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

```asm
<main>:
  endbr64
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  lea    D5(%rip),%r12
  mov    %r13,%rdi
  push   %rbp
  mov    %r12,%rsi
  lea    D33(%rip),%rbp
  call   <strlonger@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strlonger@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
data_type: string
name: disassemble_data
arguments: data_label: D5
data_type: string
name: disassemble_data
arguments: data_label: D33
data_type: string
name: disassemble_data
arguments: data_label: D12
data_type: string

tool: D7: 1234

tool: D5: 123

tool: D33: %d


tool: D12: =====================

assistant: ```c
int main() {
  char *s1 = "1234";
  char *t1 = "123";
  printf("%d\n", strlonger(s1, t1));
  char *s2 = "123";
  char *t2 = "1234";
  printf("%d\n", strlonger(s2, t2));
  printf("=====================\n");
  printf("%d\n", (1u - 2u) > 0);
  printf("%d\n", (int)(1u - 2u) > 0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  lea    D27(%rip),%r12
  mov    %r13,%rdi
  push   %rbp
  mov    %r12,%rsi
  lea    D32(%rip),%rbp
  call   <strlonger@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strlonger@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
data_type: string
name: retrieve_data
arguments: data_tag: D27
data_type: string
name: retrieve_data
arguments: data_tag: D32
data_type: string
name: retrieve_data
arguments: data_tag: D20
data_type: string

tool: D16: 1234

tool: D27: 123

tool: D32: %d


tool: D20: =====================

assistant: ```c
int main() {
  char *s1 = "1234";
  char *t1 = "123";
  printf("%d\n", strlonger(s1, t1));
  char *s2 = "123";
  char *t2 = "1234";
  printf("%d\n", strlonger(s2, t2));
  printf("=====================\n");
  printf("%d\n", (1u - 2u) > 0);
  printf("%d\n", (int)(1u - 2u) > 0);
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L8
L7:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <print_statvfs@plt>
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < argc; i++) {
    print_statvfs(argv[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L10
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x8(%rsi),%rbx
  lea    -0x2(%rdi),%eax
  lea    0x10(%rsi,%rax,8),%rbp
L4:
  mov    (%rbx),%rdi
  call   <print_statvfs@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L4
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < argc; i++) {
    print_statvfs(argv[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L16
  push   %rbp
  lea    -0x2(%rdi),%eax
  push   %rbx
  lea    0x10(%rsi,%rax,8),%rbp
  lea    0x8(%rsi),%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <print_statvfs@plt>
  cmp    %rbp,%rbx
  jne    L19
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < argc; i++) {
    print_statvfs(argv[i]);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L19
  push   %rbp
  lea    -0x2(%rdi),%eax
  push   %rbx
  lea    0x10(%rsi,%rax,8),%rbp
  lea    0x8(%rsi),%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <print_statvfs@plt>
  cmp    %rbp,%rbx
  jne    L14
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L19:
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  for (i = 1; i < argc; i++) {
    print_statvfs(argv[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<coop_array_copy_real_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L18
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void coop_array_copy_real_(double *x, double *y, int *n) {
  int i;
  for (i = 0; i < *n; i++)
    *(y + i) = *(x + i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<coop_array_copy_real_>:
  endbr64
  cmpl   $0x0,(%rdx)
  jle    L3
  mov    $0x0,%eax
L12:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %eax,(%rdx)
  jg     L12
L3:
  ret
```

assistant: ```c
void coop_array_copy_real_(double *x, double *y, int *n) {
  int i;
  for (i = 0; i < *n; i++)
    *(y + i) = *(x + i);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<coop_array_copy_real_>:
  endbr64
  movslq (%rdx),%rdx
  test   %edx,%edx
  jle    L0
  xor    %eax,%eax
  nopl   (%rax)
L6:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L6
L0:
  ret
```

assistant: ```c
void coop_array_copy_real_(double *x, double *y, int *n) {
  int i;
  for (i = 0; i < *n; i++)
    *(y + i) = *(x + i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<coop_array_copy_real_>:
  endbr64
  mov    %rdi,%rcx
  mov    (%rdx),%edi
  test   %edi,%edi
  jle    L17
  lea    -0x1(%rdi),%eax
  cmp    $0x2,%eax
  jbe    L23
  lea    0x8(%rcx),%rax
  cmp    %rax,%rsi
  je     L23
  mov    %edi,%edx
  xor    %eax,%eax
  shr    %edx
  shl    $0x4,%rdx
  nopl   0x0(%rax,%rax,1)
L4:
  movupd (%rcx,%rax,1),%xmm1
  movups %xmm1,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L4
  mov    %edi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%edi
  je     L17
  movsd  (%rcx,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L49:
  movsd  (%rcx,%rax,8),%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L49
L17:
  ret
```

assistant: ```c
void coop_array_copy_real_(double *x, double *y, int *n) {
  int i;
  for (i = 0; i < *n; i++)
    *(y + i) = *(x + i);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_native>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x3,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L16:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D8(%rip),%rax
  add    %rdx,%rax
  mov    $0x8,%edx
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L33
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  jmp    L1
L33:
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L16
  mov    $0xffffffff,%eax
L1:
  leave
  ret
```

assistant: ```c
int get_native() {
  static unsigned char float_reps[][8] = {{0x41, 0x10, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x3f, 0xf0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xf0, 0x3f}};
  static double one = 1.00;
  int i, j;
  j = sizeof(float_reps) / 8;
  for (i = 0; i < j; i++) {
    if (memcmp(&one, float_reps + i, 8) == 0)
      return (i + 1);
  }
  return (-1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_native>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D0(%rip),%rbp
  mov    $0x0,%ebx
  lea    D15(%rip),%r12
L25:
  mov    $0x8,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L24
  add    $0x1,%ebx
  add    $0x8,%rbp
  cmp    $0x3,%ebx
  jne    L25
  mov    $0xffffffff,%eax
L12:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  lea    0x1(%rbx),%eax
  jmp    L12
```

assistant: ```c
int get_native() {
  static unsigned char float_reps[][8] = {{0x41, 0x10, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x3f, 0xf0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xf0, 0x3f}};
  static double one = 1.00;
  int i, j;
  j = sizeof(float_reps) / 8;
  for (i = 0; i < j; i++) {
    if (memcmp(&one, float_reps + i, 8) == 0)
      return (i + 1);
  }
  return (-1);
}
```
user: What is the c source code of the assembly code below?

```asm
<get_native>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  lea    D13(%rip),%rbx
L7:
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x1,%r12d
  call   <memcmp@plt>
  test   %rax,%rax
  je     L12
  add    $0x8,%rbx
  cmp    $0x3,%r12d
  jne    L7
  mov    $0xffffffff,%r12d
L12:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_native() {
  static unsigned char float_reps[][8] = {{0x41, 0x10, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x3f, 0xf0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xf0, 0x3f}};
  static double one = 1.00;
  int i, j;
  j = sizeof(float_reps) / 8;
  for (i = 0; i < j; i++) {
    if (memcmp(&one, float_reps + i, 8) == 0)
      return (i + 1);
  }
  return (-1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_native>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  lea    D1(%rip),%rbx
L2:
  mov    $0x8,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x1,%r12d
  call   <memcmp@plt>
  test   %rax,%rax
  je     L5
  add    $0x8,%rbx
  cmp    $0x3,%r12d
  jne    L2
  mov    $0xffffffff,%r12d
L5:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int get_native() {
  static unsigned char float_reps[][8] = {{0x41, 0x10, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x3f, 0xf0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00}, {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xf0, 0x3f}};
  static double one = 1.00;
  int i, j;
  j = sizeof(float_reps) / 8;
  for (i = 0; i < j; i++) {
    if (memcmp(&one, float_reps + i, 8) == 0)
      return (i + 1);
  }
  return (-1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<gpu_plus3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  movss  (%rax),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0xc,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x10,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x10,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x14,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x14,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x18,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x18,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1c,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x1c,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x1c,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x20,%rax
  movss  (%rax),%xmm1
  mov    -0x10(%rbp),%rax
  add    $0x20,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x20,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void gpu_plus3(const float m[9], const float m2[9], float ans[9]) {
  ans[0] = m[0] + m2[0];
  ans[1] = m[1] + m2[1];
  ans[2] = m[2] + m2[2];
  ans[3] = m[3] + m2[3];
  ans[4] = m[4] + m2[4];
  ans[5] = m[5] + m2[5];
  ans[6] = m[6] + m2[6];
  ans[7] = m[7] + m2[7];
  ans[8] = m[8] + m2[8];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gpu_plus3>:
  endbr64
  movss  (%rdi),%xmm0
  addss  (%rsi),%xmm0
  movss  %xmm0,(%rdx)
  movss  0x4(%rdi),%xmm0
  addss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdx)
  movss  0x8(%rdi),%xmm0
  addss  0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdx)
  movss  0xc(%rdi),%xmm0
  addss  0xc(%rsi),%xmm0
  movss  %xmm0,0xc(%rdx)
  movss  0x10(%rdi),%xmm0
  addss  0x10(%rsi),%xmm0
  movss  %xmm0,0x10(%rdx)
  movss  0x14(%rdi),%xmm0
  addss  0x14(%rsi),%xmm0
  movss  %xmm0,0x14(%rdx)
  movss  0x18(%rdi),%xmm0
  addss  0x18(%rsi),%xmm0
  movss  %xmm0,0x18(%rdx)
  movss  0x1c(%rdi),%xmm0
  addss  0x1c(%rsi),%xmm0
  movss  %xmm0,0x1c(%rdx)
  movss  0x20(%rdi),%xmm0
  addss  0x20(%rsi),%xmm0
  movss  %xmm0,0x20(%rdx)
  ret
```

assistant: ```c
void gpu_plus3(const float m[9], const float m2[9], float ans[9]) {
  ans[0] = m[0] + m2[0];
  ans[1] = m[1] + m2[1];
  ans[2] = m[2] + m2[2];
  ans[3] = m[3] + m2[3];
  ans[4] = m[4] + m2[4];
  ans[5] = m[5] + m2[5];
  ans[6] = m[6] + m2[6];
  ans[7] = m[7] + m2[7];
  ans[8] = m[8] + m2[8];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gpu_plus3>:
  endbr64
  movss  (%rdi),%xmm0
  addss  (%rsi),%xmm0
  movss  %xmm0,(%rdx)
  movss  0x4(%rdi),%xmm0
  addss  0x4(%rsi),%xmm0
  movss  %xmm0,0x4(%rdx)
  movss  0x8(%rdi),%xmm0
  addss  0x8(%rsi),%xmm0
  movss  %xmm0,0x8(%rdx)
  movss  0xc(%rdi),%xmm0
  addss  0xc(%rsi),%xmm0
  movss  %xmm0,0xc(%rdx)
  movss  0x10(%rdi),%xmm0
  addss  0x10(%rsi),%xmm0
  movss  %xmm0,0x10(%rdx)
  movss  0x14(%rdi),%xmm0
  addss  0x14(%rsi),%xmm0
  movss  %xmm0,0x14(%rdx)
  movss  0x18(%rdi),%xmm0
  addss  0x18(%rsi),%xmm0
  movss  %xmm0,0x18(%rdx)
  movss  0x1c(%rdi),%xmm0
  addss  0x1c(%rsi),%xmm0
  movss  %xmm0,0x1c(%rdx)
  movss  0x20(%rdi),%xmm0
  addss  0x20(%rsi),%xmm0
  movss  %xmm0,0x20(%rdx)
  ret
```

assistant: ```c
void gpu_plus3(const float m[9], const float m2[9], float ans[9]) {
  ans[0] = m[0] + m2[0];
  ans[1] = m[1] + m2[1];
  ans[2] = m[2] + m2[2];
  ans[3] = m[3] + m2[3];
  ans[4] = m[4] + m2[4];
  ans[5] = m[5] + m2[5];
  ans[6] = m[6] + m2[6];
  ans[7] = m[7] + m2[7];
  ans[8] = m[8] + m2[8];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strToIntArray@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strToIntArray@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%ecx
  mov    -0x8(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <multiply@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_infered: string

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int len1 = strlen(argv[1]);
  int len2 = strlen(argv[2]);
  int *n1 = strToIntArray(argv[1], len1);
  int *n2 = strToIntArray(argv[2], len2);
  int i, j;
  int product = multiply(n1, len1, n2, len2);
  printf("%d\n", product);
  free(n1);
  free(n2);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    0x10(%rbx),%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    0x8(%rbx),%rdi
  mov    %r12d,%esi
  call   <strToIntArray@plt>
  mov    %rax,%rbp
  mov    0x10(%rbx),%rdi
  mov    %r13d,%esi
  call   <strToIntArray@plt>
  mov    %rax,%rbx
  mov    %r13d,%ecx
  mov    %rax,%rdx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <multiply@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: tag: D9
datatype: string

tool: D9: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int len1 = strlen(argv[1]);
  int len2 = strlen(argv[2]);
  int *n1 = strToIntArray(argv[1], len1);
  int *n2 = strToIntArray(argv[2], len2);
  int i, j;
  int product = multiply(n1, len1, n2, len2);
  printf("%d\n", product);
  free(n1);
  free(n2);
  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   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  mov    0x8(%rbx),%rdi
  mov    %r13d,%esi
  mov    %eax,%r14d
  call   <strToIntArray@plt>
  mov    0x10(%rbx),%rdi
  mov    %r14d,%esi
  mov    %rax,%r12
  call   <strToIntArray@plt>
  mov    %r14d,%ecx
  mov    %r13d,%esi
  mov    %r12,%rdi
  mov    %rax,%rdx
  mov    %rax,%rbp
  call   <multiply@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

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

tool: D7: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int len1 = strlen(argv[1]);
  int len2 = strlen(argv[2]);
  int *n1 = strToIntArray(argv[1], len1);
  int *n2 = strToIntArray(argv[2], len2);
  int i, j;
  int product = multiply(n1, len1, n2, len2);
  printf("%d\n", product);
  free(n1);
  free(n2);
  return 0;
}
```
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
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    0x10(%rbx),%rdi
  mov    %eax,%r13d
  call   <strlen@plt>
  mov    0x8(%rbx),%rdi
  mov    %r13d,%esi
  mov    %eax,%r14d
  call   <strToIntArray@plt>
  mov    0x10(%rbx),%rdi
  mov    %r14d,%esi
  mov    %rax,%r12
  call   <strToIntArray@plt>
  mov    %r14d,%ecx
  mov    %r13d,%esi
  mov    %r12,%rdi
  mov    %rax,%rdx
  mov    %rax,%rbp
  call   <multiply@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_infered: string

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int len1 = strlen(argv[1]);
  int len2 = strlen(argv[2]);
  int *n1 = strToIntArray(argv[1], len1);
  int *n2 = strToIntArray(argv[2], len2);
  int i, j;
  int product = multiply(n1, len1, n2, len2);
  printf("%d\n", product);
  free(n1);
  free(n2);
  return 0;
}
```