ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<init_newtracelogmsq>:
  endbr64
  push   %r13
  mov    $0xd05,%edi
  push   %r12
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    D23(%rip),%rax
  mov    (%rax),%esi
  or     $0x1b4,%esi
  call   <msgget@plt>
  test   %eax,%eax
  js     L7
  mov    %eax,%r12d
  mov    D18(%rip),%rax
  lea    0x4(%rsp),%r13
  mov    %r13,%rdx
  mov    %r12d,%edi
  mov    (%rax),%esi
  call   <msgctl@plt>
  mov    D27(%rip),%rax
  mov    %r13,%rdx
  mov    %r12d,%edi
  movl   $0xc800,0x4(%rsp)
  mov    (%rax),%esi
  call   <msgctl@plt>
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0xffffffff,%r12d
  jmp    L14
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int init_newtracelogmsq() {
  int msqid;
  struct msqid_ds buf;
  msqid = msgget(3333, IPC_CREAT | 0664);
  if (msqid < 0)
    return -1;
  msgctl(msqid, IPC_STAT, &buf);
  buf.msg_qbytes = 50 * 1024;
  msgctl(msqid, IPC_SET, &buf);
  return msqid;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eye>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %rax,%rcx
  mov    $0x0,%ebx
  movq   $0x0,-0x28(%rbp)
  jmp    L19
L31:
  movq   $0x0,-0x20(%rbp)
  jmp    L11
L5:
  mov    %rax,%rdx
  imul   -0x28(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0x20(%rbp),%rdx
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rcx,%rdx,8)
  addq   $0x1,-0x20(%rbp)
L11:
  mov    -0x20(%rbp),%rdx
  cmp    -0x38(%rbp),%rdx
  jb     L5
  mov    %rax,%rdx
  imul   -0x28(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0x28(%rbp),%rdx
  movsd  D3(%rip),%xmm0
  movsd  %xmm0,(%rcx,%rdx,8)
  addq   $0x1,-0x28(%rbp)
L19:
  mov    -0x28(%rbp),%rdx
  cmp    -0x38(%rbp),%rdx
  jb     L31
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_guessed: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void eye(size_t n, double X[n][n]) {
  size_t i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      X[i][j] = 0;
    X[i][i] = 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<eye>:
  endbr64
  test   %rdi,%rdi
  je     L1
  lea    0x8(,%rdi,8),%r11
  mov    %rsi,%r8
  lea    0x0(,%rdi,8),%r10
  mov    $0x0,%r9d
  movsd  D6(%rip),%xmm0
L16:
  mov    %rsi,%rdx
  mov    $0x0,%eax
L24:
  movq   $0x0,(%rdx,%rax,8)
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L24
  movsd  %xmm0,(%r8)
  lea    0x1(%r9),%rax
  add    %r11,%r8
  add    %r10,%rsi
  cmp    %rcx,%r9
  je     L1
  mov    %rax,%r9
  jmp    L16
L1:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void eye(size_t n, double X[n][n]) {
  size_t i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      X[i][j] = 0;
    X[i][i] = 1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<eye>:
  endbr64
  test   %rdi,%rdi
  je     L6
  push   %r14
  mov    %rsi,%rcx
  push   %r13
  lea    0x0(,%rdi,8),%r13
  push   %r12
  lea    0x8(%r13),%r14
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %rcx,%rdi
  mov    %r13,%rdx
  xor    %esi,%esi
  add    $0x1,%rbx
  call   <memset@plt>
  mov    %rax,%rcx
  mov    D9(%rip),%rax
  add    %r13,%rcx
  mov    %rax,0x0(%rbp)
  add    %r14,%rbp
  cmp    %rbx,%r12
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type_infered: byte[8]

tool: {"D9": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void eye(size_t n, double X[n][n]) {
  size_t i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      X[i][j] = 0;
    X[i][i] = 1;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<eye>:
  endbr64
  test   %rdi,%rdi
  je     L17
  push   %r14
  mov    %rsi,%rcx
  push   %r13
  lea    0x0(,%rdi,8),%r13
  push   %r12
  lea    0x8(%r13),%r14
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %rcx,%rdi
  mov    %r13,%rdx
  xor    %esi,%esi
  add    $0x1,%rbx
  call   <memset@plt>
  mov    %rax,%rcx
  mov    D6(%rip),%rax
  add    %r13,%rcx
  mov    %rax,0x0(%rbp)
  add    %r14,%rbp
  cmp    %rbx,%r12
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

assistant: ```c
void eye(size_t n, double X[n][n]) {
  size_t i, j;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++)
      X[i][j] = 0;
    X[i][i] = 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_zero_sized_VLA_multi2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %esi,-0x58(%rbp)
  mov    %fs:0x28,%rsi
  mov    %rsi,-0x38(%rbp)
  xor    %esi,%esi
  mov    %rsp,%rsi
  mov    %rsi,%r13
  cmpl   $0x0,-0x54(%rbp)
  je     L54
  mov    %r13,%rsp
  jmp    L47
L54:
  mov    -0x54(%rbp),%esi
  mov    -0x58(%rbp),%edi
  movslq %esi,%r12
  sub    $0x1,%r12
  mov    %r12,-0x50(%rbp)
  movslq %esi,%r12
  mov    %r12,%r14
  mov    $0x0,%r15d
  movslq %edi,%r12
  sub    $0x1,%r12
  mov    %r12,-0x48(%rbp)
  movslq %esi,%r12
  mov    %r12,%r10
  mov    $0x0,%r11d
  movslq %edi,%r12
  mov    %r12,%rax
  mov    $0x0,%edx
  mov    %r11,%r14
  imul   %rax,%r14
  mov    %rdx,%r12
  imul   %r10,%r12
  add    %r14,%r12
  mul    %r10
  lea    (%r12,%rdx,1),%r10
  mov    %r10,%rdx
  movslq %esi,%rax
  mov    %rax,%r8
  mov    $0x0,%r9d
  movslq %edi,%rax
  mov    %rax,%rcx
  mov    $0x0,%ebx
  mov    %r9,%rdx
  imul   %rcx,%rdx
  mov    %rbx,%rax
  imul   %r8,%rax
  lea    (%rdx,%rax,1),%r10
  mov    %r8,%rax
  mul    %rcx
  lea    (%r10,%rdx,1),%rcx
  mov    %rcx,%rdx
  movslq %esi,%rdx
  movslq %edi,%rax
  imul   %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L42:
  cmp    %rdx,%rsp
  je     L53
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L42
L53:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L49
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L49:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x40(%rbp)
  mov    %r13,%rsp
L47:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void check_zero_sized_VLA_multi2(int x, int y) {
  if (x)
    return;
  int vla[y][x];
}
```
user: Translate the below assembly code into C programming language:

```asm
<check_zero_sized_VLA_multi2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rsp,%rax
  test   %edi,%edi
  jne    L0
  mov    %rax,%rsp
L2:
  leave
  ret
L0:
  mov    %rax,%rsp
  jmp    L2
```

assistant: ```c
void check_zero_sized_VLA_multi2(int x, int y) {
  if (x)
    return;
  int vla[y][x];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<check_zero_sized_VLA_multi2>:
  endbr64
  ret
```

assistant: ```c
void check_zero_sized_VLA_multi2(int x, int y) {
  if (x)
    return;
  int vla[y][x];
}
```
user: Translate the below assembly code into C programming language:

```asm
<check_zero_sized_VLA_multi2>:
  endbr64
  ret
```

assistant: ```c
void check_zero_sized_VLA_multi2(int x, int y) {
  if (x)
    return;
  int vla[y][x];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ieee_p_3499444699_sub_1224999178775814671_3536714472>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdx,%r13
  mov    %rcx,%r12
  mov    0xc(%rcx),%r14d
  mov    %r14d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %rax,%rbp
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    (%r12),%ecx
  mov    0x4(%r12),%eax
  mov    0x8(%r12),%edx
  mov    %ecx,(%rbx)
  mov    %eax,0x4(%rbx)
  mov    %edx,0x8(%rbx)
  sub    %ecx,%eax
  imul   %edx,%eax
  add    $0x1,%eax
  mov    %eax,0xc(%rbx)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ieee_p_3499444699_sub_1224999178775814671_3536714472(char *t1, char *t2, char *t3, char *t4) {
  char t6[24];
  char *t0;
  char *t7;
  unsigned char t8;
  char *t9;
  char *t10;
  unsigned int t11;
  char *t12;
  int t13;
  char *t14;
  int t15;
  char *t16;
  int t17;
  char *t18;
  char *t19;
  int t20;
  unsigned int t21;
LAB0:
  t7 = (t6 + 4U);
  t8 = (t3 != 0);
  if (t8 == 1)
    goto LAB3;
LAB2:
  t9 = (t6 + 12U);
  *((char **)t9) = t4;
  t10 = (t4 + 12U);
  t11 = *((unsigned int *)t10);
  t11 = (t11 * 1U);
  t0 = xsi_get_transient_memory(t11);
  memcpy(t0, t3, t11);
  t12 = (t4 + 0U);
  t13 = *((int *)t12);
  t14 = (t4 + 4U);
  t15 = *((int *)t14);
  t16 = (t4 + 8U);
  t17 = *((int *)t16);
  t18 = (t2 + 0U);
  t19 = (t18 + 0U);
  *((int *)t19) = t13;
  t19 = (t18 + 4U);
  *((int *)t19) = t15;
  t19 = (t18 + 8U);
  *((int *)t19) = t17;
  t20 = (t15 - t13);
  t21 = (t20 * t17);
  t21 = (t21 + 1);
  t19 = (t18 + 12U);
  *((unsigned int *)t19) = t21;
LAB1:
  return t0;
LAB3:
  *((char **)t7) = t3;
  goto LAB2;
LAB4:;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ieee_p_3499444699_sub_1224999178775814671_3536714472>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    0xc(%rcx),%r14d
  mov    %rsi,%rbx
  mov    %r14d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  mov    0x0(%rbp),%ecx
  mov    0x4(%rbp),%edx
  mov    0x8(%rbp),%eax
  mov    %edx,0x4(%rbx)
  sub    %ecx,%edx
  imul   %eax,%edx
  mov    %eax,0x8(%rbx)
  mov    %r12,%rax
  mov    %ecx,(%rbx)
  add    $0x1,%edx
  mov    %edx,0xc(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ieee_p_3499444699_sub_1224999178775814671_3536714472(char *t1, char *t2, char *t3, char *t4) {
  char t6[24];
  char *t0;
  char *t7;
  unsigned char t8;
  char *t9;
  char *t10;
  unsigned int t11;
  char *t12;
  int t13;
  char *t14;
  int t15;
  char *t16;
  int t17;
  char *t18;
  char *t19;
  int t20;
  unsigned int t21;
LAB0:
  t7 = (t6 + 4U);
  t8 = (t3 != 0);
  if (t8 == 1)
    goto LAB3;
LAB2:
  t9 = (t6 + 12U);
  *((char **)t9) = t4;
  t10 = (t4 + 12U);
  t11 = *((unsigned int *)t10);
  t11 = (t11 * 1U);
  t0 = xsi_get_transient_memory(t11);
  memcpy(t0, t3, t11);
  t12 = (t4 + 0U);
  t13 = *((int *)t12);
  t14 = (t4 + 4U);
  t15 = *((int *)t14);
  t16 = (t4 + 8U);
  t17 = *((int *)t16);
  t18 = (t2 + 0U);
  t19 = (t18 + 0U);
  *((int *)t19) = t13;
  t19 = (t18 + 4U);
  *((int *)t19) = t15;
  t19 = (t18 + 8U);
  *((int *)t19) = t17;
  t20 = (t15 - t13);
  t21 = (t20 * t17);
  t21 = (t21 + 1);
  t19 = (t18 + 12U);
  *((unsigned int *)t19) = t21;
LAB1:
  return t0;
LAB3:
  *((char **)t7) = t3;
  goto LAB2;
LAB4:;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ieee_p_3499444699_sub_1224999178775814671_3536714472>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    0xc(%rcx),%r14d
  mov    %rsi,%rbx
  mov    %r14d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  movq   0x0(%rbp),%xmm0
  mov    0x8(%rbp),%eax
  movd   %xmm0,%ecx
  pshufd $0xe5,%xmm0,%xmm1
  movd   %xmm1,%edx
  mov    %eax,0x8(%rbx)
  sub    %ecx,%edx
  movq   %xmm0,(%rbx)
  imul   %eax,%edx
  mov    %r12,%rax
  add    $0x1,%edx
  mov    %edx,0xc(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *ieee_p_3499444699_sub_1224999178775814671_3536714472(char *t1, char *t2, char *t3, char *t4) {
  char t6[24];
  char *t0;
  char *t7;
  unsigned char t8;
  char *t9;
  char *t10;
  unsigned int t11;
  char *t12;
  int t13;
  char *t14;
  int t15;
  char *t16;
  int t17;
  char *t18;
  char *t19;
  int t20;
  unsigned int t21;
LAB0:
  t7 = (t6 + 4U);
  t8 = (t3 != 0);
  if (t8 == 1)
    goto LAB3;
LAB2:
  t9 = (t6 + 12U);
  *((char **)t9) = t4;
  t10 = (t4 + 12U);
  t11 = *((unsigned int *)t10);
  t11 = (t11 * 1U);
  t0 = xsi_get_transient_memory(t11);
  memcpy(t0, t3, t11);
  t12 = (t4 + 0U);
  t13 = *((int *)t12);
  t14 = (t4 + 4U);
  t15 = *((int *)t14);
  t16 = (t4 + 8U);
  t17 = *((int *)t16);
  t18 = (t2 + 0U);
  t19 = (t18 + 0U);
  *((int *)t19) = t13;
  t19 = (t18 + 4U);
  *((int *)t19) = t15;
  t19 = (t18 + 8U);
  *((int *)t19) = t17;
  t20 = (t15 - t13);
  t21 = (t20 * t17);
  t21 = (t21 + 1);
  t19 = (t18 + 12U);
  *((unsigned int *)t19) = t21;
LAB1:
  return t0;
LAB3:
  *((char **)t7) = t3;
  goto LAB2;
LAB4:;
}
```
user: What is the c source code of the assembly code below:

```asm
<multiply202>:
  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    L13
L30:
  movl   $0x0,-0x3c(%rbp)
  jmp    L10
L4:
  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)
L10:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L4
  addl   $0x1,-0x40(%rbp)
L13:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L30
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply202(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 would the C source look like for the assembly code provided here?

```asm
<multiply202>:
  endbr64
  test   %edi,%edi
  jle    L4
  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
L16:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L24:
  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    L24
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  ret
```

assistant: ```c
void multiply202(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: Translate the below assembly code into C programming language:

```asm
<multiply202>:
  endbr64
  test   %edi,%edi
  jle    L13
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L25:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  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    L17
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L25
  pop    %rbx
  ret
L13:
  ret
```

assistant: ```c
void multiply202(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: 请从给定的汇编指令推导出C源代码:

```asm
<multiply202>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L2
  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)
L11:
  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     L85
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L85
  nopl   0x0(%rax)
L70:
  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    L70
  cmp    -0x24(%rsp),%r11d
  je     L5
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L5
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L5
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L5:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L11
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L85:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L22:
  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    L22
  mov    -0x18(%rsp),%r8
  jmp    L5
```

assistant: ```c
void multiply202(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
<mem_get_be32_as_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  or     %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int mem_get_be32_as_int(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = ((unsigned int)mem[0]) << 24;
  val |= mem[1] << 16;
  val |= mem[2] << 8;
  val |= mem[3];
  return val;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mem_get_be32_as_int>:
  endbr64
  movzbl (%rdi),%edx
  shl    $0x18,%edx
  movzbl 0x3(%rdi),%eax
  or     %eax,%edx
  movzbl 0x1(%rdi),%eax
  shl    $0x10,%eax
  or     %eax,%edx
  movzbl 0x2(%rdi),%eax
  shl    $0x8,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned int mem_get_be32_as_int(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = ((unsigned int)mem[0]) << 24;
  val |= mem[1] << 16;
  val |= mem[2] << 8;
  val |= mem[3];
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mem_get_be32_as_int>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  ret
```

assistant: ```c
unsigned int mem_get_be32_as_int(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = ((unsigned int)mem[0]) << 24;
  val |= mem[1] << 16;
  val |= mem[2] << 8;
  val |= mem[3];
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mem_get_be32_as_int>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  ret
```

assistant: ```c
unsigned int mem_get_be32_as_int(const void *vmem) {
  unsigned int val;
  const unsigned char *mem = (const unsigned char *)vmem;
  val = ((unsigned int)mem[0]) << 24;
  val |= mem[1] << 16;
  val |= mem[2] << 8;
  val |= mem[3];
  return val;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rj232_send>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L15
L2:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  mov    $0x10,%edi
  call   <rj_writebyte@plt>
  mov    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L2
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int rj232_send(unsigned char *buf, int len) {
  int i, rval;
  for (i = 0; i < len; i++)
    rval = rj_writebyte(0x10, *buf++);
  return rval;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rj232_send>:
  endbr64
  test   %esi,%esi
  jle    L6
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%esi
  lea    (%rdi,%rsi,1),%rbp
L0:
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%esi
  mov    $0x10,%edi
  call   <rj_writebyte@plt>
  cmp    %rbp,%rbx
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int rj232_send(unsigned char *buf, int len) {
  int i, rval;
  for (i = 0; i < len; i++)
    rval = rj_writebyte(0x10, *buf++);
  return rval;
}
```