ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<f_to_s_max_I1>:
  endbr64
  mov    (%rdi),%rax
  mov    %rax,(%rdx)
  cmp    $0x1,%rsi
  jbe    L29
  lea    0x8(%rdi),%rax
  lea    (%rdi,%rsi,8),%rsi
  jmp    L6
L36:
  add    $0x8,%rax
  cmp    %rsi,%rax
  je     L29
L6:
  mov    (%rax),%rcx
  cmp    (%rdx),%rcx
  jle    L36
  mov    %rcx,(%rdx)
  jmp    L36
L29:
  ret
```

assistant: ```c
void f_to_s_max_I1(int8_t *X, uint64_t nR, int8_t *ptr_rslt) {
  *ptr_rslt = X[0];
  for (uint64_t i = 1; i < nR; i++) {
    if (X[i] > *ptr_rslt) {
      *ptr_rslt = X[i];
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<f_to_s_max_I1>:
  endbr64
  mov    (%rdi),%r8
  mov    %r8,(%rdx)
  cmp    $0x1,%rsi
  jbe    L14
  lea    0x8(%rdi),%rax
  lea    (%rdi,%rsi,8),%rsi
  jmp    L24
  nopw   0x0(%rax,%rax,1)
L19:
  mov    (%rdx),%r8
L24:
  mov    (%rax),%rcx
  cmp    %r8,%rcx
  jle    L13
  mov    %rcx,(%rdx)
L13:
  add    $0x8,%rax
  cmp    %rsi,%rax
  jne    L19
L14:
  ret
```

assistant: ```c
void f_to_s_max_I1(int8_t *X, uint64_t nR, int8_t *ptr_rslt) {
  *ptr_rslt = X[0];
  for (uint64_t i = 1; i < nR; i++) {
    if (X[i] > *ptr_rslt) {
      *ptr_rslt = X[i];
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f_to_s_max_I1>:
  endbr64
  mov    (%rdi),%r8
  lea    0x8(%rdi),%rax
  lea    (%rdi,%rsi,8),%rdi
  mov    %r8,(%rdx)
  cmp    $0x1,%rsi
  jbe    L5
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rax),%rcx
  cmp    %r8,%rcx
  jle    L22
  mov    %rcx,(%rdx)
  mov    %rcx,%r8
L22:
  add    $0x8,%rax
  cmp    %rdi,%rax
  jne    L19
L5:
  ret
```

assistant: ```c
void f_to_s_max_I1(int8_t *X, uint64_t nR, int8_t *ptr_rslt) {
  *ptr_rslt = X[0];
  for (uint64_t i = 1; i < nR; i++) {
    if (X[i] > *ptr_rslt) {
      *ptr_rslt = X[i];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<create_block_column>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L23
L30:
  mov    -0x30(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L13
L8:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x2c(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  sub    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,%esi
  mov    -0x4(%rbp),%eax
  sub    %esi,%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  imul   -0x28(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L8
  addl   $0x1,-0x8(%rbp)
L23:
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  imul   -0x28(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L30
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int create_block_column(int **input, int **out, int size, int param, int indexk, int indexj) {
  int i, j;
  for (i = indexk * param; i < param * (indexk + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      out[i - indexk * param][j - indexj * param] = input[i][j];
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<create_block_column>:
  endbr64
  imul   %ecx,%r8d
  lea    (%rcx,%r8,1),%edx
  cmp    %edx,%r8d
  jge    L18
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %ecx,%ebp
  mov    %r9d,%r11d
  imul   %ecx,%r11d
  lea    (%rcx,%r11,1),%r12d
  movslq %r8d,%r8
  lea    (%rdi,%r8,8),%rsi
  lea    -0x1(%r12),%eax
  sub    %r11d,%eax
  lea    0x4(,%rax,4),%r9
  movslq %r11d,%r8
  shl    $0x2,%r8
  mov    $0x0,%r10d
L30:
  cmp    %r12d,%r11d
  jge    L34
  movslq %r10d,%rax
  lea    (%rbx,%rax,8),%rdi
  mov    $0x0,%eax
L24:
  mov    %rax,%rdx
  add    (%rsi),%rdx
  mov    (%rdx,%r8,1),%ecx
  mov    (%rdi),%rdx
  mov    %ecx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %r9,%rax
  jne    L24
L34:
  add    $0x1,%r10d
  add    $0x8,%rsi
  cmp    %ebp,%r10d
  jne    L30
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int create_block_column(int **input, int **out, int size, int param, int indexk, int indexj) {
  int i, j;
  for (i = indexk * param; i < param * (indexk + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      out[i - indexk * param][j - indexj * param] = input[i][j];
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<create_block_column>:
  endbr64
  imul   %ecx,%r8d
  lea    (%rcx,%r8,1),%edx
  cmp    %edx,%r8d
  jge    L1
  imul   %ecx,%r9d
  push   %r12
  movslq %r8d,%r8
  mov    %rsi,%r10
  push   %rbp
  lea    (%rdi,%r8,8),%r12
  xor    %r8d,%r8d
  push   %rbx
  lea    -0x1(%rcx),%ebx
  movslq %r9d,%rbp
  lea    (%rcx,%r9,1),%esi
  mov    %rbp,%r11
  neg    %r11
  shl    $0x2,%r11
  nopw   0x0(%rax,%rax,1)
L3:
  cmp    %esi,%r9d
  jge    L31
  movslq %r8d,%rax
  mov    (%r12,%r8,8),%rdi
  mov    (%r10,%rax,8),%rcx
  mov    %rbp,%rax
  add    %r11,%rcx
  cs nopw 0x0(%rax,%rax,1)
L25:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L25
L31:
  lea    0x1(%r8),%rax
  cmp    %rbx,%r8
  je     L18
  mov    %rax,%r8
  jmp    L3
  nopl   0x0(%rax)
L18:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
int create_block_column(int **input, int **out, int size, int param, int indexk, int indexj) {
  int i, j;
  for (i = indexk * param; i < param * (indexk + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      out[i - indexk * param][j - indexj * param] = input[i][j];
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<create_block_column>:
  endbr64
  imul   %ecx,%r8d
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    (%rcx,%r8,1),%ebx
  cmp    %ebx,%r8d
  jge    L98
  mov    %r9d,%eax
  mov    %ecx,%r10d
  imul   %ecx,%eax
  lea    (%rcx,%rax,1),%r9d
  cmp    %r9d,%eax
  jge    L98
  mov    %rsi,%rdx
  movslq %r8d,%rsi
  mov    %rdi,%r11
  movslq %eax,%rdi
  mov    %rsi,%rcx
  mov    %rdi,-0x18(%rsp)
  lea    0x4(,%rdi,4),%rbp
  neg    %rcx
  lea    (%rdx,%rcx,8),%r12
  lea    0x0(,%rdi,4),%rdx
  mov    %r10d,%ecx
  mov    %rdx,-0x40(%rsp)
  mov    %r10d,%edx
  and    $0xfffffffc,%ecx
  shr    $0x2,%edx
  mov    %ecx,-0x44(%rsp)
  lea    -0x1(%rdx),%r8d
  lea    (%rax,%rcx,1),%edx
  movslq %ecx,%rcx
  shl    $0x2,%rcx
  movslq %edx,%r13
  add    $0x1,%r8
  mov    %rcx,-0x38(%rsp)
  lea    0x1(%rdx),%ecx
  shl    $0x4,%r8
  shl    $0x2,%r13
  movslq %ecx,%r14
  mov    %ecx,-0x30(%rsp)
  sub    %eax,%ecx
  movslq %ecx,%rcx
  shl    $0x2,%r14
  shl    $0x2,%rcx
  mov    %rcx,-0x28(%rsp)
  lea    0x2(%rdx),%ecx
  movslq %ecx,%r15
  mov    %ecx,-0x2c(%rsp)
  sub    %eax,%ecx
  movslq %ecx,%rdx
  shl    $0x2,%r15
  lea    0x0(,%rdx,4),%rax
  mov    %rax,-0x20(%rsp)
  mov    %rdi,%rax
  neg    %rax
  shl    $0x2,%rax
  mov    %rax,-0x10(%rsp)
  lea    -0x1(%r10),%eax
  mov    %eax,-0x48(%rsp)
  nopw   0x0(%rax,%rax,1)
L54:
  mov    (%r11,%rsi,8),%rcx
  mov    (%r12,%rsi,8),%rdx
  lea    (%rcx,%rbp,1),%rdi
  mov    %rdx,%rax
  sub    %rdi,%rax
  cmp    $0x8,%rax
  jbe    L32
  cmpl   $0x2,-0x48(%rsp)
  jbe    L32
  mov    -0x40(%rsp),%rax
  lea    (%rcx,%rax,1),%rdi
  xor    %eax,%eax
  xchg   %ax,%ax
L0:
  movdqu (%rdi,%rax,1),%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %r8,%rax
  jne    L0
  cmp    -0x44(%rsp),%r10d
  je     L2
  mov    (%rcx,%r13,1),%eax
  mov    -0x38(%rsp),%rdi
  mov    %eax,(%rdx,%rdi,1)
  cmp    -0x30(%rsp),%r9d
  jle    L2
  mov    (%rcx,%r14,1),%eax
  mov    -0x28(%rsp),%rdi
  mov    %eax,(%rdx,%rdi,1)
  cmp    -0x2c(%rsp),%r9d
  jle    L2
  mov    (%rcx,%r15,1),%eax
  mov    -0x20(%rsp),%rcx
  mov    %eax,(%rdx,%rcx,1)
L2:
  add    $0x1,%rsi
  cmp    %esi,%ebx
  jg     L54
L98:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L32:
  mov    -0x18(%rsp),%rax
  add    -0x10(%rsp),%rdx
  nopw   0x0(%rax,%rax,1)
L69:
  mov    (%rcx,%rax,4),%edi
  mov    %edi,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r9d
  jg     L69
  jmp    L2
```

assistant: ```c
int create_block_column(int **input, int **out, int size, int param, int indexk, int indexj) {
  int i, j;
  for (i = indexk * param; i < param * (indexk + 1); i++)
    for (j = indexj * param; j < param * (indexj + 1); j++)
      out[i - indexk * param][j - indexj * param] = input[i][j];
  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
<sleep_ms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x20,%rax
  shr    $0x6,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  imul   $0x3e8,%eax,%eax
  sub    %eax,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  imul   $0xf4240,%eax,%eax
  mov    %eax,-0x10(%rbp)
  nop
L17:
  lea    -0x14(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  cmp    $0xffffffff,%eax
  je     L17
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
void sleep_ms(unsigned int milliseconds) {
  struct timespec req = {0};
  time_t seconds = (int)(milliseconds / 1000);
  milliseconds = milliseconds - ((unsigned int)seconds * 1000);
  req.tv_sec = seconds;
  req.tv_nsec = milliseconds * 1000000L;
  while (nanosleep(&req, &req) == -1)
    ;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sleep_ms>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x14(%rsp)
  mov    %edi,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0xc(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0x10(%rsp)
  lea    0xc(%rsp),%rbx
L5:
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <nanosleep@plt>
  cmp    $0xffffffff,%eax
  je     L5
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int milliseconds) {
  struct timespec req = {0};
  time_t seconds = (int)(milliseconds / 1000);
  milliseconds = milliseconds - ((unsigned int)seconds * 1000);
  req.tv_sec = seconds;
  req.tv_nsec = milliseconds * 1000000L;
  while (nanosleep(&req, &req) == -1)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sleep_ms>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  movl   $0x0,0x14(%rsp)
  lea    0xc(%rsp),%rbx
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0xc(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0x10(%rsp)
  nopl   0x0(%rax)
L0:
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <nanosleep@plt>
  cmp    $0xffffffff,%eax
  je     L0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int milliseconds) {
  struct timespec req = {0};
  time_t seconds = (int)(milliseconds / 1000);
  milliseconds = milliseconds - ((unsigned int)seconds * 1000);
  req.tv_sec = seconds;
  req.tv_nsec = milliseconds * 1000000L;
  while (nanosleep(&req, &req) == -1)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sleep_ms>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  movl   $0x0,0x14(%rsp)
  lea    0xc(%rsp),%rbx
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0xc(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0x10(%rsp)
  nopl   0x0(%rax)
L10:
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <nanosleep@plt>
  cmp    $0xffffffff,%eax
  je     L10
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int milliseconds) {
  struct timespec req = {0};
  time_t seconds = (int)(milliseconds / 1000);
  milliseconds = milliseconds - ((unsigned int)seconds * 1000);
  req.tv_sec = seconds;
  req.tv_nsec = milliseconds * 1000000L;
  while (nanosleep(&req, &req) == -1)
    ;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0xc,%edi
  call   <my_thumb@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xc,%edi
  call   <my_thumb2@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x90,%esi
  mov    $0xc,%edi
  call   <add@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <test_branch_instructions@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
datatype: string

tool: D12: result:%08x


assistant: ```c
int main() {
  int n = my_thumb(12);
  printf("result:%08x\n", n);
  n = my_thumb2(12);
  printf("result:%08x\n", n);
  n = add(12, 144);
  printf("result:%08x\n", n);
  test_branch_instructions();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0xc,%edi
  call   <my_thumb@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0xc,%edi
  call   <my_thumb2@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x90,%esi
  mov    $0xc,%edi
  call   <add@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <test_branch_instructions@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string

tool: D3: result:%08x


assistant: ```c
int main() {
  int n = my_thumb(12);
  printf("result:%08x\n", n);
  n = my_thumb2(12);
  printf("result:%08x\n", n);
  n = add(12, 144);
  printf("result:%08x\n", n);
  test_branch_instructions();
  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   %rbp
  mov    $0xc,%edi
  lea    D0(%rip),%rbp
  call   <my_thumb@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xc,%edi
  call   <my_thumb2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x90,%esi
  mov    $0xc,%edi
  call   <add@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <test_branch_instructions@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
datatype: string

tool: D0: result:%08x


assistant: ```c
int main() {
  int n = my_thumb(12);
  printf("result:%08x\n", n);
  n = my_thumb2(12);
  printf("result:%08x\n", n);
  n = add(12, 144);
  printf("result:%08x\n", n);
  test_branch_instructions();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0xc,%edi
  lea    D8(%rip),%rbp
  call   <my_thumb@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xc,%edi
  call   <my_thumb2@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x90,%esi
  mov    $0xc,%edi
  call   <add@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <test_branch_instructions@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type_guessed: string

tool: D8: result:%08x


assistant: ```c
int main() {
  int n = my_thumb(12);
  printf("result:%08x\n", n);
  n = my_thumb2(12);
  printf("result:%08x\n", n);
  n = add(12, 144);
  printf("result:%08x\n", n);
  test_branch_instructions();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<matrix_mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L45
L19:
  mov    -0x38(%rbp),%rax
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x20(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  addq   $0x1,-0x20(%rbp)
L45:
  mov    -0x20(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jl     L19
  movq   $0x0,-0x20(%rbp)
  jmp    L27
L65:
  movq   $0x0,-0x18(%rbp)
  jmp    L40
L79:
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  movq   $0x0,-0x10(%rbp)
  jmp    L77
L50:
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rcx
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  cqto
  idivq  -0x40(%rbp)
  mov    %rdx,%rsi
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  cqto
  idivq  -0x40(%rbp)
  mov    %rdx,%rax
  imul   %rax,%rsi
  mov    %rsi,%rdx
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rsi
  mov    -0x18(%rbp),%rax
  shl    $0x3,%rax
  add    %rsi,%rax
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addq   $0x1,-0x10(%rbp)
L77:
  mov    -0x10(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jl     L50
  addq   $0x1,-0x18(%rbp)
L40:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jl     L79
  addq   $0x1,-0x20(%rbp)
L27:
  mov    -0x20(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jl     L65
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long long int **matrix_mul(long long int **m, long long int **n, long long int num, long long int mod) {
  long long int i, j, k, **result;
  result = (long long int **)malloc(sizeof(long long int) * num);
  for (i = 0; i < num; i++) {
    result[i] = (long long int *)malloc(sizeof(long long int) * num);
  }
  for (i = 0; i < num; i++) {
    for (j = 0; j < num; j++) {
      result[i][j] = 0;
      for (k = 0; k < num; k++) {
        result[i][j] += ((m[i][k]) % mod * (n[k][j] % mod));
      }
    }
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<matrix_mul>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,%r12
  mov    %rdx,%r14
  mov    %rcx,%rbx
  lea    0x0(,%rdx,8),%r15d
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  test   %r14,%r14
  jle    L55
  mov    %rax,%r13
  mov    $0x0,%ebp
L46:
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    %rbp,%r11
  add    $0x1,%rbp
  add    $0x8,%r13
  cmp    %rbp,%r14
  jne    L46
  shl    $0x3,%rbp
  mov    (%rsp),%rax
  lea    (%rax,%rbp,1),%r14
  mov    %rax,%r10
  mov    0x8(%rsp),%rdx
  sub    %rax,%rdx
  mov    %rdx,(%rsp)
  mov    $0x0,%r15d
  jmp    L67
L56:
  mov    %rax,%r13
L3:
  lea    0x0(,%r13,8),%r8
  mov    (%r9),%rax
  movq   $0x0,(%rax,%r13,8)
  mov    $0x0,%ecx
L29:
  mov    %r8,%rdi
  add    (%r9),%rdi
  mov    (%r10),%rax
  mov    (%rax,%rcx,1),%rax
  cqto
  idiv   %rbx
  mov    %rdx,%rsi
  mov    (%r12,%rcx,1),%rax
  mov    (%rax,%r8,1),%rax
  cqto
  idiv   %rbx
  imul   %rdx,%rsi
  add    %rsi,(%rdi)
  add    $0x8,%rcx
  cmp    %rcx,%rbp
  jne    L29
  lea    0x1(%r13),%rax
  cmp    %r13,%r11
  jne    L56
  add    $0x8,%r10
  cmp    %r10,%r14
  je     L55
L67:
  mov    (%rsp),%rax
  lea    (%rax,%r10,1),%r9
  mov    %r15,%r13
  jmp    L3
L55:
  mov    0x8(%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long long int **matrix_mul(long long int **m, long long int **n, long long int num, long long int mod) {
  long long int i, j, k, **result;
  result = (long long int **)malloc(sizeof(long long int) * num);
  for (i = 0; i < num; i++) {
    result[i] = (long long int *)malloc(sizeof(long long int) * num);
  }
  for (i = 0; i < num; i++) {
    for (j = 0; j < num; j++) {
      result[i][j] = 0;
      for (k = 0; k < num; k++) {
        result[i][j] += ((m[i][k]) % mod * (n[k][j] % mod));
      }
    }
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<matrix_mul>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  lea    0x0(,%rdx,8),%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  test   %r13,%r13
  jle    L14
  mov    %rax,%r15
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %ebp,%edi
  add    $0x8,%r15
  call   <malloc@plt>
  mov    %r14,%r11
  lea    0x1(%r14),%r14
  mov    %rax,-0x8(%r15)
  cmp    %r14,%r13
  jne    L33
  mov    0x10(%rsp),%rax
  shl    $0x3,%r13
  mov    0x18(%rsp),%r15
  lea    (%rax,%r13,1),%rdi
  mov    %rax,%r14
  sub    %rax,%r15
  mov    %rdi,0x8(%rsp)
L29:
  mov    (%r15,%r14,1),%rbp
  mov    (%r14),%r10
  xor    %r9d,%r9d
  nopw   0x0(%rax,%rax,1)
L40:
  movq   $0x0,(%r10)
  xor    %edi,%edi
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L35:
  mov    (%r12,%rcx,8),%rax
  mov    %rcx,%r8
  mov    (%rax,%r9,1),%rax
  cqto
  idiv   %rbx
  mov    0x0(%rbp,%rcx,8),%rax
  add    $0x1,%rcx
  mov    %rdx,%rsi
  cqto
  idiv   %rbx
  imul   %rdx,%rsi
  add    %rsi,%rdi
  mov    %rdi,(%r10)
  cmp    %r8,%r11
  jne    L35
  add    $0x8,%r9
  add    $0x8,%r10
  cmp    %r9,%r13
  jne    L40
  add    $0x8,%r14
  cmp    %r14,0x8(%rsp)
  jne    L29
L14:
  mov    0x10(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long long int **matrix_mul(long long int **m, long long int **n, long long int num, long long int mod) {
  long long int i, j, k, **result;
  result = (long long int **)malloc(sizeof(long long int) * num);
  for (i = 0; i < num; i++) {
    result[i] = (long long int *)malloc(sizeof(long long int) * num);
  }
  for (i = 0; i < num; i++) {
    for (j = 0; j < num; j++) {
      result[i][j] = 0;
      for (k = 0; k < num; k++) {
        result[i][j] += ((m[i][k]) % mod * (n[k][j] % mod));
      }
    }
  }
  return result;
}
```