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

```asm
<hash>:
  endbr64
  movsbl (%rdi),%eax
  lea    0x1(%rdi),%rcx
  test   %al,%al
  je     L16
  mov    $0x1505,%edx
  nopl   0x0(%rax)
L15:
  mov    %edx,%esi
  add    $0x1,%rcx
  shl    $0x5,%esi
  add    %esi,%edx
  add    %eax,%edx
  movsbl -0x1(%rcx),%eax
  test   %al,%al
  jne    L15
  mov    %edx,%ecx
  mov    %edx,%eax
  imul   $0x64ef7a75,%rcx,%rcx
  shr    $0x20,%rcx
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x10,%eax
  imul   $0x16f37,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
L16:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int hash(char *str) {
  unsigned int hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash % 94007;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash>:
  endbr64
  movsbl (%rdi),%eax
  lea    0x1(%rdi),%rcx
  test   %al,%al
  je     L19
  mov    $0x1505,%edx
  nopl   0x0(%rax)
L2:
  mov    %edx,%esi
  add    $0x1,%rcx
  shl    $0x5,%esi
  add    %esi,%edx
  add    %eax,%edx
  movsbl -0x1(%rcx),%eax
  test   %al,%al
  jne    L2
  mov    %edx,%ecx
  mov    %edx,%eax
  imul   $0x64ef7a75,%rcx,%rcx
  shr    $0x20,%rcx
  sub    %ecx,%eax
  shr    %eax
  add    %ecx,%eax
  shr    $0x10,%eax
  imul   $0x16f37,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  ret
L19:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned int hash(char *str) {
  unsigned int hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash % 94007;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<testassign>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  fldt   0xe81(%rip)        # 2000 <_fini+0xe18>
  fstpt  -0x40(%rbp)
  fldt   -0x40(%rbp)
  fstpt  -0x60(%rbp)
  mov    -0x60(%rbp),%rax
  mov    -0x58(%rbp),%rdx
  mov    %rax,%rcx
  mov    %rdx,%rbx
  mov    %rcx,%rsi
  mov    -0x48(%rbp),%rax
  mov    $0x5,%ecx
  mov    %rax,%rdi
  call   <__atomic_store_16@plt>
  mov    -0x48(%rbp),%rax
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <__atomic_load_16@plt>
  mov    %rax,-0x70(%rbp)
  mov    %rdx,-0x68(%rbp)
  fldt   -0x70(%rbp)
  fstpt  -0x30(%rbp)
  fldt   -0x30(%rbp)
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  fstp   %st(0)
  call   <__stack_chk_fail@plt>
L7:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long double testassign(_Atomic long double *addr) {
  *addr = 115;
  return *addr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<testassign>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    $0x5,%ecx
  movabs $0xe600000000000000,%rsi
  mov    $0x4005,%edx
  call   <__atomic_store_16@plt>
  mov    $0x5,%esi
  mov    %rbx,%rdi
  call   <__atomic_load_16@plt>
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  fldt   (%rsp)
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
long double testassign(_Atomic long double *addr) {
  *addr = 115;
  return *addr;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<testassign>:
  endbr64
  push   %rbp
  mov    $0x5,%ecx
  mov    %rdi,%rbp
  mov    $0x4005,%edx
  movabs $0xe600000000000000,%rsi
  sub    $0x10,%rsp
  call   <__atomic_store_16@plt>
  mov    %rbp,%rdi
  mov    $0x5,%esi
  call   <__atomic_load_16@plt>
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  fldt   (%rsp)
  add    $0x10,%rsp
  pop    %rbp
  ret
```

assistant: ```c
long double testassign(_Atomic long double *addr) {
  *addr = 115;
  return *addr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<testassign>:
  endbr64
  push   %rbp
  mov    $0x5,%ecx
  mov    %rdi,%rbp
  mov    $0x4005,%edx
  movabs $0xe600000000000000,%rsi
  sub    $0x10,%rsp
  call   <__atomic_store_16@plt>
  mov    %rbp,%rdi
  mov    $0x5,%esi
  call   <__atomic_load_16@plt>
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  fldt   (%rsp)
  add    $0x10,%rsp
  pop    %rbp
  ret
```

assistant: ```c
long double testassign(_Atomic long double *addr) {
  *addr = 115;
  return *addr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L9:
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jbe    L12
  mov    -0x18(%rbp),%rax
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
L12:
  jmp    L9
```

tool_calls: name: read_data
arguments: data_tag: D3
data_type: string
name: read_data
arguments: data_tag: D15
data_type: string
name: read_data
arguments: data_tag: D35
data_type: string
name: read_data
arguments: data_tag: D24
data_type: string

tool: D3: Calculate square root of a number

tool: D15: > 

tool: D35: %lf

tool: D24:   sqrt(%f) = %f



assistant: ```c
int main() {
  double n, result;
  printf("Calculate square root of a number\n");
  while (1) {
    printf("> ");
    scanf("%lf", &n);
    result = sqrt(n);
    if (result > 0)
      printf("  sqrt(%f) = %f\n\n", n, result);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <puts@plt>
  lea    D6(%rip),%r12
  lea    D5(%rip),%rbp
L8:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    (%rsp),%rbx
  pxor   %xmm2,%xmm2
  movq   %rbx,%xmm3
  ucomisd %xmm3,%xmm2
  ja     L15
  movq   %rbx,%xmm1
  sqrtsd %xmm1,%xmm1
L21:
  pxor   %xmm4,%xmm4
  comisd %xmm4,%xmm1
  jbe    L8
  movq   %rbx,%xmm0
  lea    D35(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L8
L15:
  movq   %rbx,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  jmp    L21
```

tool_calls: name: read_data
arguments: label: D24
type_infered: string
name: read_data
arguments: label: D6
type_infered: string
name: read_data
arguments: label: D5
type_infered: string
name: read_data
arguments: label: D35
type_infered: string

tool: D24: Calculate square root of a number

tool: D6: > 

tool: D5: %lf

tool: D35:   sqrt(%f) = %f



assistant: ```c
int main() {
  double n, result;
  printf("Calculate square root of a number\n");
  while (1) {
    printf("> ");
    scanf("%lf", &n);
    result = sqrt(n);
    if (result > 0)
      printf("  sqrt(%f) = %f\n\n", n, result);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  lea    D12(%rip),%rdi
  lea    D3(%rip),%r13
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D19(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <puts@plt>
  nopl   0x0(%rax)
L19:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm2
  pxor   %xmm3,%xmm3
  ucomisd %xmm2,%xmm3
  ja     L22
  movapd %xmm2,%xmm1
  sqrtsd %xmm1,%xmm1
L12:
  pxor   %xmm4,%xmm4
  comisd %xmm4,%xmm1
  jbe    L19
  movapd %xmm2,%xmm0
  mov    %r13,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L19
L22:
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm2
  movapd %xmm0,%xmm1
  jmp    L12
```

tool_calls: name: parse_data
arguments: tag: D12
data_type: string
name: parse_data
arguments: tag: D3
data_type: string
name: parse_data
arguments: tag: D18
data_type: string
name: parse_data
arguments: tag: D19
data_type: string

tool: D12: Calculate square root of a number

tool: D3:   sqrt(%f) = %f



tool: D18: > 

tool: D19: %lf

assistant: ```c
int main() {
  double n, result;
  printf("Calculate square root of a number\n");
  while (1) {
    printf("> ");
    scanf("%lf", &n);
    result = sqrt(n);
    if (result > 0)
      printf("  sqrt(%f) = %f\n\n", n, result);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%rdi
  lea    D34(%rip),%r13
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D10(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <puts@plt>
  nopl   0x0(%rax)
L16:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm2
  pxor   %xmm3,%xmm3
  ucomisd %xmm2,%xmm3
  ja     L13
  movapd %xmm2,%xmm1
  sqrtsd %xmm1,%xmm1
  comisd %xmm3,%xmm1
  jbe    L16
L4:
  movapd %xmm2,%xmm0
  mov    %r13,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L16
L13:
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  pxor   %xmm5,%xmm5
  movsd  0x8(%rsp),%xmm2
  comisd %xmm5,%xmm0
  movapd %xmm0,%xmm1
  jbe    L16
  jmp    L4
```

tool_calls: name: disassemble_data
arguments: label: D9
type: string
name: disassemble_data
arguments: label: D34
type: string
name: disassemble_data
arguments: label: D4
type: string
name: disassemble_data
arguments: label: D10
type: string

tool: D9: Calculate square root of a number

tool: D34:   sqrt(%f) = %f



tool: D4: > 

tool: D10: %lf

assistant: ```c
int main() {
  double n, result;
  printf("Calculate square root of a number\n");
  while (1) {
    printf("> ");
    scanf("%lf", &n);
    result = sqrt(n);
    if (result > 0)
      printf("  sqrt(%f) = %f\n\n", n, result);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ctrl_t_day_original>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ctrl_t_day_common@plt>
  mov    $0x0,%eax
  call   <goto_check@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ctrl_t_day_original() {
  ctrl_t_day_common();
  goto_check();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ctrl_t_day_original>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ctrl_t_day_common@plt>
  mov    $0x0,%eax
  call   <goto_check@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void ctrl_t_day_original() {
  ctrl_t_day_common();
  goto_check();
}
```
user: What is the c source code of the assembly code below:

```asm
<ctrl_t_day_original>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ctrl_t_day_common@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <goto_check@plt>
```

assistant: ```c
void ctrl_t_day_original() {
  ctrl_t_day_common();
  goto_check();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ctrl_t_day_original>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ctrl_t_day_common@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <goto_check@plt>
```

assistant: ```c
void ctrl_t_day_original() {
  ctrl_t_day_common();
  goto_check();
}
```
user: Translate the below assembly code into C programming language:

```asm
<setup_altstack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <sigaltstack@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int setup_altstack(void *start, unsigned long size) {
  stack_t ss;
  memset(&ss, 0, sizeof(ss));
  ss.ss_size = size;
  ss.ss_sp = start;
  return sigaltstack(&ss, NULL);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<setup_altstack>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memset@plt>
  mov    %rbp,(%rsp)
  mov    %rbx,0x8(%rsp)
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <sigaltstack@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int setup_altstack(void *start, unsigned long size) {
  stack_t ss;
  memset(&ss, 0, sizeof(ss));
  ss.ss_size = size;
  ss.ss_sp = start;
  return sigaltstack(&ss, NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<setup_altstack>:
  endbr64
  push   %r12
  mov    $0x10,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  mov    %rbp,(%rsp)
  mov    %rbx,0x8(%rsp)
  call   <sigaltstack@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int setup_altstack(void *start, unsigned long size) {
  stack_t ss;
  memset(&ss, 0, sizeof(ss));
  ss.ss_size = size;
  ss.ss_sp = start;
  return sigaltstack(&ss, NULL);
}
```
user: What is the c source code of the assembly code below:

```asm
<setup_altstack>:
  endbr64
  push   %r12
  mov    $0x10,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <memset@plt>
  xor    %esi,%esi
  mov    %r12,%rdi
  mov    %rbp,(%rsp)
  mov    %rbx,0x8(%rsp)
  call   <sigaltstack@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int setup_altstack(void *start, unsigned long size) {
  stack_t ss;
  memset(&ss, 0, sizeof(ss));
  ss.ss_size = size;
  ss.ss_sp = start;
  return sigaltstack(&ss, NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<i_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  D44(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  D42(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D42(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  D42(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
L22:
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  divsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x20(%rbp),%xmm1
  movsd  D42(%rip),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x18(%rbp),%xmm0
  jp     L21
  pxor   %xmm0,%xmm0
  ucomisd -0x18(%rbp),%xmm0
  je     L2
L21:
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <_finite@plt>
  test   %rax,%rax
  jne    L22
L2:
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D44
type_infered: f64
name: unpack_data
arguments: data_label: D42
type_infered: f64

tool: D44: 2

tool: D42: 1

assistant: ```c
double i_zero(double x) {
  double n, a, halfx, sum;
  halfx = x / 2.0;
  sum = 1.0;
  a = 1.0;
  n = 1.0;
  do {
    a *= halfx;
    a /= n;
    sum += a * a;
    n += 1.0;
  } while (a != 0.0 && _finite(sum));
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<i_zero>:
  endbr64
  sub    $0x28,%rsp
  mulsd  D17(%rip),%xmm0
  movsd  %xmm0,0x18(%rsp)
  movsd  D5(%rip),%xmm7
  movsd  %xmm7,(%rsp)
  movsd  %xmm7,0x10(%rsp)
  movsd  %xmm7,0x8(%rsp)
L11:
  movsd  0x10(%rsp),%xmm0
  mulsd  0x18(%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  divsd  %xmm2,%xmm0
  movapd %xmm0,%xmm1
  movsd  %xmm0,0x10(%rsp)
  mulsd  %xmm0,%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  movsd  D5(%rip),%xmm3
  addsd  %xmm2,%xmm3
  movsd  %xmm3,0x8(%rsp)
  pxor   %xmm6,%xmm6
  ucomisd %xmm6,%xmm1
  jp     L1
  je     L5
L1:
  movsd  (%rsp),%xmm0
  call   <_finite@plt>
  test   %rax,%rax
  jne    L11
L5:
  movsd  (%rsp),%xmm0
  add    $0x28,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D17
type: f32[2]
name: read_data
arguments: data_label: D5
type: f64

tool: {"D17": [0.0,1.75]}

tool: D5: 1

assistant: ```c
double i_zero(double x) {
  double n, a, halfx, sum;
  halfx = x / 2.0;
  sum = 1.0;
  a = 1.0;
  n = 1.0;
  do {
    a *= halfx;
    a /= n;
    sum += a * a;
    n += 1.0;
  } while (a != 0.0 && _finite(sum));
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<i_zero>:
  endbr64
  sub    $0x28,%rsp
  mov    D20(%rip),%rax
  mulsd  D23(%rip),%xmm0
  movq   %rax,%xmm1
  movq   %rax,%xmm2
  movsd  %xmm0,0x18(%rsp)
  movq   %rax,%xmm0
  jmp    L20
  xchg   %ax,%ax
L28:
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm2,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <_finite@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  test   %rax,%rax
  movsd  0x10(%rsp),%xmm1
  je     L15
L20:
  mulsd  0x18(%rsp),%xmm1
  pxor   %xmm5,%xmm5
  divsd  %xmm2,%xmm1
  addsd  D20(%rip),%xmm2
  movapd %xmm1,%xmm3
  ucomisd %xmm5,%xmm1
  mulsd  %xmm1,%xmm3
  addsd  %xmm3,%xmm0
  jp     L28
  jne    L28
L15:
  add    $0x28,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D20
data_type: f64
name: unpack_data
arguments: data_label: D23
data_type: f32[2]

tool: D20: 1

tool: {"D23": [0.0,1.75]}

assistant: ```c
double i_zero(double x) {
  double n, a, halfx, sum;
  halfx = x / 2.0;
  sum = 1.0;
  a = 1.0;
  n = 1.0;
  do {
    a *= halfx;
    a /= n;
    sum += a * a;
    n += 1.0;
  } while (a != 0.0 && _finite(sum));
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<i_zero>:
  endbr64
  sub    $0x28,%rsp
  mov    D24(%rip),%rax
  mulsd  D21(%rip),%xmm0
  movq   %rax,%xmm1
  movq   %rax,%xmm2
  movsd  %xmm0,0x18(%rsp)
  movq   %rax,%xmm0
  jmp    L29
  xchg   %ax,%ax
L28:
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm2,0x8(%rsp)
  movsd  %xmm0,(%rsp)
  call   <_finite@plt>
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm2
  test   %rax,%rax
  movsd  0x10(%rsp),%xmm1
  je     L35
L29:
  mulsd  0x18(%rsp),%xmm1
  pxor   %xmm5,%xmm5
  divsd  %xmm2,%xmm1
  addsd  D24(%rip),%xmm2
  movapd %xmm1,%xmm3
  ucomisd %xmm5,%xmm1
  mulsd  %xmm1,%xmm3
  addsd  %xmm3,%xmm0
  jp     L28
  jne    L28
L35:
  add    $0x28,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D24
type: f64
name: read_data
arguments: data_tag: D21
type: f32[2]

tool: D24: 1

tool: {"D21": [0.0,1.75]}

assistant: ```c
double i_zero(double x) {
  double n, a, halfx, sum;
  halfx = x / 2.0;
  sum = 1.0;
  a = 1.0;
  n = 1.0;
  do {
    a *= halfx;
    a /= n;
    sum += a * a;
    n += 1.0;
  } while (a != 0.0 && _finite(sum));
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_tokens>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
L24:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,(%rdx)
  jmp    L13
L34:
  addq   $0x1,-0x20(%rbp)
L13:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L34
L22:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L24
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void parse_tokens(char **tokens, char *response) {
  int tok = 0;
  do {
    tokens[tok++] = response;
    while (*response != '\0' && *response != ' ') {
      response++;
    }
  } while (*response++);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_tokens>:
  endbr64
L22:
  mov    %rsi,(%rdi)
  movzbl (%rsi),%eax
  test   $0xdf,%al
  je     L16
L20:
  add    $0x1,%rsi
  movzbl (%rsi),%eax
  test   $0xdf,%al
  jne    L20
L16:
  add    $0x1,%rsi
  add    $0x8,%rdi
  test   %al,%al
  jne    L22
  ret
```

assistant: ```c
void parse_tokens(char **tokens, char *response) {
  int tok = 0;
  do {
    tokens[tok++] = response;
    while (*response != '\0' && *response != ' ') {
      response++;
    }
  } while (*response++);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_tokens>:
  endbr64
  nopl   0x0(%rax)
L3:
  mov    %rsi,(%rdi)
  movzbl (%rsi),%eax
  test   $0xdf,%al
  je     L25
  nopw   0x0(%rax,%rax,1)
L28:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  test   $0xdf,%al
  jne    L28
L25:
  add    $0x1,%rsi
  add    $0x8,%rdi
  test   %al,%al
  jne    L3
  ret
```

assistant: ```c
void parse_tokens(char **tokens, char *response) {
  int tok = 0;
  do {
    tokens[tok++] = response;
    while (*response != '\0' && *response != ' ') {
      response++;
    }
  } while (*response++);
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_tokens>:
  endbr64
  nopl   0x0(%rax)
L28:
  mov    %rsi,(%rdi)
  movzbl (%rsi),%eax
  test   $0xdf,%al
  je     L19
  nopw   0x0(%rax,%rax,1)
L17:
  movzbl 0x1(%rsi),%eax
  add    $0x1,%rsi
  test   $0xdf,%al
  jne    L17
L19:
  add    $0x1,%rsi
  add    $0x8,%rdi
  test   %al,%al
  jne    L28
  ret
```

assistant: ```c
void parse_tokens(char **tokens, char *response) {
  int tok = 0;
  do {
    tokens[tok++] = response;
    while (*response != '\0' && *response != ' ') {
      response++;
    }
  } while (*response++);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<usb_debug_get_file_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L6
L21:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L11
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
L11:
  addl   $0x1,-0x10(%rbp)
L6:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L21
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *usb_debug_get_file_name(char *p_filename) {
  int pos, len, i;
  char *p;
  pos = 0;
  p = p_filename;
  len = strlen(p);
  for (i = 0; i < len; i++)
    if (p[i] == '/')
      pos = i + 1;
  p = p + pos;
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<usb_debug_get_file_name>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L19
  lea    0x1(%rax),%ecx
  mov    $0x1,%eax
  mov    $0x0,%edx
L2:
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  cmove  %eax,%edx
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L2
L15:
  movslq %edx,%rdx
  lea    (%rbx,%rdx,1),%rax
  pop    %rbx
  ret
L19:
  mov    $0x0,%edx
  jmp    L15
```

assistant: ```c
char *usb_debug_get_file_name(char *p_filename) {
  int pos, len, i;
  char *p;
  pos = 0;
  p = p_filename;
  len = strlen(p);
  for (i = 0; i < len; i++)
    if (p[i] == '/')
      pos = i + 1;
  p = p + pos;
  return p;
}
```