc_page_aligned>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  shl    $0xc,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  shr    $0xc,%eax
  add    $0x1,%eax
  shl    $0xc,%eax
  mov    %eax,%eax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned int *malloc_page_aligned(unsigned int pages) {
  unsigned int *temp = malloc((pages + 1) * 4096);
  unsigned int *data = (unsigned int *)(((unsigned int)temp / 4096 + 1) * 4096);
  return data;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<malloc_page_aligned>:
  endbr64
  sub    $0x8,%rsp
  add    $0x1,%edi
  shl    $0xc,%edi
  call   <malloc@plt>
  and    $0xfffff000,%eax
  lea    0x1000(%rax),%eax
  mov    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
unsigned int *malloc_page_aligned(unsigned int pages) {
  unsigned int *temp = malloc((pages + 1) * 4096);
  unsigned int *data = (unsigned int *)(((unsigned int)temp / 4096 + 1) * 4096);
  return data;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<malloc_page_aligned>:
  endbr64
  add    $0x1,%edi
  sub    $0x8,%rsp
  shl    $0xc,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  and    $0xfffff000,%eax
  add    $0x1000,%eax
  ret
```

assistant: ```c
unsigned int *malloc_page_aligned(unsigned int pages) {
  unsigned int *temp = malloc((pages + 1) * 4096);
  unsigned int *data = (unsigned int *)(((unsigned int)temp / 4096 + 1) * 4096);
  return data;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<malloc_page_aligned>:
  endbr64
  add    $0x1,%edi
  sub    $0x8,%rsp
  shl    $0xc,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  and    $0xfffff000,%eax
  add    $0x1000,%eax
  ret
```

assistant: ```c
unsigned int *malloc_page_aligned(unsigned int pages) {
  unsigned int *temp = malloc((pages + 1) * 4096);
  unsigned int *data = (unsigned int *)(((unsigned int)temp / 4096 + 1) * 4096);
  return data;
}
```
user: Translate the below assembly code into C programming language:

```asm
<HORSPOOL>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x450,%rsp
  mov    %rdi,-0x438(%rbp)
  mov    %rsi,-0x440(%rbp)
  mov    %rdx,-0x448(%rbp)
  mov    %ecx,-0x44c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x420(%rbp)
  jmp    L13
L52:
  mov    -0x420(%rbp),%eax
  cltq
  mov    -0x44c(%rbp),%edx
  mov    %edx,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x420(%rbp)
L13:
  cmpl   $0xff,-0x420(%rbp)
  jle    L52
  movl   $0x0,-0x41c(%rbp)
  jmp    L75
L12:
  mov    -0x44c(%rbp),%eax
  sub    -0x41c(%rbp),%eax
  mov    %eax,%edx
  mov    -0x41c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x440(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  sub    $0x1,%edx
  cltq
  mov    %edx,-0x410(%rbp,%rax,4)
  addl   $0x1,-0x41c(%rbp)
L75:
  mov    -0x44c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x41c(%rbp)
  jl     L12
  mov    -0x44c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x440(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x422(%rbp)
  movq   $0x0,-0x418(%rbp)
  jmp    L61
L91:
  mov    -0x44c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x418(%rbp),%rax
  add    %rdx,%rax
  lea    -0x1(%rax),%rdx
  mov    -0x438(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x421(%rbp)
  movzbl -0x421(%rbp),%eax
  cmp    -0x422(%rbp),%al
  jne    L81
  mov    -0x44c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x418(%rbp),%rcx
  mov    -0x438(%rbp),%rax
  add    %rax,%rcx
  mov    -0x440(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L81
  mov    -0x418(%rbp),%rax
  jmp    L38
L81:
  movzbl -0x421(%rbp),%eax
  cltq
  mov    -0x410(%rbp,%rax,4),%eax
  cltq
  add    %rax,-0x418(%rbp)
L61:
  mov    -0x44c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x448(%rbp),%rax
  sub    %rdx,%rax
  cmp    %rax,-0x418(%rbp)
  jle    L91
  mov    $0xffffffffffffffff,%rax
L38:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

assistant: ```c
long HORSPOOL(char *y, char *x, long n, int m) {
  long i;
  int a, j, bm_bc[256];
  unsigned char ch, lastch;
  for (a = 0; a < 256; a++)
    bm_bc[a] = m;
  for (j = 0; j < m - 1; j++)
    bm_bc[(unsigned char)x[j]] = m - j - 1;
  lastch = x[m - 1];
  i = 0;
  while (i <= n - m) {
    ch = y[i + m - 1];
    if (ch == lastch)
      if (memcmp(&y[i], x, m - 1) == 0)
        return (i);
    i += bm_bc[ch];
  }
  return (-1);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HORSPOOL>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,%r15
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  lea    0x410(%rsp),%rdx
L50:
  mov    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L50
  lea    -0x1(%rcx),%eax
  mov    %eax,0xc(%rsp)
  test   %eax,%eax
  jle    L62
  lea    -0x2(%rcx),%eax
  lea    0x1(%r15,%rax,1),%r8
  mov    %r15,%rax
  lea    -0x1(%rcx,%r15,1),%edi
L11:
  movzbl (%rax),%edx
  mov    %edi,%esi
  sub    %eax,%esi
  mov    %esi,0x10(%rsp,%rdx,4)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L11
L62:
  movslq %ecx,%rcx
  movzbl -0x1(%r15,%rcx,1),%r14d
  sub    %rcx,%r13
  js     L67
  mov    $0x0,%r12d
  mov    (%rsp),%rax
  lea    (%rax,%rcx,1),%rbx
  jmp    L73
L46:
  movzbl %bpl,%ebp
  movslq 0x10(%rsp,%rbp,4),%rax
  add    %rax,%r12
  cmp    %r13,%r12
  jg     L7
L73:
  movzbl -0x1(%r12,%rbx,1),%ebp
  cmp    %bpl,%r14b
  jne    L46
  mov    (%rsp),%rax
  lea    (%rax,%r12,1),%rdi
  mov    0xc(%rsp),%edx
  mov    %r15,%rsi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L46
  jmp    L103
L7:
  mov    $0xffffffffffffffff,%r12
L103:
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    %r12,%rax
  add    $0x428,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L67:
  mov    $0xffffffffffffffff,%r12
  jmp    L103
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long HORSPOOL(char *y, char *x, long n, int m) {
  long i;
  int a, j, bm_bc[256];
  unsigned char ch, lastch;
  for (a = 0; a < 256; a++)
    bm_bc[a] = m;
  for (j = 0; j < m - 1; j++)
    bm_bc[(unsigned char)x[j]] = m - j - 1;
  lastch = x[m - 1];
  i = 0;
  while (i <= n - m) {
    ch = y[i + m - 1];
    if (ch == lastch)
      if (memcmp(&y[i], x, m - 1) == 0)
        return (i);
    i += bm_bc[ch];
  }
  return (-1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<HORSPOOL>:
  endbr64
  push   %r15
  mov    %rdx,%r8
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x410(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
L7:
  mov    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L7
  lea    -0x1(%rcx),%r14d
  test   %r14d,%r14d
  jle    L93
  lea    -0x2(%rcx),%eax
  lea    -0x1(%rcx,%r13,1),%edi
  lea    0x1(%r13,%rax,1),%r9
  mov    %r13,%rax
  nopl   0x0(%rax)
L84:
  movzbl (%rax),%edx
  mov    %edi,%esi
  sub    %eax,%esi
  add    $0x1,%rax
  mov    %esi,0x10(%rsp,%rdx,4)
  cmp    %r9,%rax
  jne    L84
L93:
  movslq %ecx,%rcx
  movzbl -0x1(%r13,%rcx,1),%ebx
  sub    %rcx,%r8
  js     L27
  mov    0x8(%rsp),%rax
  xor    %r12d,%r12d
  lea    (%rax,%rcx,1),%r15
  jmp    L96
  nopl   0x0(%rax,%rax,1)
L87:
  movslq 0x10(%rsp,%rbp,4),%rax
  add    %rax,%r12
  cmp    %r8,%r12
  jg     L27
L96:
  movzbl -0x1(%r12,%r15,1),%ebp
  cmp    %bpl,%bl
  jne    L87
  mov    0x8(%rsp),%rax
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %r8,(%rsp)
  lea    (%rax,%r12,1),%rdi
  call   <memcmp@plt>
  mov    (%rsp),%r8
  test   %rax,%rax
  jne    L87
L25:
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x428,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  mov    $0xffffffffffffffff,%r12
  jmp    L25
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long HORSPOOL(char *y, char *x, long n, int m) {
  long i;
  int a, j, bm_bc[256];
  unsigned char ch, lastch;
  for (a = 0; a < 256; a++)
    bm_bc[a] = m;
  for (j = 0; j < m - 1; j++)
    bm_bc[(unsigned char)x[j]] = m - j - 1;
  lastch = x[m - 1];
  i = 0;
  while (i <= n - m) {
    ch = y[i + m - 1];
    if (ch == lastch)
      if (memcmp(&y[i], x, m - 1) == 0)
        return (i);
    i += bm_bc[ch];
  }
  return (-1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<HORSPOOL>:
  endbr64
  push   %r15
  movd   %ecx,%xmm1
  mov    %rdx,%r15
  push   %r14
  pshufd $0x0,%xmm1,%xmm0
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x410(%rsp),%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x418(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rax
  nopl   0x0(%rax)
L86:
  movaps %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L86
  lea    -0x1(%rcx),%r8d
  test   %r8d,%r8d
  jle    L54
  lea    -0x2(%rcx),%eax
  lea    -0x1(%rcx,%r13,1),%edi
  lea    0x1(%r13,%rax,1),%r9
  mov    %r13,%rax
  nopl   (%rax)
L50:
  movzbl (%rax),%edx
  mov    %edi,%esi
  sub    %eax,%esi
  add    $0x1,%rax
  mov    %esi,0x10(%rsp,%rdx,4)
  cmp    %rax,%r9
  jne    L50
L54:
  movslq %ecx,%rax
  movzbl -0x1(%r13,%rax,1),%ebp
  sub    %rax,%r15
  js     L16
  mov    0x8(%rsp),%rcx
  xor    %r12d,%r12d
  lea    (%rcx,%rax,1),%r14
  jmp    L53
  nopl   0x0(%rax,%rax,1)
L45:
  movslq 0x10(%rsp,%rbx,4),%rax
  add    %rax,%r12
  cmp    %r15,%r12
  jg     L16
L53:
  movzbl -0x1(%r12,%r14,1),%ebx
  cmp    %bl,%bpl
  jne    L45
  mov    0x8(%rsp),%rax
  mov    %r8d,%edx
  mov    %r13,%rsi
  mov    %r8d,0x4(%rsp)
  lea    (%rax,%r12,1),%rdi
  call   <memcmp@plt>
  mov    0x4(%rsp),%r8d
  test   %rax,%rax
  jne    L45
L49:
  mov    0x418(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  add    $0x428,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  mov    $0xffffffffffffffff,%r12
  jmp    L49
L70:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long HORSPOOL(char *y, char *x, long n, int m) {
  long i;
  int a, j, bm_bc[256];
  unsigned char ch, lastch;
  for (a = 0; a < 256; a++)
    bm_bc[a] = m;
  for (j = 0; j < m - 1; j++)
    bm_bc[(unsigned char)x[j]] = m - j - 1;
  lastch = x[m - 1];
  i = 0;
  while (i <= n - m) {
    ch = y[i + m - 1];
    if (ch == lastch)
      if (memcmp(&y[i], x, m - 1) == 0)
        return (i);
    i += bm_bc[ch];
  }
  return (-1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xb0(%rbp)
  jmp    L93
L87:
  lea    -0x70(%rbp),%rcx
  mov    -0xb0(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  lea    (%rcx,%rax,1),%rdx
  mov    -0xb0(%rbp),%eax
  cltq
  mov    %rdx,-0xa0(%rbp,%rax,8)
  addl   $0x1,-0xb0(%rbp)
L93:
  cmpl   $0x4,-0xb0(%rbp)
  jle    L87
  lea    -0xa0(%rbp),%rax
  mov    %rax,-0xa8(%rbp)
  movl   $0x0,-0xb0(%rbp)
  jmp    L88
L75:
  movl   $0x0,-0xac(%rbp)
  jmp    L109
L66:
  mov    -0xb0(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0xa8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0xac(%rbp)
L109:
  cmpl   $0x4,-0xac(%rbp)
  jle    L66
  addl   $0x1,-0xb0(%rbp)
L88:
  cmpl   $0x4,-0xb0(%rbp)
  jle    L75
  movl   $0x0,-0xb0(%rbp)
  jmp    L24
L27:
  movl   $0x0,-0xac(%rbp)
  jmp    L105
L69:
  mov    -0xb0(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0xa8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xac(%rbp)
L105:
  cmpl   $0x4,-0xac(%rbp)
  jle    L69
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0xb0(%rbp)
L24:
  cmpl   $0x4,-0xb0(%rbp)
  jle    L27
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
data_type: string
name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D19: %d

tool: D6: %7d

assistant: ```c
main() {
  int i, j;
  int a[5][5];
  int *p[5];
  int **q;
  for (i = 0; i < 5; i++)
    p[i] = a[i];
  q = p;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      scanf("%d", *(q + i) + j);
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++)
      printf("%7d", *(*(q + i) + j));
    printf("\n");
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rax,(%rsp)
  lea    0x44(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    0x58(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x6c(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x80(%rsp),%rax
  mov    %rax,0x20(%rsp)
  mov    %rsp,%rbp
  lea    0x28(%rsp),%r12
  mov    %rbp,%r13
  lea    D5(%rip),%r14
L8:
  mov    $0x0,%ebx
L14:
  mov    %rbx,%rsi
  add    0x0(%r13),%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    $0x14,%rbx
  jne    L14
  add    $0x8,%r13
  cmp    %r13,%r12
  jne    L8
  lea    D10(%rip),%r13
L30:
  mov    $0x0,%ebx
L18:
  mov    0x0(%rbp),%rax
  mov    (%rax,%rbx,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    $0x14,%rbx
  jne    L18
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rbp
  cmp    %rbp,%r12
  jne    L30
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D5
data_type: string
name: parse_data
arguments: data_tag: D10
data_type: string

tool: D5: %d

tool: D10: %7d

assistant: ```c
main() {
  int i, j;
  int a[5][5];
  int *p[5];
  int **q;
  for (i = 0; i < 5; i++)
    p[i] = a[i];
  q = p;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      scanf("%d", *(q + i) + j);
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++)
      printf("%7d", *(*(q + i) + j));
    printf("\n");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xa8(%rsp)
  xor    %eax,%eax
  lea    0x54(%rsp),%rax
  lea    0x40(%rsp),%rdx
  mov    %rax,0x18(%rsp)
  lea    0x68(%rsp),%rax
  lea    0x10(%rsp),%r12
  mov    %rax,0x20(%rsp)
  lea    0x7c(%rsp),%rax
  lea    0x38(%rsp),%r13
  mov    %r12,%r15
  mov    %rax,0x28(%rsp)
  lea    0x90(%rsp),%rax
  mov    %rdx,0x8(%rsp)
  mov    %rdx,0x10(%rsp)
  mov    %rax,0x30(%rsp)
L42:
  mov    %rdx,%rbx
  lea    0x14(%rdx),%rbp
L12:
  mov    %rbx,%rsi
  mov    %r14,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L12
  add    $0x8,%r15
  cmp    %r13,%r15
  je     L24
  mov    (%r15),%rdx
  jmp    L42
  nop
L24:
  mov    0x8(%rsp),%rbx
  lea    D1(%rip),%r14
L9:
  lea    0x14(%rbx),%rbp
L40:
  mov    (%rbx),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L40
  mov    $0xa,%edi
  add    $0x8,%r12
  call   <putchar@plt>
  cmp    %r13,%r12
  je     L27
  mov    (%r12),%rbx
  jmp    L9
  xchg   %ax,%ax
L27:
  mov    0xa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0xb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_infered: string
name: retrieve_data
arguments: data_tag: D1
type_infered: string

tool: D4: %d

tool: D1: %7d

assistant: ```c
main() {
  int i, j;
  int a[5][5];
  int *p[5];
  int **q;
  for (i = 0; i < 5; i++)
    p[i] = a[i];
  q = p;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      scanf("%d", *(q + i) + j);
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++)
      printf("%7d", *(*(q + i) + j));
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D0(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rsp,%rbp
  lea    0x28(%rsp),%r12
  movq   %rax,%xmm0
  lea    0x44(%rsp),%rax
  mov    %rbp,%r13
  movq   %rax,%xmm1
  lea    0x58(%rsp),%rax
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  movq   %rax,%xmm0
  lea    0x6c(%rsp),%rax
  movq   %rax,%xmm2
  lea    0x80(%rsp),%rax
  mov    %rax,0x20(%rsp)
  punpcklqdq %xmm2,%xmm0
  movaps %xmm0,0x10(%rsp)
L1:
  xor    %ebx,%ebx
L49:
  mov    0x0(%r13),%rsi
  mov    %r14,%rdi
  add    %rbx,%rsi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    $0x14,%rbx
  jne    L49
  add    $0x8,%r13
  cmp    %r13,%r12
  jne    L1
  lea    D14(%rip),%r13
L29:
  xor    %ebx,%ebx
L4:
  mov    0x0(%rbp),%rax
  mov    %r13,%rdi
  mov    (%rax,%rbx,1),%esi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    $0x14,%rbx
  jne    L4
  mov    $0xa,%edi
  add    $0x8,%rbp
  call   <putchar@plt>
  cmp    %r12,%rbp
  jne    L29
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: string
name: retrieve_data
arguments: data_label: D14
data_type: string

tool: D0: %d

tool: D14: %7d

assistant: ```c
main() {
  int i, j;
  int a[5][5];
  int *p[5];
  int **q;
  for (i = 0; i < 5; i++)
    p[i] = a[i];
  q = p;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      scanf("%d", *(q + i) + j);
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++)
      printf("%7d", *(*(q + i) + j));
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<eprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L20
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L20:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <vfprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xc8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L16
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0xc8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3a,%al
  jne    L16
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  mov    $0x20,%edi
  call   <fputc@plt>
  mov    $0x0,%edi
  call   <perror@plt>
L16:
  mov    D21(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void eprintf(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vfprintf(stderr, fmt, ap);
  va_end(ap);
  if (fmt[0] != '\0' && fmt[strlen(fmt) - 1] == ':') {
    fputc(' ', stderr);
    perror(NULL);
  }
  exit(EXIT_FAILURE);
}
```
user: What is the c source code of the assembly code below:

```asm
<eprintf>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    D18(%rip),%rax
  mov    (%rax),%edi
  call   <vfprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  cmpb   $0x0,(%rbx)
  jne    L3
L9:
  mov    D28(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  cmpb   $0x3a,-0x1(%rbx,%rax,1)
  jne    L9
  mov    D18(%rip),%rax
  mov    (%rax),%esi
  mov    $0x20,%edi
  call   <fputc@plt>
  mov    $0x0,%edi
  call   <perror@plt>
  jmp    L9
```

assistant: ```c
void eprintf(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vfprintf(stderr, fmt, ap);
  va_end(ap);
  if (fmt[0] != '\0' && fmt[strlen(fmt) - 1] == ':') {
    fputc(' ', stderr);
    perror(NULL);
  }
  exit(EXIT_FAILURE);
}
```