ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<dosys>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <metas@plt>
  test   %eax,%eax
  je     L2
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <doshell@plt>
L2:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1090 <doexec@plt>
```

assistant: ```c
int dosys(comstring, nohalt)
register char *comstring;
int nohalt;
{
  register int status;
  if (metas(comstring))
    status = doshell(comstring, nohalt);
  else
    status = doexec(comstring);
  return (status);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dosys>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <metas@plt>
  test   %eax,%eax
  je     L1
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1080 <doshell@plt>
L1:
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    1090 <doexec@plt>
```

assistant: ```c
int dosys(comstring, nohalt)
register char *comstring;
int nohalt;
{
  register int status;
  if (metas(comstring))
    status = doshell(comstring, nohalt);
  else
    status = doexec(comstring);
  return (status);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x3,-0x8(%rbp)
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = 3;
  y = ++x;
  return y;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = 3;
  y = ++x;
  return y;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = 3;
  y = ++x;
  return y;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  mov    $0x4,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = 3;
  y = ++x;
  return y;
}
```
user: What is the c source code of the assembly code below?

```asm
<releaseAccBuf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L12:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L12
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void releaseAccBuf(int **p, int len) {
  int i;
  for (i = 0; i < len; i++)
    free(p[i]);
  free(p);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<releaseAccBuf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  test   %esi,%esi
  jle    L16
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
L14:
  mov    (%rbx),%rdi
  call   <free@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L14
L16:
  mov    %r12,%rdi
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void releaseAccBuf(int **p, int len) {
  int i;
  for (i = 0; i < len; i++)
    free(p[i]);
  free(p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<releaseAccBuf>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L0
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nopw   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbp,%rbx
  jne    L19
L0:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <free@plt>
```

assistant: ```c
void releaseAccBuf(int **p, int len) {
  int i;
  for (i = 0; i < len; i++)
    free(p[i]);
  free(p);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<releaseAccBuf>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nopw   0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbp,%rbx
  jne    L14
L18:
  pop    %rbx
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1050 <free@plt>
```

assistant: ```c
void releaseAccBuf(int **p, int len) {
  int i;
  for (i = 0; i < len; i++)
    free(p[i]);
  free(p);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<image_rgb24_from_rgb96f>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %rdx,%r12
  mov    %edi,%eax
  imul   %esi,%eax
  lea    (%rax,%rax,2),%edi
  movslq %edi,%rdi
  call   <malloc@plt>
  mov    %rax,(%rsp)
  test   %ebp,%ebp
  jle    L24
  lea    (%rbx,%rbx,2),%r14d
  mov    $0x0,%r13d
  mov    $0x0,%r11d
  mov    $0x0,%r10d
  lea    -0x1(%rbx),%eax
  mov    %rax,0x8(%rsp)
  lea    0xc(%r12),%r15
  movss  D5(%rip),%xmm0
  jmp    L58
L8:
  movslq %r13d,%rdi
  lea    (%r12,%rdi,4),%rsi
  add    (%rsp),%rdi
  movslq %r11d,%rax
  add    0x8(%rsp),%rax
  lea    (%rax,%rax,2),%rax
  lea    (%r15,%rax,4),%r9
L5:
  movaps %xmm0,%xmm1
  mulss  0x4(%rsi),%xmm1
  cvttss2si %xmm1,%edx
  movaps %xmm0,%xmm1
  mulss  0x8(%rsi),%xmm1
  cvttss2si %xmm1,%eax
  movaps %xmm0,%xmm1
  mulss  (%rsi),%xmm1
  cvttss2si %xmm1,%ecx
  test   %ecx,%ecx
  mov    $0x0,%r8d
  cmovs  %r8d,%ecx
  mov    $0xff,%r8d
  cmp    %r8d,%ecx
  cmovg  %r8d,%ecx
  mov    %cl,(%rdi)
  test   %edx,%edx
  mov    $0x0,%ecx
  cmovs  %ecx,%edx
  mov    $0xff,%ecx
  cmp    %ecx,%edx
  cmovg  %ecx,%edx
  mov    %dl,0x1(%rdi)
  test   %eax,%eax
  mov    $0x0,%edx
  cmovs  %edx,%eax
  mov    $0xff,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %al,0x2(%rdi)
  add    $0xc,%rsi
  add    $0x3,%rdi
  cmp    %r9,%rsi
  jne    L5
L30:
  add    $0x1,%r10d
  add    %ebx,%r11d
  add    %r14d,%r13d
  cmp    %r10d,%ebp
  je     L24
L58:
  test   %ebx,%ebx
  jg     L8
  jmp    L30
L24:
  mov    (%rsp),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: byte[4]

tool: {"D5": [0x00,0x00,0x7f,0x43]}

assistant: ```c
unsigned char *image_rgb24_from_rgb96f(signed int xres, signed int yres, float *fimg) {
  unsigned char *img;
  signed int x;
  signed int y;
  signed int R;
  signed int G;
  signed int B;
  void *return_value_malloc$1;
  return_value_malloc$1 = malloc((unsigned long int)(xres * yres * 3));
  img = (unsigned char *)return_value_malloc$1;
  y = 0;
  while (!(y >= yres)) {
    x = 0;
    while (!(x >= xres)) {
      signed int addr = (xres * y + x) * 3;
      R = (signed int)(fimg[(signed long int)addr] * 255.000000f);
      G = (signed int)(fimg[(signed long int)(addr + 1)] * 255.000000f);
      B = (signed int)(fimg[(signed long int)(addr + 2)] * 255.000000f);
      if (R > 255)
        R = 255;
      if (R < 0)
        R = 0;
      img[(signed long int)addr] = (unsigned char)R;
      if (G > 255)
        G = 255;
      if (G < 0)
        G = 0;
      img[(signed long int)(addr + 1)] = (unsigned char)G;
      if (B > 255)
        B = 255;
      if (B < 0)
        B = 0;
      img[(signed long int)(addr + 2)] = (unsigned char)B;
      x = x + 1;
    }
    y = y + 1;
  }
  return img;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<image_rgb24_from_rgb96f>:
  endbr64
  mov    %edi,%eax
  push   %r15
  imul   %esi,%eax
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  lea    (%rax,%rax,2),%edi
  push   %rbx
  movslq %edi,%rdi
  sub    $0x18,%rsp
  call   <malloc@plt>
  mov    %rax,%r14
  test   %r12d,%r12d
  jle    L12
  lea    -0x1(%rbp),%eax
  movss  D3(%rip),%xmm0
  xor    %ebx,%ebx
  xor    %r11d,%r11d
  mov    %rax,(%rsp)
  lea    0xc(%r13),%rax
  lea    0x0(%rbp,%rbp,2),%r15d
  xor    %r10d,%r10d
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L2:
  test   %ebp,%ebp
  jle    L19
  mov    0x8(%rsp),%rdi
  movslq %r11d,%rax
  add    (%rsp),%rax
  movslq %ebx,%rsi
  lea    (%rax,%rax,2),%rax
  lea    0x0(%r13,%rsi,4),%rcx
  add    %r14,%rsi
  lea    (%rdi,%rax,4),%r9
  nopw   0x0(%rax,%rax,1)
L25:
  movss  0x4(%rcx),%xmm1
  xor    %r8d,%r8d
  mulss  %xmm0,%xmm1
  cvttss2si %xmm1,%edx
  movss  0x8(%rcx),%xmm1
  mulss  %xmm0,%xmm1
  cvttss2si %xmm1,%eax
  movss  (%rcx),%xmm1
  mulss  %xmm0,%xmm1
  cvttss2si %xmm1,%edi
  test   %edi,%edi
  cmovs  %r8d,%edi
  mov    $0xff,%r8d
  cmp    %r8d,%edi
  cmovg  %r8d,%edi
  mov    %dil,(%rsi)
  xor    %edi,%edi
  test   %edx,%edx
  cmovs  %edi,%edx
  mov    $0xff,%edi
  cmp    %edi,%edx
  cmovg  %edi,%edx
  mov    %dl,0x1(%rsi)
  xor    %edx,%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  mov    $0xff,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0xc,%rcx
  add    $0x3,%rsi
  mov    %al,-0x1(%rsi)
  cmp    %r9,%rcx
  jne    L25
L19:
  add    $0x1,%r10d
  add    %ebp,%r11d
  add    %r15d,%ebx
  cmp    %r10d,%r12d
  jne    L2
L12:
  add    $0x18,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: read_data
arguments: data_tag: D3
datatype: byte[4]

tool: {"D3": [0x00,0x00,0x7f,0x43]}

assistant: ```c
unsigned char *image_rgb24_from_rgb96f(signed int xres, signed int yres, float *fimg) {
  unsigned char *img;
  signed int x;
  signed int y;
  signed int R;
  signed int G;
  signed int B;
  void *return_value_malloc$1;
  return_value_malloc$1 = malloc((unsigned long int)(xres * yres * 3));
  img = (unsigned char *)return_value_malloc$1;
  y = 0;
  while (!(y >= yres)) {
    x = 0;
    while (!(x >= xres)) {
      signed int addr = (xres * y + x) * 3;
      R = (signed int)(fimg[(signed long int)addr] * 255.000000f);
      G = (signed int)(fimg[(signed long int)(addr + 1)] * 255.000000f);
      B = (signed int)(fimg[(signed long int)(addr + 2)] * 255.000000f);
      if (R > 255)
        R = 255;
      if (R < 0)
        R = 0;
      img[(signed long int)addr] = (unsigned char)R;
      if (G > 255)
        G = 255;
      if (G < 0)
        G = 0;
      img[(signed long int)(addr + 1)] = (unsigned char)G;
      if (B > 255)
        B = 255;
      if (B < 0)
        B = 0;
      img[(signed long int)(addr + 2)] = (unsigned char)B;
      x = x + 1;
    }
    y = y + 1;
  }
  return img;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<purchases_input>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <double_input@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
type_infered: string

tool: D1: Total Purchase Price? 

assistant: ```c
double purchases_input(void) {
  double purchases = 0.0;
  printf("Total Purchase Price? ");
  purchases = double_input();
  return purchases;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<purchases_input>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <double_input@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
type: string

tool: D5: Total Purchase Price? 

assistant: ```c
double purchases_input(void) {
  double purchases = 0.0;
  printf("Total Purchase Price? ");
  purchases = double_input();
  return purchases;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<purchases_input>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <double_input@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_guessed: string

tool: D7: Total Purchase Price? 

assistant: ```c
double purchases_input(void) {
  double purchases = 0.0;
  printf("Total Purchase Price? ");
  purchases = double_input();
  return purchases;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<purchases_input>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <double_input@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
datatype: string

tool: D4: Total Purchase Price? 

assistant: ```c
double purchases_input(void) {
  double purchases = 0.0;
  printf("Total Purchase Price? ");
  purchases = double_input();
  return purchases;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<byteorder>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x102,-0xa(%rbp)
  movzbl -0xa(%rbp),%eax
  cmp    $0x1,%al
  jne    L40
  movzbl -0x9(%rbp),%eax
  cmp    $0x2,%al
  jne    L40
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L29
L40:
  movzbl -0xa(%rbp),%eax
  cmp    $0x2,%al
  jne    L65
  movzbl -0x9(%rbp),%eax
  cmp    $0x1,%al
  jne    L65
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L29
L65:
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L29:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D22
data_type: string
name: parse_data
arguments: data_tag: D14
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string

tool: D22: Big-endian


tool: D14: Little-endian


tool: D4: Unkown...


assistant: ```c
void byteorder() {
  union {
    short value;
    char union_bytes[sizeof(short)];
  } test;
  test.value = 0x0102;
  if (test.union_bytes[0] == 1 && test.union_bytes[1] == 2) {
    printf("Big-endian\n");
  } else if (test.union_bytes[0] == 2 && test.union_bytes[1] == 1) {
    printf("Little-endian\n");
  } else {
    printf("Unkown...\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<byteorder>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D2
type_guessed: string

tool: D2: Little-endian


assistant: ```c
void byteorder() {
  union {
    short value;
    char union_bytes[sizeof(short)];
  } test;
  test.value = 0x0102;
  if (test.union_bytes[0] == 1 && test.union_bytes[1] == 2) {
    printf("Big-endian\n");
  } else if (test.union_bytes[0] == 2 && test.union_bytes[1] == 1) {
    printf("Little-endian\n");
  } else {
    printf("Unkown...\n");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<byteorder>:
  endbr64
  lea    D9(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_label: D9
data_type: string

tool: D9: Little-endian


assistant: ```c
void byteorder() {
  union {
    short value;
    char union_bytes[sizeof(short)];
  } test;
  test.value = 0x0102;
  if (test.union_bytes[0] == 1 && test.union_bytes[1] == 2) {
    printf("Big-endian\n");
  } else if (test.union_bytes[0] == 2 && test.union_bytes[1] == 1) {
    printf("Little-endian\n");
  } else {
    printf("Unkown...\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<byteorder>:
  endbr64
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D8: Little-endian


assistant: ```c
void byteorder() {
  union {
    short value;
    char union_bytes[sizeof(short)];
  } test;
  test.value = 0x0102;
  if (test.union_bytes[0] == 1 && test.union_bytes[1] == 2) {
    printf("Big-endian\n");
  } else if (test.union_bytes[0] == 2 && test.union_bytes[1] == 1) {
    printf("Little-endian\n");
  } else {
    printf("Unkown...\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_iterative_factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L19
  mov    $0x0,%eax
  jmp    L3
L19:
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L27
L25:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L27:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  cmp    %eax,-0x14(%rbp)
  jg     L25
  mov    -0x8(%rbp),%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_iterative_factorial>:
  endbr64
  mov    $0x0,%edx
  test   %edi,%edi
  js     L37
  cmp    $0x1,%edi
  jle    L38
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%edx
L24:
  imul   %eax,%edx
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L24
L37:
  mov    %edx,%eax
  ret
L38:
  mov    $0x1,%edx
  jmp    L37
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_iterative_factorial>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  js     L0
  cmp    $0x1,%edi
  jle    L35
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%r8d
  xchg   %ax,%ax
L33:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L33
L0:
  mov    %r8d,%eax
  ret
  nop
L35:
  mov    $0x1,%r8d
  jmp    L0
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_iterative_factorial>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  js     L17
  cmp    $0x1,%edi
  jle    L22
  add    $0x1,%edi
  mov    $0x2,%eax
  mov    $0x1,%r8d
  xchg   %ax,%ax
L8:
  imul   %eax,%r8d
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L8
L17:
  mov    %r8d,%eax
  ret
  nop
L22:
  mov    $0x1,%r8d
  jmp    L17
```

assistant: ```c
int ft_iterative_factorial(int nb) {
  int out;
  int i;
  if (nb < 0)
    return (0);
  out = 1;
  i = 1;
  while (i++ < nb)
    out *= i;
  return (out);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L5:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *strdup(s) const char *s;
{
  char *dup;
  dup = (char *)malloc(strlen(s) + 1);
  if (dup)
    strcpy(dup, s);
  return dup;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L5
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L5:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strdup(s) const char *s;
{
  char *dup;
  dup = (char *)malloc(strlen(s) + 1);
  if (dup)
    strcpy(dup, s);
  return dup;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L0
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rax,%r8
L0:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(s) const char *s;
{
  char *dup;
  dup = (char *)malloc(strlen(s) + 1);
  if (dup)
    strcpy(dup, s);
  return dup;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L7
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rax,%r8
L7:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strdup(s) const char *s;
{
  char *dup;
  dup = (char *)malloc(strlen(s) + 1);
  if (dup)
    strcpy(dup, s);
  return dup;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<unsigned_high_prod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x1f,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  shr    $0x1f,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <signed_high_prod@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  imul   -0x18(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
unsigned unsigned_high_prod(unsigned x, unsigned y) {
  int sig_x = x >> 31;
  int sig_y = y >> 31;
  int signed_prod = signed_high_prod(x, y);
  return signed_prod + x * sig_y + y * sig_x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<unsigned_high_prod>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <signed_high_prod@plt>
  mov    %eax,%edx
  mov    %ebp,%eax
  shr    $0x1f,%eax
  imul   %ebx,%eax
  shr    $0x1f,%ebx
  imul   %ebp,%ebx
  add    %ebx,%eax
  add    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned unsigned_high_prod(unsigned x, unsigned y) {
  int sig_x = x >> 31;
  int sig_y = y >> 31;
  int signed_prod = signed_high_prod(x, y);
  return signed_prod + x * sig_y + y * sig_x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<unsigned_high_prod>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <signed_high_prod@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebp,%eax
  shr    $0x1f,%eax
  imul   %ebx,%eax
  shr    $0x1f,%ebx
  imul   %ebp,%ebx
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  add    %r8d,%eax
  ret
```

assistant: ```c
unsigned unsigned_high_prod(unsigned x, unsigned y) {
  int sig_x = x >> 31;
  int sig_y = y >> 31;
  int signed_prod = signed_high_prod(x, y);
  return signed_prod + x * sig_y + y * sig_x;
}
```
user: What is the c source code of the assembly code below?

```asm
<unsigned_high_prod>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <signed_high_prod@plt>
  add    $0x8,%rsp
  mov    %eax,%r8d
  mov    %ebp,%eax
  shr    $0x1f,%eax
  imul   %ebx,%eax
  shr    $0x1f,%ebx
  imul   %ebp,%ebx
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  add    %r8d,%eax
  ret
```

assistant: ```c
unsigned unsigned_high_prod(unsigned x, unsigned y) {
  int sig_x = x >> 31;
  int sig_y = y >> 31;
  int signed_prod = signed_high_prod(x, y);
  return signed_prod + x * sig_y + y * sig_x;
}
```