ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x1e,%r12d
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  movss  D51(%rip),%xmm4
  nopl   0x0(%rax)
L22:
  movaps %xmm4,%xmm6
  movss  D15(%rip),%xmm3
  mov    $0x3d,%ebx
  mulss  %xmm4,%xmm6
  movss  %xmm6,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L3:
  movaps %xmm3,%xmm0
  movss  0xc(%rsp),%xmm2
  pxor   %xmm5,%xmm5
  mov    $0x20,%edi
  mulss  %xmm3,%xmm0
  addss  %xmm0,%xmm2
  mulss  %xmm4,%xmm0
  subss  D8(%rip),%xmm2
  movaps %xmm2,%xmm1
  mulss  %xmm2,%xmm1
  mulss  %xmm4,%xmm0
  mulss  %xmm2,%xmm1
  mulss  %xmm4,%xmm0
  subss  %xmm0,%xmm1
  comiss %xmm1,%xmm5
  jb     L29
  mulss  D19(%rip),%xmm1
  cvttss2si %xmm1,%eax
  cltq
  movsbl 0x0(%rbp,%rax,1),%edi
L29:
  movss  %xmm3,0x8(%rsp)
  movss  %xmm4,0x4(%rsp)
  call   <putchar@plt>
  sub    $0x1,%ebx
  movss  0x8(%rsp),%xmm3
  movss  0x4(%rsp),%xmm4
  addss  D37(%rip),%xmm3
  jne    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  movss  0x4(%rsp),%xmm4
  subss  D23(%rip),%xmm4
  sub    $0x1,%r12d
  jne    L22
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type_infered: string
name: read_data
arguments: data_label: D51
type_infered: byte[4]
name: read_data
arguments: data_label: D15
type_infered: byte[4]
name: read_data
arguments: data_label: D8
type_infered: f32
name: read_data
arguments: data_label: D19
type_infered: byte[4]
name: read_data
arguments: data_label: D37
type_infered: byte[4]
name: read_data
arguments: data_label: D23
type_infered: byte[4]

tool: D7: .:-=+*#%@

tool: {"D51": [0x00,0x00,0xc0,0x3f]}

tool: {"D15": [0x00,0x00,0xc0,0xbf]}

tool: D8: 1

tool: {"D19": [0x00,0x00,0x00,0xc1]}

tool: {"D37": [0xcd,0xcc,0x4c,0x3d]}

tool: {"D23": [0xcd,0xcc,0xcc,0x3d]}

assistant: ```c
int main() {
  for (float y = 1.5f; y > -1.5f; y -= 0.1f) {
    for (float x = -1.5f; x < 1.5f; x += 0.05f) {
      float z = x * x + y * y - 1;
      float f = z * z * z - x * x * y * y * y;
      putchar(f <= 0.0f ? ".:-=+*#%@"[(int)(f * -8.0f)] : ' ');
    }
    putchar('\n');
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x1e,%r12d
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  movss  D46(%rip),%xmm4
  nopl   0x0(%rax)
L27:
  movaps %xmm4,%xmm6
  movss  D59(%rip),%xmm3
  mov    $0x3d,%ebx
  mulss  %xmm4,%xmm6
  movss  %xmm6,0xc(%rsp)
  nopw   0x0(%rax,%rax,1)
L7:
  movaps %xmm3,%xmm0
  movss  0xc(%rsp),%xmm2
  pxor   %xmm5,%xmm5
  movss  %xmm3,0x8(%rsp)
  mulss  %xmm3,%xmm0
  movss  %xmm4,0x4(%rsp)
  mov    $0x20,%edi
  addss  %xmm0,%xmm2
  mulss  %xmm4,%xmm0
  subss  D42(%rip),%xmm2
  movaps %xmm2,%xmm1
  mulss  %xmm2,%xmm1
  mulss  %xmm4,%xmm0
  mulss  %xmm2,%xmm1
  mulss  %xmm4,%xmm0
  subss  %xmm0,%xmm1
  comiss %xmm1,%xmm5
  jb     L15
  mulss  D40(%rip),%xmm1
  cvttss2si %xmm1,%eax
  cltq
  movsbl 0x0(%rbp,%rax,1),%edi
L15:
  call   <putchar@plt>
  sub    $0x1,%ebx
  movss  0x8(%rsp),%xmm3
  movss  0x4(%rsp),%xmm4
  addss  D45(%rip),%xmm3
  jne    L7
  mov    $0xa,%edi
  movss  %xmm4,0x4(%rsp)
  call   <putchar@plt>
  movss  0x4(%rsp),%xmm4
  subss  D37(%rip),%xmm4
  sub    $0x1,%r12d
  jne    L27
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: string
name: parse_data
arguments: data_label: D46
datatype: byte[4]
name: parse_data
arguments: data_label: D59
datatype: byte[4]
name: parse_data
arguments: data_label: D42
datatype: f32
name: parse_data
arguments: data_label: D40
datatype: byte[4]
name: parse_data
arguments: data_label: D45
datatype: byte[4]
name: parse_data
arguments: data_label: D37
datatype: byte[4]

tool: D7: .:-=+*#%@

tool: {"D46": [0x00,0x00,0xc0,0x3f]}

tool: {"D59": [0x00,0x00,0xc0,0xbf]}

tool: D42: 1

tool: {"D40": [0x00,0x00,0x00,0xc1]}

tool: {"D45": [0xcd,0xcc,0x4c,0x3d]}

tool: {"D37": [0xcd,0xcc,0xcc,0x3d]}

assistant: ```c
int main() {
  for (float y = 1.5f; y > -1.5f; y -= 0.1f) {
    for (float x = -1.5f; x < 1.5f; x += 0.05f) {
      float z = x * x + y * y - 1;
      float f = z * z * z - x * x * y * y * y;
      putchar(f <= 0.0f ? ".:-=+*#%@"[(int)(f * -8.0f)] : ' ');
    }
    putchar('\n');
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <intro@plt>
  mov    $0x0,%eax
  call   <mainLoop@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  intro();
  mainLoop();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <intro@plt>
  mov    $0x0,%eax
  call   <mainLoop@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  intro();
  mainLoop();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <intro@plt>
  xor    %eax,%eax
  call   <mainLoop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  intro();
  mainLoop();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <intro@plt>
  xor    %eax,%eax
  call   <mainLoop@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  intro();
  mainLoop();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getraw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int getraw(uint8_t *byte, uint8_t *istream) {
  *byte = *istream;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getraw>:
  endbr64
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned int getraw(uint8_t *byte, uint8_t *istream) {
  *byte = *istream;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getraw>:
  endbr64
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned int getraw(uint8_t *byte, uint8_t *istream) {
  *byte = *istream;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<udelay>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L16
  cmpq   $0x4e20,-0x8(%rbp)
  jbe    L16
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__udelay_bad@plt>
L16:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__udelay@plt>
  nop
  leave
  ret
```

assistant: ```c
void udelay(unsigned long usecs) {
  if (__builtin_constant_p(usecs) && (usecs) > 20000) {
    __udelay_bad(usecs);
  }
  __udelay(usecs);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<udelay>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <__builtin_constant_p@plt>
  cmp    $0x4e20,%rbx
  jbe    L24
  test   %rax,%rax
  jne    L29
L24:
  mov    %rbx,%rdi
  call   <__udelay@plt>
  pop    %rbx
  ret
L29:
  mov    %rbx,%rdi
  call   <__udelay_bad@plt>
  jmp    L24
```

assistant: ```c
void udelay(unsigned long usecs) {
  if (__builtin_constant_p(usecs) && (usecs) > 20000) {
    __udelay_bad(usecs);
  }
  __udelay(usecs);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<udelay>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <__builtin_constant_p@plt>
  cmp    $0x4e20,%rbp
  jbe    L5
  test   %rax,%rax
  jne    L11
L5:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <__udelay@plt>
  nopl   0x0(%rax)
L11:
  mov    %rbp,%rdi
  call   <__udelay_bad@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <__udelay@plt>
```

assistant: ```c
void udelay(unsigned long usecs) {
  if (__builtin_constant_p(usecs) && (usecs) > 20000) {
    __udelay_bad(usecs);
  }
  __udelay(usecs);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<udelay>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <__builtin_constant_p@plt>
  cmp    $0x4e20,%rbp
  jbe    L18
  test   %rax,%rax
  jne    L0
L18:
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <__udelay@plt>
  nopl   0x0(%rax)
L0:
  mov    %rbp,%rdi
  call   <__udelay_bad@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <__udelay@plt>
```

assistant: ```c
void udelay(unsigned long usecs) {
  if (__builtin_constant_p(usecs) && (usecs) > 20000) {
    __udelay_bad(usecs);
  }
  __udelay(usecs);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%esi
  mov    -0x18(%rbp),%eax
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  mov    %eax,%edi
  call   <reduce@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D26
type: string
name: unpack_data
arguments: data_tag: D9
type: string
name: unpack_data
arguments: data_tag: D6
type: string

tool: D26: Enter a fraction (x/y): 

tool: D9: %d/%d

tool: D6: In lowest terms: %d/%d


assistant: ```c
int main(void) {
  int numerator, denominator;
  int reduced_numerator, reduced_denominator;
  printf("Enter a fraction (x/y): ");
  scanf("%d/%d", &numerator, &denominator);
  reduce(numerator, denominator, &reduced_numerator, &reduced_denominator);
  printf("In lowest terms: %d/%d\n", reduced_numerator, reduced_denominator);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  call   <reduce@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
data_type: string
name: extract_symbolic_data
arguments: label: D24
data_type: string
name: extract_symbolic_data
arguments: label: D5
data_type: string

tool: D17: Enter a fraction (x/y): 

tool: D24: %d/%d

tool: D5: In lowest terms: %d/%d


assistant: ```c
int main(void) {
  int numerator, denominator;
  int reduced_numerator, reduced_denominator;
  printf("Enter a fraction (x/y): ");
  scanf("%d/%d", &numerator, &denominator);
  reduce(numerator, denominator, &reduced_numerator, &reduced_denominator);
  printf("In lowest terms: %d/%d\n", reduced_numerator, reduced_denominator);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  lea    0x10(%rsp),%rdx
  lea    0x14(%rsp),%rcx
  call   <reduce@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  xor    %eax,%eax
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D19
type_infered: string
name: retrieve_data
arguments: label: D18
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string

tool: D19: Enter a fraction (x/y): 

tool: D18: %d/%d

tool: D12: In lowest terms: %d/%d


assistant: ```c
int main(void) {
  int numerator, denominator;
  int reduced_numerator, reduced_denominator;
  printf("Enter a fraction (x/y): ");
  scanf("%d/%d", &numerator, &denominator);
  reduce(numerator, denominator, &reduced_numerator, &reduced_denominator);
  printf("In lowest terms: %d/%d\n", reduced_numerator, reduced_denominator);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x8(%rsp),%edi
  lea    0x10(%rsp),%rdx
  lea    0x14(%rsp),%rcx
  call   <reduce@plt>
  mov    0x14(%rsp),%edx
  mov    0x10(%rsp),%esi
  xor    %eax,%eax
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D26
data_type: string
name: retrieve_data
arguments: data_label: D20
data_type: string

tool: D19: Enter a fraction (x/y): 

tool: D26: %d/%d

tool: D20: In lowest terms: %d/%d


assistant: ```c
int main(void) {
  int numerator, denominator;
  int reduced_numerator, reduced_denominator;
  printf("Enter a fraction (x/y): ");
  scanf("%d/%d", &numerator, &denominator);
  reduce(numerator, denominator, &reduced_numerator, &reduced_denominator);
  printf("In lowest terms: %d/%d\n", reduced_numerator, reduced_denominator);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ilog2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0xffff0000,-0x18(%rbp)
  movl   $0xff00,-0x14(%rbp)
  movl   $0xf0,-0x10(%rbp)
  movl   $0xc,-0xc(%rbp)
  movl   $0x2,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0x18(%rbp),%eax
  test   %eax,%eax
  je     L35
  mov    $0x10,%eax
  jmp    L37
L35:
  mov    $0x0,%eax
L37:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L17
  mov    $0x8,%eax
  jmp    L38
L17:
  mov    $0x0,%eax
L38:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0x10(%rbp),%eax
  test   %eax,%eax
  je     L18
  mov    $0x4,%eax
  jmp    L32
L18:
  mov    $0x0,%eax
L32:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0xc(%rbp),%eax
  test   %eax,%eax
  je     L71
  mov    $0x2,%eax
  jmp    L8
L71:
  mov    $0x0,%eax
L8:
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  and    -0x8(%rbp),%eax
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sarl   %cl,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int ilog2(int x) {
  int shamt = 0;
  int curr_shamt;
  int mask_1 = ~0 << 16;
  int mask_2 = 0xff << 8;
  int mask_3 = 0xf0;
  int mask_4 = 0x0c;
  int mask_5 = 0x02;
  curr_shamt = ((!!(x & mask_1)) << 31 >> 31) & 16;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_2)) << 31 >> 31) & 8;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_3)) << 31 >> 31) & 4;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_4)) << 31 >> 31) & 2;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_5)) << 31 >> 31) & 1;
  shamt += curr_shamt;
  x >>= curr_shamt;
  shamt += x;
  return shamt + (~1 + 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ilog2>:
  endbr64
  test   $0xffff0000,%edi
  setne  %al
  movzbl %al,%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  sar    %cl,%edi
  test   $0xff00,%edi
  setne  %cl
  movzbl %cl,%ecx
  shl    $0x3,%ecx
  add    %ecx,%eax
  sar    %cl,%edi
  test   $0xf0,%dil
  setne  %cl
  movzbl %cl,%ecx
  shl    $0x2,%ecx
  add    %ecx,%eax
  sar    %cl,%edi
  test   $0xc,%dil
  setne  %cl
  movzbl %cl,%ecx
  add    %ecx,%ecx
  add    %ecx,%eax
  sar    %cl,%edi
  mov    %edi,%ecx
  shr    %ecx
  and    $0x1,%ecx
  add    %ecx,%eax
  sar    %cl,%edi
  lea    -0x1(%rax,%rdi,1),%eax
  ret
```

assistant: ```c
int ilog2(int x) {
  int shamt = 0;
  int curr_shamt;
  int mask_1 = ~0 << 16;
  int mask_2 = 0xff << 8;
  int mask_3 = 0xf0;
  int mask_4 = 0x0c;
  int mask_5 = 0x02;
  curr_shamt = ((!!(x & mask_1)) << 31 >> 31) & 16;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_2)) << 31 >> 31) & 8;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_3)) << 31 >> 31) & 4;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_4)) << 31 >> 31) & 2;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_5)) << 31 >> 31) & 1;
  shamt += curr_shamt;
  x >>= curr_shamt;
  shamt += x;
  return shamt + (~1 + 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ilog2>:
  endbr64
  xor    %eax,%eax
  test   $0xffff0000,%edi
  setne  %al
  shl    $0x4,%eax
  mov    %eax,%ecx
  sar    %cl,%edi
  xor    %ecx,%ecx
  test   $0xff00,%edi
  setne  %cl
  shl    $0x3,%ecx
  sar    %cl,%edi
  add    %ecx,%eax
  xor    %ecx,%ecx
  test   $0xf0,%dil
  setne  %cl
  shl    $0x2,%ecx
  sar    %cl,%edi
  add    %ecx,%eax
  xor    %ecx,%ecx
  test   $0xc,%dil
  setne  %cl
  add    %ecx,%ecx
  sar    %cl,%edi
  add    %ecx,%eax
  mov    %edi,%ecx
  shr    %ecx
  and    $0x1,%ecx
  add    %ecx,%eax
  sar    %cl,%edi
  lea    -0x1(%rax,%rdi,1),%eax
  ret
```

assistant: ```c
int ilog2(int x) {
  int shamt = 0;
  int curr_shamt;
  int mask_1 = ~0 << 16;
  int mask_2 = 0xff << 8;
  int mask_3 = 0xf0;
  int mask_4 = 0x0c;
  int mask_5 = 0x02;
  curr_shamt = ((!!(x & mask_1)) << 31 >> 31) & 16;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_2)) << 31 >> 31) & 8;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_3)) << 31 >> 31) & 4;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_4)) << 31 >> 31) & 2;
  shamt += curr_shamt;
  x >>= curr_shamt;
  curr_shamt = ((!!(x & mask_5)) << 31 >> 31) & 1;
  shamt += curr_shamt;
  x >>= curr_shamt;
  shamt += x;
  return shamt + (~1 + 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<enum_table_val>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L6
L10:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L29
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    $0x3,%rax
  jmp    L17
L29:
  addq   $0x8,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L10
  mov    $0x0,%eax
L17:
  leave
  ret
```

assistant: ```c
int enum_table_val(const char *table[], const char *key) {
  for (const char **i = table + 1; *i; i++)
    if (!strcmp(key, *i))
      return i - table;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<enum_table_val>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  lea    0x8(%rdi),%rbx
  mov    0x8(%rdi),%rsi
  test   %rsi,%rsi
  je     L22
L6:
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L39
  add    $0x8,%rbx
  mov    (%rbx),%rsi
  test   %rsi,%rsi
  jne    L6
  mov    $0x0,%eax
  jmp    L16
L39:
  mov    %rbx,%rax
  sub    %r12,%rax
  sar    $0x3,%rax
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  mov    $0x0,%eax
  jmp    L16
```

assistant: ```c
int enum_table_val(const char *table[], const char *key) {
  for (const char **i = table + 1; *i; i++)
    if (!strcmp(key, *i))
      return i - table;
  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
<enum_table_val>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    0x8(%rdi),%rsi
  test   %rsi,%rsi
  je     L15
  mov    %rdi,%r12
  lea    0x8(%rdi),%rbx
  jmp    L34
  nopl   (%rax)
L22:
  mov    0x8(%rbx),%rsi
  add    $0x8,%rbx
  test   %rsi,%rsi
  je     L15
L34:
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L22
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sub    %r12,%rax
  pop    %r12
  sar    $0x3,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int enum_table_val(const char *table[], const char *key) {
  for (const char **i = table + 1; *i; i++)
    if (!strcmp(key, *i))
      return i - table;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<enum_table_val>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    0x8(%rdi),%rsi
  test   %rsi,%rsi
  je     L9
  mov    %rdi,%r12
  lea    0x8(%rdi),%rbx
  jmp    L36
  nopl   (%rax)
L6:
  mov    0x8(%rbx),%rsi
  add    $0x8,%rbx
  test   %rsi,%rsi
  je     L9
L36:
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L6
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  sub    %r12,%rax
  pop    %r12
  sar    $0x3,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int enum_table_val(const char *table[], const char *key) {
  for (const char **i = table + 1; *i; i++)
    if (!strcmp(key, *i))
      return i - table;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L18
L1:
  addq   $0x1,-0x8(%rbp)
L18:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L20
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
L20:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t strspn(const char *s, const char *charset) {
  size_t i;
  for (i = 0; strchr(charset, s[i]) != NULL && s[i] != '\0'; i++)
    ;
  return i;
}
```