ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: unpack_data
arguments: tag: D17
type_infered: string
name: unpack_data
arguments: tag: D18
type_infered: string

tool: D17: usage: %s [-cCltxVvh?]


tool: D18: 
   -c | --console:                   print cbmem console
   -C | --coverage:                  dump coverage information
   -l | --list:                      print cbmem table of contents
   -x | --hexdump:                   print hexdump of cbmem area
   -t | --timestamps:                print timestamp information
   -V | --verbose:                   verbose (debugging) output
   -v | --version:                   print the version
   -h | --help:                      print this help


assistant: ```c
void print_usage(const char *name) {
  printf("usage: %s [-cCltxVvh?]\n", name);
  printf("\n"
         "   -c | --console:                   print cbmem console\n"
         "   -C | --coverage:                  dump coverage information\n"
         "   -l | --list:                      print cbmem table of contents\n"
         "   -x | --hexdump:                   print hexdump of cbmem area\n"
         "   -t | --timestamps:                print timestamp information\n"
         "   -V | --verbose:                   verbose (debugging) output\n"
         "   -v | --version:                   print the version\n"
         "   -h | --help:                      print this help\n"
         "\n");
  exit(1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_usage>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: parse_data
arguments: tag: D16
type_infered: string
name: parse_data
arguments: tag: D6
type_infered: string

tool: D16: usage: %s [-cCltxVvh?]


tool: D6: 
   -c | --console:                   print cbmem console
   -C | --coverage:                  dump coverage information
   -l | --list:                      print cbmem table of contents
   -x | --hexdump:                   print hexdump of cbmem area
   -t | --timestamps:                print timestamp information
   -V | --verbose:                   verbose (debugging) output
   -v | --version:                   print the version
   -h | --help:                      print this help


assistant: ```c
void print_usage(const char *name) {
  printf("usage: %s [-cCltxVvh?]\n", name);
  printf("\n"
         "   -c | --console:                   print cbmem console\n"
         "   -C | --coverage:                  dump coverage information\n"
         "   -l | --list:                      print cbmem table of contents\n"
         "   -x | --hexdump:                   print hexdump of cbmem area\n"
         "   -t | --timestamps:                print timestamp information\n"
         "   -V | --verbose:                   verbose (debugging) output\n"
         "   -v | --version:                   print the version\n"
         "   -h | --help:                      print this help\n"
         "\n");
  exit(1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L8
L27:
  mov    -0x18(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jne    L27
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D33
type_guessed: string
name: read_data
arguments: data_label: D12
type_guessed: string
name: read_data
arguments: data_label: D48
type_guessed: string
name: read_data
arguments: data_label: D35
type_guessed: string

tool: D33: Note: Mutiply two negative numbers may take long time but it works i dont know why?.

tool: D12: First Number MULTIPLY: 

tool: D48: %d

tool: D35: Second Number MULTIPLY: 

assistant: ```c
int main() {
  int num1 = 0;
  int num2 = 0;
  int Counter = 0;
  int Product = 0;
  printf("Note: Mutiply two negative numbers may take long time but it works i dont know why?.\n");
  printf("First Number MULTIPLY: ");
  scanf("%d", &num1);
  printf("Second Number MULTIPLY: ");
  scanf("%d", &num2);
  while (num2 != Counter) {
    Product = Product + num1;
    Counter++;
  }
  printf("%d", Product);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  lea    D11(%rip),%rdi
  call   <puts@plt>
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L5
  mov    (%rsp),%edx
  mov    $0x0,%eax
L29:
  add    $0x1,%eax
  cmp    %esi,%eax
  jne    L29
  imul   %edx,%esi
L5:
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D11
data_type: string
name: parse_data
arguments: data_tag: D7
data_type: string
name: parse_data
arguments: data_tag: D2
data_type: string
name: parse_data
arguments: data_tag: D38
data_type: string

tool: D11: Note: Mutiply two negative numbers may take long time but it works i dont know why?.

tool: D7: First Number MULTIPLY: 

tool: D2: %d

tool: D38: Second Number MULTIPLY: 

assistant: ```c
int main() {
  int num1 = 0;
  int num2 = 0;
  int Counter = 0;
  int Product = 0;
  printf("Note: Mutiply two negative numbers may take long time but it works i dont know why?.\n");
  printf("First Number MULTIPLY: ");
  scanf("%d", &num1);
  printf("Second Number MULTIPLY: ");
  scanf("%d", &num2);
  while (num2 != Counter) {
    Product = Product + num1;
    Counter++;
  }
  printf("%d", Product);
}
```
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
  lea    D11(%rip),%rdi
  lea    D16(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <puts@plt>
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L6
  imul   (%rsp),%esi
L6:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D11
type_infered: string
name: read_data
arguments: tag: D16
type_infered: string
name: read_data
arguments: tag: D23
type_infered: string
name: read_data
arguments: tag: D0
type_infered: string

tool: D11: Note: Mutiply two negative numbers may take long time but it works i dont know why?.

tool: D16: %d

tool: D23: First Number MULTIPLY: 

tool: D0: Second Number MULTIPLY: 

assistant: ```c
int main() {
  int num1 = 0;
  int num2 = 0;
  int Counter = 0;
  int Product = 0;
  printf("Note: Mutiply two negative numbers may take long time but it works i dont know why?.\n");
  printf("First Number MULTIPLY: ");
  scanf("%d", &num1);
  printf("Second Number MULTIPLY: ");
  scanf("%d", &num2);
  while (num2 != Counter) {
    Product = Product + num1;
    Counter++;
  }
  printf("%d", Product);
}
```
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
  lea    D27(%rip),%rdi
  lea    D19(%rip),%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <puts@plt>
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  je     L6
  imul   (%rsp),%esi
L6:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D27
type: string
name: read_data
arguments: label: D19
type: string
name: read_data
arguments: label: D20
type: string
name: read_data
arguments: label: D18
type: string

tool: D27: Note: Mutiply two negative numbers may take long time but it works i dont know why?.

tool: D19: %d

tool: D20: First Number MULTIPLY: 

tool: D18: Second Number MULTIPLY: 

assistant: ```c
int main() {
  int num1 = 0;
  int num2 = 0;
  int Counter = 0;
  int Product = 0;
  printf("Note: Mutiply two negative numbers may take long time but it works i dont know why?.\n");
  printf("First Number MULTIPLY: ");
  scanf("%d", &num1);
  printf("Second Number MULTIPLY: ");
  scanf("%d", &num2);
  while (num2 != Counter) {
    Product = Product + num1;
    Counter++;
  }
  printf("%d", Product);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<_dist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movss  %xmm0,-0x2c(%rbp)
  movss  %xmm1,-0x30(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L3
L21:
  movq   $0x0,-0x8(%rbp)
  jmp    L37
L19:
  pxor   %xmm2,%xmm2
  cvtsi2ssq -0x10(%rbp),%xmm2
  movss  -0x2c(%rbp),%xmm0
  movaps %xmm0,%xmm1
  subss  %xmm2,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2ssq -0x10(%rbp),%xmm2
  movss  -0x2c(%rbp),%xmm0
  subss  %xmm2,%xmm0
  mulss  %xmm0,%xmm1
  pxor   %xmm3,%xmm3
  cvtsi2ssq -0x8(%rbp),%xmm3
  movss  -0x30(%rbp),%xmm0
  movaps %xmm0,%xmm2
  subss  %xmm3,%xmm2
  pxor   %xmm3,%xmm3
  cvtsi2ssq -0x8(%rbp),%xmm3
  movss  -0x30(%rbp),%xmm0
  subss  %xmm3,%xmm0
  mulss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  movd   %xmm1,%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %rdx,%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0x8(%rbp)
L37:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L19
  addq   $0x1,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jl     L21
  nop
  nop
  leave
  ret
```

assistant: ```c
void _dist(float *d, long dimx, long dimy, float xc, float yc) {
  long i, j;
  for (i = 0; i < dimx; ++i) {
    for (j = 0; j < dimy; ++j) {
      d[i + j * dimx] = (float)sqrt((xc - (float)i) * (xc - (float)i) + (yc - (float)j) * (yc - (float)j));
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_dist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  movss  %xmm0,0x1c(%rsp)
  movss  %xmm1,0x18(%rsp)
  test   %rsi,%rsi
  jle    L43
  mov    %rsi,%r15
  mov    %rdx,%r12
  lea    0x0(,%rsi,4),%r13
  mov    $0x0,%r14d
  jmp    L14
L37:
  pxor   %xmm1,%xmm1
  cvtsi2ss %r14,%xmm1
  movss  0x1c(%rsp),%xmm0
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  movss  %xmm0,0xc(%rsp)
  mov    0x10(%rsp),%rax
  lea    (%rax,%r14,4),%rbp
  mov    $0x0,%ebx
L5:
  pxor   %xmm1,%xmm1
  cvtsi2ss %rbx,%xmm1
  movss  0x18(%rsp),%xmm0
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  addss  0xc(%rsp),%xmm0
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,0x0(%rbp)
  add    $0x1,%rbx
  add    %r13,%rbp
  cmp    %rbx,%r12
  jne    L5
L19:
  add    $0x1,%r14
  cmp    %r14,%r15
  je     L43
L14:
  test   %r12,%r12
  jg     L37
  jmp    L19
L43:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void _dist(float *d, long dimx, long dimy, float xc, float yc) {
  long i, j;
  for (i = 0; i < dimx; ++i) {
    for (j = 0; j < dimy; ++j) {
      d[i + j * dimx] = (float)sqrt((xc - (float)i) * (xc - (float)i) + (yc - (float)j) * (yc - (float)j));
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_dist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  movss  %xmm0,0x1c(%rsp)
  movss  %xmm1,0x18(%rsp)
  test   %rsi,%rsi
  jle    L22
  mov    %rsi,%r13
  mov    %rdx,%rbx
  lea    0x0(,%rsi,4),%rbp
  xor    %r12d,%r12d
  nopl   (%rax)
L31:
  test   %rbx,%rbx
  jle    L30
  pxor   %xmm1,%xmm1
  movss  0x1c(%rsp),%xmm0
  mov    0x10(%rsp),%rax
  xor    %r15d,%r15d
  cvtsi2ss %r12,%xmm1
  lea    (%rax,%r12,4),%r14
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  movss  %xmm0,0xc(%rsp)
  xchg   %ax,%ax
L35:
  pxor   %xmm1,%xmm1
  movss  0x18(%rsp),%xmm0
  cvtsi2ss %r15,%xmm1
  add    $0x1,%r15
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  addss  0xc(%rsp),%xmm0
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,(%r14)
  add    %rbp,%r14
  cmp    %r15,%rbx
  jne    L35
L30:
  add    $0x1,%r12
  cmp    %r12,%r13
  jne    L31
L22:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void _dist(float *d, long dimx, long dimy, float xc, float yc) {
  long i, j;
  for (i = 0; i < dimx; ++i) {
    for (j = 0; j < dimy; ++j) {
      d[i + j * dimx] = (float)sqrt((xc - (float)i) * (xc - (float)i) + (yc - (float)j) * (yc - (float)j));
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_dist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  movss  %xmm0,0x1c(%rsp)
  movss  %xmm1,0x18(%rsp)
  test   %rsi,%rsi
  jle    L28
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jle    L28
  mov    %rsi,%r13
  lea    0x0(,%rsi,4),%rbp
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L23:
  pxor   %xmm1,%xmm1
  movss  0x1c(%rsp),%xmm0
  mov    0x10(%rsp),%rax
  xor    %r15d,%r15d
  cvtsi2ss %r12,%xmm1
  lea    (%rax,%r12,4),%r14
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  movss  %xmm0,0xc(%rsp)
  nopl   0x0(%rax)
L7:
  pxor   %xmm1,%xmm1
  movss  0x18(%rsp),%xmm0
  cvtsi2ss %r15,%xmm1
  add    $0x1,%r15
  subss  %xmm1,%xmm0
  mulss  %xmm0,%xmm0
  addss  0xc(%rsp),%xmm0
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ss %rax,%xmm0
  movss  %xmm0,(%r14)
  add    %rbp,%r14
  cmp    %r15,%rbx
  jne    L7
  add    $0x1,%r12
  cmp    %r12,%r13
  jne    L23
L28:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void _dist(float *d, long dimx, long dimy, float xc, float yc) {
  long i, j;
  for (i = 0; i < dimx; ++i) {
    for (j = 0; j < dimy; ++j) {
      d[i + j * dimx] = (float)sqrt((xc - (float)i) * (xc - (float)i) + (yc - (float)j) * (yc - (float)j));
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<buildassignmentvector_lng>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L8
L40:
  movq   $0x0,-0x8(%rbp)
  jmp    L56
L43:
  mov    -0x28(%rbp),%rax
  imul   -0x8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L22
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
  jmp    L51
L22:
  addq   $0x1,-0x8(%rbp)
L56:
  mov    -0x8(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jl     L43
L51:
  addq   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L40
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void buildassignmentvector_lng(long *assignment, bool *starMatrix, long nOfRows, long nOfColumns) {
  long row, col;
  for (row = 0; row < nOfRows; row++)
    for (col = 0; col < nOfColumns; col++)
      if (starMatrix[row + nOfRows * col]) {
        assignment[row] = col;
        break;
      }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<buildassignmentvector_lng>:
  endbr64
  test   %rdx,%rdx
  jle    L22
  lea    0x0(,%rdx,4),%r9
  mov    $0x0,%r10d
  jmp    L4
L56:
  mov    %rax,(%rdi,%r10,8)
L67:
  add    $0x1,%r10
  add    $0x4,%rsi
  cmp    %r10,%rdx
  je     L22
L4:
  mov    %rsi,%r8
  mov    $0x0,%eax
  test   %rcx,%rcx
  jle    L67
L65:
  cmpl   $0x0,(%r8)
  jne    L56
  add    $0x1,%rax
  add    %r9,%r8
  cmp    %rax,%rcx
  jne    L65
  jmp    L67
L22:
  ret
```

assistant: ```c
void buildassignmentvector_lng(long *assignment, bool *starMatrix, long nOfRows, long nOfColumns) {
  long row, col;
  for (row = 0; row < nOfRows; row++)
    for (col = 0; col < nOfColumns; col++)
      if (starMatrix[row + nOfRows * col]) {
        assignment[row] = col;
        break;
      }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<buildassignmentvector_lng>:
  endbr64
  test   %rdx,%rdx
  jle    L0
  lea    0x0(,%rdx,4),%r9
  xor    %r10d,%r10d
  nopl   0x0(%rax)
L24:
  mov    %rsi,%r8
  xor    %eax,%eax
  test   %rcx,%rcx
  jg     L37
  jmp    L47
  nopl   0x0(%rax)
L46:
  add    $0x1,%rax
  add    %r9,%r8
  cmp    %rax,%rcx
  je     L47
L37:
  mov    (%r8),%r11d
  test   %r11d,%r11d
  je     L46
  mov    %rax,(%rdi,%r10,8)
L47:
  add    $0x1,%r10
  add    $0x4,%rsi
  cmp    %r10,%rdx
  jne    L24
L0:
  ret
```

assistant: ```c
void buildassignmentvector_lng(long *assignment, bool *starMatrix, long nOfRows, long nOfColumns) {
  long row, col;
  for (row = 0; row < nOfRows; row++)
    for (col = 0; col < nOfColumns; col++)
      if (starMatrix[row + nOfRows * col]) {
        assignment[row] = col;
        break;
      }
}
```
user: What is the c source code of the assembly code below:

```asm
<buildassignmentvector_lng>:
  endbr64
  test   %rdx,%rdx
  jle    L7
  lea    0x0(,%rdx,4),%r9
  xor    %r10d,%r10d
  nopl   0x0(%rax)
L11:
  mov    %rsi,%r8
  xor    %eax,%eax
  test   %rcx,%rcx
  jg     L50
  jmp    L44
  nopl   0x0(%rax)
L56:
  add    $0x1,%rax
  add    %r9,%r8
  cmp    %rax,%rcx
  je     L44
L50:
  mov    (%r8),%r11d
  test   %r11d,%r11d
  je     L56
  mov    %rax,(%rdi,%r10,8)
L44:
  add    $0x1,%r10
  add    $0x4,%rsi
  cmp    %r10,%rdx
  jne    L11
L7:
  ret
```

assistant: ```c
void buildassignmentvector_lng(long *assignment, bool *starMatrix, long nOfRows, long nOfColumns) {
  long row, col;
  for (row = 0; row < nOfRows; row++)
    for (col = 0; col < nOfColumns; col++)
      if (starMatrix[row + nOfRows * col]) {
        assignment[row] = col;
        break;
      }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<compute_inverse_key>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x0,%edi
  call   <perror@plt>
  mov    $0x0,%eax
  jmp    L25
L16:
  movq   $0x0,-0x10(%rbp)
  jmp    L21
L23:
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x30(%rax),%ecx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
L21:
  mov    -0x10(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L23
  mov    -0x8(%rbp),%rax
L25:
  leave
  ret
```

assistant: ```c
char *compute_inverse_key(char *key, size_t key_len) {
  char *inverse_key = malloc(key_len + 1);
  if (NULL == inverse_key) {
    perror(NULL);
    return NULL;
  }
  for (size_t idx = 0; idx < key_len; ++idx) {
    int digit = (int)(key[idx] - '0');
    inverse_key[digit] = (char)(idx + '0');
  }
  return inverse_key;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<compute_inverse_key>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  lea    0x1(%rsi),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L25
  lea    (%r12,%rbp,1),%rdi
  mov    %r12,%rdx
  mov    $0x30,%esi
  sub    %r12d,%esi
  test   %rbp,%rbp
  je     L29
L21:
  movsbl (%rdx),%eax
  sub    $0x30,%eax
  cltq
  lea    (%rsi,%rdx,1),%ecx
  mov    %cl,(%rbx,%rax,1)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L21
L29:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  mov    $0x0,%edi
  call   <perror@plt>
  jmp    L29
```

assistant: ```c
char *compute_inverse_key(char *key, size_t key_len) {
  char *inverse_key = malloc(key_len + 1);
  if (NULL == inverse_key) {
    perror(NULL);
    return NULL;
  }
  for (size_t idx = 0; idx < key_len; ++idx) {
    int digit = (int)(key[idx] - '0');
    inverse_key[digit] = (char)(idx + '0');
  }
  return inverse_key;
}
```
user: Translate the below assembly code into C programming language:

```asm
<compute_inverse_key>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x1(%rsi),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L25
  mov    $0x30,%ecx
  lea    (%rbx,%rbp,1),%r8
  mov    %rbx,%rdi
  sub    %ebx,%ecx
  test   %rbp,%rbp
  je     L35
  nopw   0x0(%rax,%rax,1)
L4:
  movsbl (%rdi),%eax
  lea    (%rcx,%rdi,1),%edx
  add    $0x1,%rdi
  sub    $0x30,%eax
  cltq
  mov    %dl,(%r12,%rax,1)
  cmp    %r8,%rdi
  jne    L4
L35:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  xor    %edi,%edi
  call   <perror@plt>
  jmp    L35
```

assistant: ```c
char *compute_inverse_key(char *key, size_t key_len) {
  char *inverse_key = malloc(key_len + 1);
  if (NULL == inverse_key) {
    perror(NULL);
    return NULL;
  }
  for (size_t idx = 0; idx < key_len; ++idx) {
    int digit = (int)(key[idx] - '0');
    inverse_key[digit] = (char)(idx + '0');
  }
  return inverse_key;
}
```