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

```asm
<substr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x30(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rbx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %rax,%rbx
  jbe    L5
  mov    -0x30(%rbp),%ecx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L5:
  movl   $0x0,-0x1c(%rbp)
  jmp    L1
L17:
  mov    -0x2c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L1:
  mov    -0x1c(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L17
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D0
type: string

tool: D0: Error: Substring (%s, %d, %d) will read beyond string boundary.


assistant: ```c
char *substr(S, pos, len)
char *S;
int pos, len;
{
  int i;
  char *SubS;
  SubS = malloc(len + 1);
  if ((pos + len) > strlen(S)) {
    printf("Error: Substring (%s, %d, %d) will read beyond string boundary.\n", S, pos, len);
    exit(1);
  }
  for (i = 0; i < len; i++)
    SubS[i] = S[pos + i];
  SubS[len] = '\0';
  return SubS;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<substr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebx
  mov    %edx,%r12d
  lea    0x1(%rdx),%edi
  movslq %edi,%rdi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  lea    (%r12,%rbx,1),%eax
  cltq
  cmp    %rdx,%rax
  ja     L14
  mov    %r12d,%esi
  mov    $0x0,%eax
  movslq %ebx,%rcx
  add    %r13,%rcx
  test   %r12d,%r12d
  jle    L21
L15:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L15
L21:
  movslq %r12d,%r12
  movb   $0x0,0x0(%rbp,%r12,1)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %r13,%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
```

tool_calls: name: parse_data
arguments: data_label: D1
type_infered: string

tool: D1: Error: Substring (%s, %d, %d) will read beyond string boundary.


assistant: ```c
char *substr(S, pos, len)
char *S;
int pos, len;
{
  int i;
  char *SubS;
  SubS = malloc(len + 1);
  if ((pos + len) > strlen(S)) {
    printf("Error: Substring (%s, %d, %d) will read beyond string boundary.\n", S, pos, len);
    exit(1);
  }
  for (i = 0; i < len; i++)
    SubS[i] = S[pos + i];
  SubS[len] = '\0';
  return SubS;
}
```
user: What is the c source code of the assembly code below:

```asm
<substr>:
  endbr64
  push   %r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%r12),%edi
  movslq %edi,%rdi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r14
  call   <strlen@plt>
  mov    %rax,%r8
  lea    (%r12,%r13,1),%eax
  cltq
  cmp    %r8,%rax
  ja     L18
  test   %r12d,%r12d
  jle    L1
  lea    0x0(%rbp,%r13,1),%rsi
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <memcpy@plt>
L1:
  movb   $0x0,(%r14,%r12,1)
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  lea    D9(%rip),%rdi
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %rbp,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D9
datatype: string

tool: D9: Error: Substring (%s, %d, %d) will read beyond string boundary.


assistant: ```c
char *substr(S, pos, len)
char *S;
int pos, len;
{
  int i;
  char *SubS;
  SubS = malloc(len + 1);
  if ((pos + len) > strlen(S)) {
    printf("Error: Substring (%s, %d, %d) will read beyond string boundary.\n", S, pos, len);
    exit(1);
  }
  for (i = 0; i < len; i++)
    SubS[i] = S[pos + i];
  SubS[len] = '\0';
  return SubS;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<substr>:
  endbr64
  push   %r14
  push   %r13
  movslq %esi,%r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%r12),%edi
  movslq %edi,%rdi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,%r14
  call   <strlen@plt>
  mov    %rax,%r8
  lea    (%r12,%r13,1),%eax
  cltq
  cmp    %r8,%rax
  ja     L8
  test   %r12d,%r12d
  jle    L19
  lea    0x0(%rbp,%r13,1),%rsi
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <memcpy@plt>
L19:
  movb   $0x0,(%r14,%r12,1)
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  lea    D9(%rip),%rdi
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %rbp,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
```

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

tool: D9: Error: Substring (%s, %d, %d) will read beyond string boundary.


assistant: ```c
char *substr(S, pos, len)
char *S;
int pos, len;
{
  int i;
  char *SubS;
  SubS = malloc(len + 1);
  if ((pos + len) > strlen(S)) {
    printf("Error: Substring (%s, %d, %d) will read beyond string boundary.\n", S, pos, len);
    exit(1);
  }
  for (i = 0; i < len; i++)
    SubS[i] = S[pos + i];
  SubS[len] = '\0';
  return SubS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zcall>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <ot@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <outname@plt>
  mov    $0x0,%eax
  call   <nl@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D1
type: string

tool: D1: call 

assistant: ```c
zcall(sname) char *sname;
{
  ot("call ");
  outname(sname);
  nl();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<zcall>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <ot@plt>
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <outname@plt>
  mov    $0x0,%eax
  call   <nl@plt>
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
datatype: string

tool: D6: call 

assistant: ```c
zcall(sname) char *sname;
{
  ot("call ");
  outname(sname);
  nl();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<zcall>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D7(%rip),%rdi
  call   <ot@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <outname@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <nl@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type: string

tool: D7: call 

assistant: ```c
zcall(sname) char *sname;
{
  ot("call ");
  outname(sname);
  nl();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<zcall>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  lea    D6(%rip),%rdi
  call   <ot@plt>
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <outname@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <nl@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D6
data_type: string

tool: D6: call 

assistant: ```c
zcall(sname) char *sname;
{
  ot("call ");
  outname(sname);
  nl();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  sub    $0xa4d,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  nop
  movzwl -0xa(%rbp),%eax
  lea    0x13eb(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x99e,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0x99e,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_mosquitto_malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *_mosquitto_malloc(size_t size) {
  void *mem = malloc(size);
  return mem;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<_mosquitto_malloc>:
  endbr64
  sub    $0x8,%rsp
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *_mosquitto_malloc(size_t size) {
  void *mem = malloc(size);
  return mem;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_mosquitto_malloc>:
  endbr64
  jmp    1050 <malloc@plt>
```

assistant: ```c
void *_mosquitto_malloc(size_t size) {
  void *mem = malloc(size);
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_mosquitto_malloc>:
  endbr64
  jmp    1050 <malloc@plt>
```

assistant: ```c
void *_mosquitto_malloc(size_t size) {
  void *mem = malloc(size);
  return mem;
}
```
user: What is the c source code of the assembly code below:

```asm
<shdn8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rax
  add    $0x4,%rax
  mov    $0x0,%r12d
  movl   $0x0,-0x14(%rbp)
  jmp    L2
L8:
  movzwl (%rax),%edx
  mov    %edx,%ebx
  shl    $0x8,%ebx
  movzwl (%rax),%edx
  shr    $0x8,%dx
  mov    %dx,(%rax)
  movzwl (%rax),%edx
  or     %r12d,%edx
  mov    %dx,(%rax)
  mov    %ebx,%r12d
  add    $0x2,%rax
  addl   $0x1,-0x14(%rbp)
L2:
  cmpl   $0xc,-0x14(%rbp)
  jle    L8
  mov    $0x0,%eax
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int shdn8(x)
register unsigned short *x;
{
  register unsigned short newbyt, oldbyt;
  int i;
  x += 2;
  oldbyt = 0;
  for (i = 0; i < 14 - 1; i++) {
    newbyt = *x << 8;
    *x >>= 8;
    *x |= oldbyt;
    oldbyt = newbyt;
    ++x;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<shdn8>:
  endbr64
  lea    0x4(%rdi),%rdx
  add    $0x1e,%rdi
  mov    $0x0,%ecx
L6:
  movzwl (%rdx),%eax
  mov    %ecx,%esi
  mov    %eax,%ecx
  shl    $0x8,%ecx
  shr    $0x8,%ax
  or     %esi,%eax
  mov    %ax,(%rdx)
  add    $0x2,%rdx
  cmp    %rdx,%rdi
  jne    L6
  mov    $0x0,%eax
  ret
```

assistant: ```c
int shdn8(x)
register unsigned short *x;
{
  register unsigned short newbyt, oldbyt;
  int i;
  x += 2;
  oldbyt = 0;
  for (i = 0; i < 14 - 1; i++) {
    newbyt = *x << 8;
    *x >>= 8;
    *x |= oldbyt;
    oldbyt = newbyt;
    ++x;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<shdn8>:
  endbr64
  lea    0x4(%rdi),%rdx
  xor    %ecx,%ecx
  add    $0x1e,%rdi
  xchg   %ax,%ax
L4:
  movzwl (%rdx),%eax
  mov    %ecx,%esi
  add    $0x2,%rdx
  mov    %eax,%ecx
  shr    $0x8,%ax
  or     %esi,%eax
  shl    $0x8,%ecx
  mov    %ax,-0x2(%rdx)
  cmp    %rdx,%rdi
  jne    L4
  xor    %eax,%eax
  ret
```

assistant: ```c
int shdn8(x)
register unsigned short *x;
{
  register unsigned short newbyt, oldbyt;
  int i;
  x += 2;
  oldbyt = 0;
  for (i = 0; i < 14 - 1; i++) {
    newbyt = *x << 8;
    *x >>= 8;
    *x |= oldbyt;
    oldbyt = newbyt;
    ++x;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<shdn8>:
  endbr64
  movzwl 0x4(%rdi),%edx
  mov    %edx,%ecx
  shl    $0x8,%edx
  shr    $0x8,%cx
  mov    %cx,0x4(%rdi)
  movzwl 0x6(%rdi),%ecx
  mov    %ecx,%esi
  shl    $0x8,%ecx
  shr    $0x8,%si
  or     %esi,%edx
  mov    %dx,0x6(%rdi)
  movzwl 0x8(%rdi),%edx
  mov    %edx,%esi
  shl    $0x8,%edx
  shr    $0x8,%si
  or     %esi,%ecx
  mov    %cx,0x8(%rdi)
  movzwl 0xa(%rdi),%ecx
  mov    %ecx,%esi
  shl    $0x8,%ecx
  shr    $0x8,%si
  or     %esi,%edx
  mov    %dx,0xa(%rdi)
  movzwl 0xc(%rdi),%edx
  mov    %edx,%esi
  shl    $0x8,%edx
  shr    $0x8,%si
  or     %esi,%ecx
  mov    %cx,0xc(%rdi)
  movzwl 0xe(%rdi),%ecx
  mov    %ecx,%esi
  shl    $0x8,%ecx
  shr    $0x8,%si
  or     %esi,%edx
  mov    %dx,0xe(%rdi)
  movzwl 0x10(%rdi),%edx
  mov    %edx,%esi
  shl    $0x8,%edx
  shr    $0x8,%si
  or     %esi,%ecx
  mov    %cx,0x10(%rdi)
  movzwl 0x12(%rdi),%ecx
  mov    %ecx,%esi
  shl    $0x8,%ecx
  shr    $0x8,%si
  or     %esi,%edx
  mov    %dx,0x12(%rdi)
  movzwl 0x14(%rdi),%edx
  mov    %edx,%esi
  shl    $0x8,%edx
  shr    $0x8,%si
  or     %esi,%ecx
  mov    %cx,0x14(%rdi)
  movzwl 0x16(%rdi),%ecx
  mov    %ecx,%esi
  shl    $0x8,%ecx
  shr    $0x8,%si
  or     %esi,%edx
  mov    %dx,0x16(%rdi)
  movzwl 0x18(%rdi),%edx
  mov    %edx,%esi
  shl    $0x8,%edx
  shr    $0x8,%si
  or     %esi,%ecx
  mov    %cx,0x18(%rdi)
  movzwl 0x1a(%rdi),%ecx
  mov    %ecx,%esi
  shr    $0x8,%si
  shl    $0x8,%ecx
  xor    %eax,%eax
  or     %esi,%edx
  mov    %dx,0x1a(%rdi)
  movzbl 0x1d(%rdi),%edx
  or     %ecx,%edx
  mov    %dx,0x1c(%rdi)
  ret
```

assistant: ```c
int shdn8(x)
register unsigned short *x;
{
  register unsigned short newbyt, oldbyt;
  int i;
  x += 2;
  oldbyt = 0;
  for (i = 0; i < 14 - 1; i++) {
    newbyt = *x << 8;
    *x >>= 8;
    *x |= oldbyt;
    oldbyt = newbyt;
    ++x;
  }
  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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x65747361,-0x1e(%rbp)
  movw   $0x72,-0x1a(%rbp)
  movw   $0x7461,-0x25(%rbp)
  movb   $0x0,-0x23(%rbp)
  movl   $0x657761,-0x22(%rbp)
  lea    -0x22(%rbp),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %eax,%r12d
  lea    -0x25(%rbp),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %eax,%ebx
  lea    -0x1e(%rbp),%rax
  mov    %rax,%rdi
  call   <stringLength@plt>
  mov    %r12d,%ecx
  mov    %ebx,%edx
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
data_type: string

tool: D3: %i %i %i


assistant: ```c
int main(void) {
  int stringLength(const char string[]);
  const char word1[] = {'a', 's', 't', 'e', 'r', '\0'};
  const char word2[] = {'a', 't', '\0'};
  const char word3[] = {'a', 'w', 'e', '\0'};
  printf("%i %i %i\n", stringLength(word1), stringLength(word2), stringLength(word3));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x65747361,0x12(%rsp)
  movw   $0x72,0x16(%rsp)
  movw   $0x7461,0xb(%rsp)
  movb   $0x0,0xd(%rsp)
  movl   $0x657761,0xe(%rsp)
  lea    0xe(%rsp),%rdi
  call   <stringLength@plt>
  mov    %eax,%ebp
  lea    0xb(%rsp),%rdi
  call   <stringLength@plt>
  mov    %eax,%ebx
  lea    0x12(%rsp),%rdi
  call   <stringLength@plt>
  mov    %eax,%esi
  mov    %ebp,%ecx
  mov    %ebx,%edx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_infered: string

tool: D0: %i %i %i


assistant: ```c
int main(void) {
  int stringLength(const char string[]);
  const char word1[] = {'a', 's', 't', 'e', 'r', '\0'};
  const char word2[] = {'a', 't', '\0'};
  const char word3[] = {'a', 'w', 'e', '\0'};
  printf("%i %i %i\n", stringLength(word1), stringLength(word2), stringLength(word3));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x7461,%edx
  push   %r12
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  mov    $0x72,%eax
  mov    %dx,0xb(%rsp)
  movl   $0x65747361,0x12(%rsp)
  mov    %ax,0x16(%rsp)
  movb   $0x0,0xd(%rsp)
  movl   $0x657761,0xe(%rsp)
  call   <stringLength@plt>
  lea    0xb(%rsp),%rdi
  mov    %eax,%r13d
  call   <stringLength@plt>
  lea    0x12(%rsp),%rdi
  mov    %eax,%r12d
  call   <stringLength@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string

tool: D4: %i %i %i


assistant: ```c
int main(void) {
  int stringLength(const char string[]);
  const char word1[] = {'a', 's', 't', 'e', 'r', '\0'};
  const char word2[] = {'a', 't', '\0'};
  const char word3[] = {'a', 'w', 'e', '\0'};
  printf("%i %i %i\n", stringLength(word1), stringLength(word2), stringLength(word3));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x7461,%edx
  push   %r12
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  mov    $0x72,%eax
  mov    %dx,0xb(%rsp)
  movl   $0x65747361,0x12(%rsp)
  mov    %ax,0x16(%rsp)
  movb   $0x0,0xd(%rsp)
  movl   $0x657761,0xe(%rsp)
  call   <stringLength@plt>
  lea    0xb(%rsp),%rdi
  mov    %eax,%r13d
  call   <stringLength@plt>
  lea    0x12(%rsp),%rdi
  mov    %eax,%r12d
  call   <stringLength@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_guessed: string

tool: D4: %i %i %i


assistant: ```c
int main(void) {
  int stringLength(const char string[]);
  const char word1[] = {'a', 's', 't', 'e', 'r', '\0'};
  const char word2[] = {'a', 't', '\0'};
  const char word3[] = {'a', 'w', 'e', '\0'};
  printf("%i %i %i\n", stringLength(word1), stringLength(word2), stringLength(word3));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<f2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L19
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  jmp    L24
L19:
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rsi
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    %rcx,%rdi
  call   <f2>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jl     L6
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0xa(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
L24:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string

tool: D9: %c

assistant: ```c
int f2(char *pcA, char *pcB, int *num) {
  int valore = 0;
  *num = *num + 1;
  if (pcA >= pcB)
    return *num;
  else
    valore = f2(pcA + 1, pcB - 1, num);
  if (*pcA >= *pcB) {
    pcA = pcB;
    *num = *num + 10;
  }
  printf("%c", *pcA);
  return *num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<f2>:
  endbr64
  mov    (%rdx),%eax
  add    $0x1,%eax
  mov    %eax,(%rdx)
  cmp    %rsi,%rdi
  jb     L2
  ret
L2:
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  lea    -0x1(%rsi),%rsi
  lea    0x1(%rdi),%rdi
  call   <f2>
  movzbl 0x0(%rbp),%eax
  cmp    %al,(%r12)
  jl     L15
  addl   $0xa,(%rbx)
  mov    %rbp,%r12
L15:
  movsbl (%r12),%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    (%rbx),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_guessed: string

tool: D5: %c

assistant: ```c
int f2(char *pcA, char *pcB, int *num) {
  int valore = 0;
  *num = *num + 1;
  if (pcA >= pcB)
    return *num;
  else
    valore = f2(pcA + 1, pcB - 1, num);
  if (*pcA >= *pcB) {
    pcA = pcB;
    *num = *num + 10;
  }
  printf("%c", *pcA);
  return *num;
}
```