ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <create@plt>
  mov    $0x1,%edi
  mov    %rax,%rbp
  call   <node_malloc@plt>
  mov    $0x2,%edi
  mov    %rax,%rbx
  call   <node_malloc@plt>
  mov    $0x3,%edi
  mov    %rax,%r15
  call   <node_malloc@plt>
  mov    $0x4,%edi
  mov    %rax,%r14
  call   <node_malloc@plt>
  mov    $0x5,%edi
  mov    %rax,%r13
  call   <node_malloc@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <insertHead@plt>
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main(void) {
  struct dll *dll = create();
  struct node *n1 = node_malloc(1);
  struct node *n2 = node_malloc(2);
  struct node *n3 = node_malloc(3);
  struct node *n4 = node_malloc(4);
  struct node *n5 = node_malloc(5);
  insertHead(dll, n1);
  insertHead(dll, n2);
  insertHead(dll, n3);
  insertHead(dll, n4);
  insertHead(dll, n5);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r15
  xor    %eax,%eax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <create@plt>
  mov    $0x1,%edi
  mov    %rax,%rbp
  call   <node_malloc@plt>
  mov    $0x2,%edi
  mov    %rax,%rbx
  call   <node_malloc@plt>
  mov    $0x3,%edi
  mov    %rax,%r15
  call   <node_malloc@plt>
  mov    $0x4,%edi
  mov    %rax,%r14
  call   <node_malloc@plt>
  mov    $0x5,%edi
  mov    %rax,%r13
  call   <node_malloc@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <insertHead@plt>
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <insertHead@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int main(void) {
  struct dll *dll = create();
  struct node *n1 = node_malloc(1);
  struct node *n2 = node_malloc(2);
  struct node *n3 = node_malloc(3);
  struct node *n4 = node_malloc(4);
  struct node *n5 = node_malloc(5);
  insertHead(dll, n1);
  insertHead(dll, n2);
  insertHead(dll, n3);
  insertHead(dll, n4);
  insertHead(dll, n5);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decompress>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %edx,-0x64(%rbp)
  mov    %ecx,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x64(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  mov    -0x60(%rbp),%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x68(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <inflateInit@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%edx
  lea    -0x40(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <inflate@plt>
  mov    %eax,-0x44(%rbp)
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <inflateEnd@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void decompress(char *p, char *p2, int len, int uLen) {
  int status;
  z_stream d_stream;
  d_stream.zalloc = (alloc_func)0;
  d_stream.zfree = (free_func)0;
  d_stream.opaque = (voidpf)0;
  d_stream.next_in = p;
  d_stream.avail_in = len;
  d_stream.next_out = p2;
  d_stream.avail_out = uLen;
  inflateInit(&d_stream);
  status = inflate(&d_stream, Z_NO_FLUSH);
  inflateEnd(&d_stream);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<decompress>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x20(%rsp)
  mov    %rdi,(%rsp)
  mov    %edx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %ecx,0x18(%rsp)
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <inflateInit@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  mov    %rbx,%rdi
  call   <inflate@plt>
  mov    %rbx,%rdi
  call   <inflateEnd@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x40,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void decompress(char *p, char *p2, int len, int uLen) {
  int status;
  z_stream d_stream;
  d_stream.zalloc = (alloc_func)0;
  d_stream.zfree = (free_func)0;
  d_stream.opaque = (voidpf)0;
  d_stream.next_in = p;
  d_stream.avail_in = len;
  d_stream.next_out = p2;
  d_stream.avail_out = uLen;
  inflateInit(&d_stream);
  status = inflate(&d_stream, Z_NO_FLUSH);
  inflateEnd(&d_stream);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<decompress>:
  endbr64
  push   %rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rdi,(%rsp)
  mov    %rbp,%rdi
  mov    %edx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %ecx,0x18(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x20(%rsp)
  call   <inflateInit@plt>
  mov    D6(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <inflate@plt>
  mov    %rbp,%rdi
  call   <inflateEnd@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x40,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void decompress(char *p, char *p2, int len, int uLen) {
  int status;
  z_stream d_stream;
  d_stream.zalloc = (alloc_func)0;
  d_stream.zfree = (free_func)0;
  d_stream.opaque = (voidpf)0;
  d_stream.next_in = p;
  d_stream.avail_in = len;
  d_stream.next_out = p2;
  d_stream.avail_out = uLen;
  inflateInit(&d_stream);
  status = inflate(&d_stream, Z_NO_FLUSH);
  inflateEnd(&d_stream);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decompress>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rdi,(%rsp)
  mov    %rbp,%rdi
  mov    %edx,0x8(%rsp)
  mov    %rsi,0x10(%rsp)
  mov    %ecx,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  call   <inflateInit@plt>
  mov    D5(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <inflate@plt>
  mov    %rbp,%rdi
  call   <inflateEnd@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x40,%rsp
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void decompress(char *p, char *p2, int len, int uLen) {
  int status;
  z_stream d_stream;
  d_stream.zalloc = (alloc_func)0;
  d_stream.zfree = (free_func)0;
  d_stream.opaque = (voidpf)0;
  d_stream.next_in = p;
  d_stream.avail_in = len;
  d_stream.next_out = p2;
  d_stream.avail_out = uLen;
  inflateInit(&d_stream);
  status = inflate(&d_stream, Z_NO_FLUSH);
  inflateEnd(&d_stream);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<updateUC_ContinuationAndDependency>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printk@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printk@plt>
  mov    $0x0,%edi
  call   <show_stack@plt>
  mov    $0x0,%eax
  call   <breakpoint@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: string
name: extract_symbolic_data
arguments: label: D4
type_infered: string
name: extract_symbolic_data
arguments: label: D12
type_infered: string

tool: D2: Dependency management doesn't work yet


tool: D4: BUG

tool: D12: kgdb assertion failed: %s


assistant: ```c
void updateUC_ContinuationAndDependency(struct UC_Continuation *cont, struct UC_DependencyNode *dep) {
  printk("Dependency management doesn't work yet\n");
  do {
    if (!(0)) {
      printk("kgdb assertion failed: %s\n", "BUG");
      show_stack(((void *)0));
      breakpoint();
    }
  } while (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<updateUC_ContinuationAndDependency>:
  endbr64
  sub    $0x8,%rsp
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printk@plt>
  lea    D20(%rip),%rsi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printk@plt>
  mov    $0x0,%edi
  call   <show_stack@plt>
  mov    $0x0,%eax
  call   <breakpoint@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D27
data_type: string
name: parse_data
arguments: label: D20
data_type: string
name: parse_data
arguments: label: D13
data_type: string

tool: D27: Dependency management doesn't work yet


tool: D20: BUG

tool: D13: kgdb assertion failed: %s


assistant: ```c
void updateUC_ContinuationAndDependency(struct UC_Continuation *cont, struct UC_DependencyNode *dep) {
  printk("Dependency management doesn't work yet\n");
  do {
    if (!(0)) {
      printk("kgdb assertion failed: %s\n", "BUG");
      show_stack(((void *)0));
      breakpoint();
    }
  } while (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<updateUC_ContinuationAndDependency>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printk@plt>
  lea    D7(%rip),%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printk@plt>
  xor    %edi,%edi
  call   <show_stack@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <breakpoint@plt>
```

tool_calls: name: parse_data
arguments: data_label: D12
type_infered: string
name: parse_data
arguments: data_label: D7
type_infered: string
name: parse_data
arguments: data_label: D5
type_infered: string

tool: D12: Dependency management doesn't work yet


tool: D7: BUG

tool: D5: kgdb assertion failed: %s


assistant: ```c
void updateUC_ContinuationAndDependency(struct UC_Continuation *cont, struct UC_DependencyNode *dep) {
  printk("Dependency management doesn't work yet\n");
  do {
    if (!(0)) {
      printk("kgdb assertion failed: %s\n", "BUG");
      show_stack(((void *)0));
      breakpoint();
    }
  } while (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<updateUC_ContinuationAndDependency>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printk@plt>
  lea    D11(%rip),%rsi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printk@plt>
  xor    %edi,%edi
  call   <show_stack@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <breakpoint@plt>
```

tool_calls: name: read_data
arguments: tag: D20
type: string
name: read_data
arguments: tag: D11
type: string
name: read_data
arguments: tag: D27
type: string

tool: D20: Dependency management doesn't work yet


tool: D11: BUG

tool: D27: kgdb assertion failed: %s


assistant: ```c
void updateUC_ContinuationAndDependency(struct UC_Continuation *cont, struct UC_DependencyNode *dep) {
  printk("Dependency management doesn't work yet\n");
  do {
    if (!(0)) {
      printk("kgdb assertion failed: %s\n", "BUG");
      show_stack(((void *)0));
      breakpoint();
    }
  } while (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  movabs $0xfffffc0200000000,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  and    $0x3,%eax
  shl    $0x3,%eax
  mov    %eax,%ecx
  sarq   %cl,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
extern unsigned long readb(unsigned long addr) {
  long result = *(volatile int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x200000000UL) + 0x00);
  result >>= (addr & 3) * 8;
  return 0xffUL & result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readb>:
  endbr64
  mov    %rdi,%rax
  shl    $0x5,%rax
  movabs $0xfffffc0200000000,%rdx
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  and    $0x3,%edi
  lea    0x0(,%rdi,8),%ecx
  sar    %cl,%rax
  movzbl %al,%eax
  ret
```

assistant: ```c
extern unsigned long readb(unsigned long addr) {
  long result = *(volatile int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x200000000UL) + 0x00);
  result >>= (addr & 3) * 8;
  return 0xffUL & result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readb>:
  endbr64
  movabs $0xfffffc0200000000,%rdx
  mov    %rdi,%rax
  and    $0x3,%edi
  shl    $0x5,%rax
  lea    0x0(,%rdi,8),%ecx
  add    %rdx,%rax
  movslq (%rax),%rax
  sar    %cl,%rax
  movzbl %al,%eax
  ret
```

assistant: ```c
extern unsigned long readb(unsigned long addr) {
  long result = *(volatile int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x200000000UL) + 0x00);
  result >>= (addr & 3) * 8;
  return 0xffUL & result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readb>:
  endbr64
  movabs $0xfffffc0200000000,%rdx
  mov    %rdi,%rax
  and    $0x3,%edi
  shl    $0x5,%rax
  lea    0x0(,%rdi,8),%ecx
  add    %rdx,%rax
  movslq (%rax),%rax
  sar    %cl,%rax
  movzbl %al,%eax
  ret
```

assistant: ```c
extern unsigned long readb(unsigned long addr) {
  long result = *(volatile int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x200000000UL) + 0x00);
  result >>= (addr & 3) * 8;
  return 0xffUL & result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<EstaOrdemCrescente>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L3
L7:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L32
  mov    $0x0,%eax
  jmp    L22
L32:
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L7
  mov    $0x1,%eax
L22:
  pop    %rbp
  ret
```

assistant: ```c
int EstaOrdemCrescente(int vetor[], int tamanho) {
  for (int i = 1; i < tamanho; i++)
    if (vetor[i - 1] > vetor[i])
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<EstaOrdemCrescente>:
  endbr64
  cmp    $0x1,%esi
  jle    L18
  mov    %rdi,%rax
  lea    -0x2(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L16:
  mov    0x4(%rax),%ecx
  cmp    %ecx,(%rax)
  jg     L29
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L16
  mov    $0x1,%eax
  ret
L18:
  mov    $0x1,%eax
  ret
L29:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int EstaOrdemCrescente(int vetor[], int tamanho) {
  for (int i = 1; i < tamanho; i++)
    if (vetor[i - 1] > vetor[i])
      return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<EstaOrdemCrescente>:
  endbr64
  cmp    $0x1,%esi
  jle    L19
  lea    -0x2(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rax
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L27:
  add    $0x4,%rdi
  cmp    %rax,%rdi
  je     L19
L29:
  mov    0x4(%rdi),%edx
  cmp    %edx,(%rdi)
  jle    L27
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int EstaOrdemCrescente(int vetor[], int tamanho) {
  for (int i = 1; i < tamanho; i++)
    if (vetor[i - 1] > vetor[i])
      return 0;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<EstaOrdemCrescente>:
  endbr64
  cmp    $0x1,%esi
  jle    L23
  lea    -0x2(%rsi),%ecx
  mov    (%rdi),%edx
  lea    0x4(%rdi),%rax
  lea    0x8(%rdi,%rcx,4),%rsi
  jmp    L30
  nopl   0x0(%rax)
L2:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L23
L30:
  mov    %edx,%ecx
  mov    (%rax),%edx
  cmp    %ecx,%edx
  jge    L2
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L23:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int EstaOrdemCrescente(int vetor[], int tamanho) {
  for (int i = 1; i < tamanho; i++)
    if (vetor[i - 1] > vetor[i])
      return 0;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GetNameFromPath>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L87
L54:
  addl   $0x1,-0x8(%rbp)
L87:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L54
  jmp    L84
L63:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  je     L33
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L33
  subl   $0x1,-0x8(%rbp)
L84:
  cmpl   $0x0,-0x8(%rbp)
  jns    L63
L33:
  movl   $0x0,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
  jmp    L8
L57:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L16
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L57
L16:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void GetNameFromPath(char *path, char *name) {
  int i, j;
  for (i = 0; path[i] != 0; i++)
    ;
  for (; i >= 0; i--) {
    if ((path[i] == '\\') || (path[i] == '/'))
      break;
  }
  for (j = 0, i++; (path[i] != '.') && (path[i] != 0); i++, j++)
    name[j] = path[i];
  name[j] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GetNameFromPath>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L65
  mov    $0x1,%eax
L1:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L1
  test   %edx,%edx
  js     L91
L59:
  movslq %edx,%rax
  add    %rdi,%rax
L68:
  movzbl (%rax),%ecx
  cmp    $0x5c,%cl
  je     L91
  cmp    $0x2f,%cl
  je     L91
  sub    $0x1,%rax
  sub    $0x1,%edx
  jns    L68
L91:
  lea    0x1(%rdx),%eax
  cltq
  movzbl (%rdi,%rax,1),%ecx
  cmp    $0x2e,%cl
  je     L52
  test   %cl,%cl
  je     L52
  mov    $0x1,%eax
  movslq %edx,%rdx
  add    %rdx,%rdi
L6:
  mov    %cl,-0x1(%rsi,%rax,1)
  mov    %eax,%edx
  movzbl 0x1(%rdi,%rax,1),%ecx
  add    $0x1,%rax
  cmp    $0x2e,%cl
  je     L94
  test   %cl,%cl
  jne    L6
L94:
  movslq %edx,%rdx
  movb   $0x0,(%rsi,%rdx,1)
  ret
L65:
  mov    $0x0,%edx
  jmp    L59
L52:
  mov    $0x0,%edx
  jmp    L94
```

assistant: ```c
void GetNameFromPath(char *path, char *name) {
  int i, j;
  for (i = 0; path[i] != 0; i++)
    ;
  for (; i >= 0; i--) {
    if ((path[i] == '\\') || (path[i] == '/'))
      break;
  }
  for (j = 0, i++; (path[i] != '.') && (path[i] != 0); i++, j++)
    name[j] = path[i];
  name[j] = 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<GetNameFromPath>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L10
  mov    $0x1,%ecx
  nopl   0x0(%rax)
L23:
  movslq %ecx,%rax
  add    $0x1,%rcx
  cmpb   $0x0,-0x1(%rdi,%rcx,1)
  jne    L23
L44:
  xor    %ecx,%ecx
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl (%rdi,%rax,1),%ecx
L37:
  mov    %eax,%r8d
  cmp    $0x5c,%cl
  je     L111
  cmp    $0x2f,%cl
  je     L111
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L15
  xor    %r8d,%r8d
L2:
  test   %dl,%dl
  je     L86
  cmp    $0x2e,%dl
  je     L86
  movslq %r8d,%rcx
  mov    $0x1,%eax
  add    %rcx,%rdi
  nopl   0x0(%rax)
L85:
  mov    %dl,-0x1(%rsi,%rax,1)
  movzbl (%rdi,%rax,1),%edx
  movslq %eax,%rcx
  add    $0x1,%rax
  cmp    $0x2e,%dl
  je     L70
  test   %dl,%dl
  jne    L85
L70:
  add    %rcx,%rsi
L86:
  movb   $0x0,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L111:
  add    $0x1,%r8d
  movslq %r8d,%rcx
  movzbl (%rdi,%rcx,1),%edx
  jmp    L2
  nopl   (%rax)
L10:
  xor    %eax,%eax
  jmp    L44
```

assistant: ```c
void GetNameFromPath(char *path, char *name) {
  int i, j;
  for (i = 0; path[i] != 0; i++)
    ;
  for (; i >= 0; i--) {
    if ((path[i] == '\\') || (path[i] == '/'))
      break;
  }
  for (j = 0, i++; (path[i] != '.') && (path[i] != 0); i++, j++)
    name[j] = path[i];
  name[j] = 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GetNameFromPath>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L102
  mov    $0x1,%ecx
  nopl   0x0(%rax)
L14:
  movslq %ecx,%rax
  add    $0x1,%rcx
  cmpb   $0x0,-0x1(%rdi,%rcx,1)
  jne    L14
  cs nopw 0x0(%rax,%rax,1)
L32:
  movzbl (%rdi,%rax,1),%ecx
  mov    %eax,%r8d
  cmp    $0x5c,%cl
  je     L0
  cmp    $0x2f,%cl
  je     L0
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  jne    L32
  xor    %r8d,%r8d
L38:
  test   %dl,%dl
  je     L91
  cmp    $0x2e,%dl
  je     L91
  movslq %r8d,%rcx
  mov    $0x1,%eax
  add    %rcx,%rdi
  nopl   0x0(%rax)
L50:
  mov    %dl,-0x1(%rsi,%rax,1)
  movzbl (%rdi,%rax,1),%edx
  movslq %eax,%rcx
  add    $0x1,%rax
  cmp    $0x2e,%dl
  je     L86
  test   %dl,%dl
  jne    L50
L86:
  add    %rcx,%rsi
L91:
  movb   $0x0,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x1,%r8d
  movslq %r8d,%rcx
  movzbl (%rdi,%rcx,1),%edx
  jmp    L38
  nopl   (%rax)
L102:
  xor    %eax,%eax
  jmp    L32
```

assistant: ```c
void GetNameFromPath(char *path, char *name) {
  int i, j;
  for (i = 0; path[i] != 0; i++)
    ;
  for (; i >= 0; i--) {
    if ((path[i] == '\\') || (path[i] == '/'))
      break;
  }
  for (j = 0, i++; (path[i] != '.') && (path[i] != 0); i++, j++)
    name[j] = path[i];
  name[j] = 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_until>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,%eax
  mov    %al,-0xc(%rbp)
  jmp    L2
L0:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  mov    $0x0,%eax
  jmp    L7
L15:
  addq   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0xc(%rbp)
  jne    L0
  mov    -0x8(%rbp),%rax
L7:
  pop    %rbp
  ret
```

assistant: ```c
const char *read_until(const char *s, char tk) {
  while (*s != tk)
    if (!*s)
      return NULL;
    else
      ++s;
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_until>:
  endbr64
  mov    %esi,%edx
  movzbl (%rdi),%eax
  cmp    %al,%sil
  je     L16
L9:
  test   %al,%al
  je     L13
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  cmp    %dl,%al
  jne    L9
  mov    %rdi,%rax
  ret
L16:
  mov    %rdi,%rax
  ret
L13:
  mov    $0x0,%eax
  ret
```

assistant: ```c
const char *read_until(const char *s, char tk) {
  while (*s != tk)
    if (!*s)
      return NULL;
    else
      ++s;
  return s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_until>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  cmp    %al,%sil
  jne    L20
  jmp    L12
L30:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  cmp    %dl,%al
  je     L12
L20:
  test   %al,%al
  jne    L30
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rdi,%rax
  ret
```

assistant: ```c
const char *read_until(const char *s, char tk) {
  while (*s != tk)
    if (!*s)
      return NULL;
    else
      ++s;
  return s;
}
```