ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fn_al_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121>:
  endbr64
  jmp    1050 <fn_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121@plt>
```

assistant: ```c
int fn_al_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121(const char *text) {
  int pos = 0;
  pos = fn_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121(text);
  if (pos != -1)
    return pos;
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fn_al_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121>:
  endbr64
  jmp    1050 <fn_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121@plt>
```

assistant: ```c
int fn_al_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121(const char *text) {
  int pos = 0;
  pos = fn_sq_ch_79_ch_98_ch_106_ch_101_ch_99_ch_116_ch_69_ch_120_ch_97_ch_99_ch_116_ch_67_ch_97_ch_114_ch_100_ch_105_ch_110_ch_97_ch_108_ch_105_ch_116_ch_121(text);
  if (pos != -1)
    return pos;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<filter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L1
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    $0x2e,%esi
  mov    %eax,%edi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L1
  mov    -0x8(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L1
  mov    $0x1,%eax
  jmp    L37
L1:
  mov    $0x0,%eax
L37:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D6: .jar

assistant: ```c
int filter(const struct dirent *dir) {
  if (dir) {
    char *last = strrchr(dir->d_name, '.');
    if (last) {
      if (!strcmp(last, ".jar")) {
        return 1;
      }
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<filter>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L12
  sub    $0x8,%rsp
  mov    $0x2e,%esi
  mov    (%rdi),%edi
  call   <strrchr@plt>
  mov    %rax,%rdi
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L6
  lea    D4(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L6:
  add    $0x8,%rsp
  ret
L12:
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: string

tool: D4: .jar

assistant: ```c
int filter(const struct dirent *dir) {
  if (dir) {
    char *last = strrchr(dir->d_name, '.');
    if (last) {
      if (!strcmp(last, ".jar")) {
        return 1;
      }
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<filter>:
  endbr64
  test   %rdi,%rdi
  je     L5
  sub    $0x8,%rsp
  mov    (%rdi),%edi
  mov    $0x2e,%esi
  call   <strrchr@plt>
  mov    %rax,%rdi
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L8
  lea    D1(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L8:
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  ret
```

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

tool: D1: .jar

assistant: ```c
int filter(const struct dirent *dir) {
  if (dir) {
    char *last = strrchr(dir->d_name, '.');
    if (last) {
      if (!strcmp(last, ".jar")) {
        return 1;
      }
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<filter>:
  endbr64
  test   %rdi,%rdi
  je     L5
  sub    $0x8,%rsp
  mov    (%rdi),%edi
  mov    $0x2e,%esi
  call   <strrchr@plt>
  mov    %rax,%rdi
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L8
  lea    D6(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L8:
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax)
L5:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
data_type: string

tool: D6: .jar

assistant: ```c
int filter(const struct dirent *dir) {
  if (dir) {
    char *last = strrchr(dir->d_name, '.');
    if (last) {
      if (!strcmp(last, ".jar")) {
        return 1;
      }
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<does_nameHaveUppers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L25
L35:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L3
  mov    $0x1,%eax
  jmp    L34
L3:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L35
  mov    $0x0,%eax
L34:
  leave
  ret
```

assistant: ```c
int does_nameHaveUppers(char *suspect) {
  int i;
  for (i = 0; i < strlen(suspect); i++) {
    if (isupper(suspect[i]))
      return (1);
  }
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<does_nameHaveUppers>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
L20:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L28
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isupper@plt>
  add    $0x1,%rbx
  test   %rax,%rax
  je     L20
  mov    $0x1,%eax
  jmp    L11
L28:
  mov    $0x0,%eax
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int does_nameHaveUppers(char *suspect) {
  int i;
  for (i = 0; i < strlen(suspect); i++) {
    if (isupper(suspect[i]))
      return (1);
  }
  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
<does_nameHaveUppers>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L8
  nopl   0x0(%rax)
L0:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isupper@plt>
  test   %rax,%rax
  jne    L20
L8:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L0
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L20:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int does_nameHaveUppers(char *suspect) {
  int i;
  for (i = 0; i < strlen(suspect); i++) {
    if (isupper(suspect[i]))
      return (1);
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<does_nameHaveUppers>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L15
  nopl   0x0(%rax)
L23:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isupper@plt>
  test   %rax,%rax
  jne    L0
L15:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L23
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int does_nameHaveUppers(char *suspect) {
  int i;
  for (i = 0; i < strlen(suspect); i++) {
    if (isupper(suspect[i]))
      return (1);
  }
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x3c(%rbp)
  movl   $0x0,-0x40(%rbp)
  jmp    L49
L27:
  lea    -0x30(%rbp),%rdx
  mov    -0x40(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x40(%rbp),%eax
  cltq
  movss  -0x30(%rbp,%rax,4),%xmm0
  movss  -0x3c(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x3c(%rbp)
  addl   $0x1,-0x40(%rbp)
L49:
  cmpl   $0x9,-0x40(%rbp)
  jle    L27
  movss  -0x3c(%rbp),%xmm0
  movss  D27(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x38(%rbp)
  movss  -0x38(%rbp),%xmm0
  cvttss2si %xmm0,%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x40(%rbp)
  jmp    L41
L17:
  mov    -0x40(%rbp),%eax
  cltq
  movss  -0x30(%rbp,%rax,4),%xmm0
  movaps %xmm0,%xmm1
  subss  -0x38(%rbp),%xmm1
  mov    -0x40(%rbp),%eax
  cltq
  movss  -0x30(%rbp,%rax,4),%xmm0
  subss  -0x38(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x34(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x34(%rbp),%xmm2
  movq   %xmm2,%rax
  mov    -0x40(%rbp),%edx
  add    $0x1,%edx
  movq   %rax,%xmm0
  mov    %edx,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x40(%rbp)
L41:
  cmpl   $0x9,-0x40(%rbp)
  jle    L17
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D22
datatype: string
name: read_data
arguments: tag: D27
datatype: f32
name: read_data
arguments: tag: D19
datatype: string
name: read_data
arguments: tag: D20
datatype: string

tool: D22: %f

tool: D27: 10

tool: D19: Average = %.2f


tool: D20: Distributed[%d] = %.2f


assistant: ```c
int main() {
  int i;
  float ave, dis, sum = 0.0, data[10];
  for (i = 0; i < 10; i++) {
    scanf("%f", &data[i]);
    sum += data[i];
  }
  ave = sum / 10.0;
  printf("Average = %.2f\n", ave);
  for (i = 0; i < 10; i++) {
    dis = (data[i] - ave) * (data[i] - ave);
    printf("Distributed[%d] = %.2f\n", i + 1, dis);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0x38(%rsp),%r13
  mov    D37(%rip),%ebp
  lea    D32(%rip),%r12
L23:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movd   %ebp,%xmm1
  addss  (%rbx),%xmm1
  movd   %xmm1,%ebp
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L23
  movaps %xmm1,%xmm3
  divss  D16(%rip),%xmm3
  movss  %xmm3,0xc(%rsp)
  cvttss2si %xmm3,%esi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebx
  lea    D13(%rip),%rbp
L27:
  movss  0xc(%rsp,%rbx,4),%xmm0
  subss  0xc(%rsp),%xmm0
  mulss  %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0xb,%rbx
  jne    L27
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D37
type: f32
name: parse_data
arguments: tag: D32
type: string
name: parse_data
arguments: tag: D16
type: f32
name: parse_data
arguments: tag: D24
type: string
name: parse_data
arguments: tag: D13
type: string

tool: D37: 0

tool: D32: %f

tool: D16: 10

tool: D24: Average = %.2f


tool: D13: Distributed[%d] = %.2f


assistant: ```c
int main() {
  int i;
  float ave, dis, sum = 0.0, data[10];
  for (i = 0; i < 10; i++) {
    scanf("%f", &data[i]);
    sum += data[i];
  }
  ave = sum / 10.0;
  printf("Average = %.2f\n", ave);
  for (i = 0; i < 10; i++) {
    dis = (data[i] - ave) * (data[i] - ave);
    printf("Distributed[%d] = %.2f\n", i + 1, dis);
  }
  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   %r13
  pxor   %xmm0,%xmm0
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x38(%rsp),%r13
  mov    %rbp,%rbx
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  movss  %xmm0,0xc(%rsp)
  add    $0x4,%rbx
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  addss  -0x4(%rbx),%xmm0
  cmp    %r13,%rbx
  jne    L27
  divss  D5(%rip),%xmm0
  cvttss2si %xmm0,%esi
  xor    %eax,%eax
  lea    D29(%rip),%rdi
  movss  %xmm0,0xc(%rsp)
  mov    $0x1,%ebx
  lea    D0(%rip),%r12
  call   <printf@plt>
  nopl   0x0(%rax)
L18:
  movss  -0x4(%rbp,%rbx,4),%xmm0
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x1,%eax
  subss  0xc(%rsp),%xmm0
  add    $0x1,%rbx
  mulss  %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0xb,%rbx
  jne    L18
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D13
data_type: string
name: read_data
arguments: data_tag: D5
data_type: f32
name: read_data
arguments: data_tag: D29
data_type: string
name: read_data
arguments: data_tag: D0
data_type: string

tool: D13: %f

tool: D5: 10

tool: D29: Average = %.2f


tool: D0: Distributed[%d] = %.2f


assistant: ```c
int main() {
  int i;
  float ave, dis, sum = 0.0, data[10];
  for (i = 0; i < 10; i++) {
    scanf("%f", &data[i]);
    sum += data[i];
  }
  ave = sum / 10.0;
  printf("Average = %.2f\n", ave);
  for (i = 0; i < 10; i++) {
    dis = (data[i] - ave) * (data[i] - ave);
    printf("Distributed[%d] = %.2f\n", i + 1, dis);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0x38(%rsp),%r13
  mov    %rbp,%rbx
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  movss  %xmm0,0xc(%rsp)
  add    $0x4,%rbx
  call   <scanf@plt>
  movss  0xc(%rsp),%xmm0
  addss  -0x4(%rbx),%xmm0
  cmp    %r13,%rbx
  jne    L14
  divss  D28(%rip),%xmm0
  cvttss2si %xmm0,%esi
  xor    %eax,%eax
  lea    D36(%rip),%rdi
  movss  %xmm0,0xc(%rsp)
  mov    $0x1,%ebx
  lea    D25(%rip),%r12
  call   <printf@plt>
  nopl   0x0(%rax)
L21:
  movss  -0x4(%rbp,%rbx,4),%xmm0
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x1,%eax
  subss  0xc(%rsp),%xmm0
  add    $0x1,%rbx
  mulss  %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    $0xb,%rbx
  jne    L21
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string
name: unpack_data
arguments: tag: D28
type: f32
name: unpack_data
arguments: tag: D36
type: string
name: unpack_data
arguments: tag: D25
type: string

tool: D2: %f

tool: D28: 10

tool: D36: Average = %.2f


tool: D25: Distributed[%d] = %.2f


assistant: ```c
int main() {
  int i;
  float ave, dis, sum = 0.0, data[10];
  for (i = 0; i < 10; i++) {
    scanf("%f", &data[i]);
    sum += data[i];
  }
  ave = sum / 10.0;
  printf("Average = %.2f\n", ave);
  for (i = 0; i < 10; i++) {
    dis = (data[i] - ave) * (data[i] - ave);
    printf("Distributed[%d] = %.2f\n", i + 1, dis);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<malloc_pasl_report>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <malloc_count_total@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <malloc_count_peak@plt>
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <malloc_count_current@plt>
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
type: string
name: retrieve_data
arguments: tag: D13
type: string
name: retrieve_data
arguments: tag: D17
type: string

tool: D5: malloc_total	%lld


tool: D13: malloc_peak	%lld


tool: D17: malloc_current	%lld


assistant: ```c
extern void malloc_pasl_report(void) {
  printf("malloc_total\t%lld\n", malloc_count_total());
  printf("malloc_peak\t%lld\n", malloc_count_peak());
  printf("malloc_current\t%lld\n", malloc_count_current());
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<malloc_pasl_report>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <malloc_count_total@plt>
  mov    %eax,%esi
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <malloc_count_peak@plt>
  mov    %eax,%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <malloc_count_current@plt>
  mov    %eax,%esi
  lea    D22(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D29
data_type: string
name: parse_data
arguments: tag: D14
data_type: string
name: parse_data
arguments: tag: D22
data_type: string

tool: D29: malloc_total	%lld


tool: D14: malloc_peak	%lld


tool: D22: malloc_current	%lld


assistant: ```c
extern void malloc_pasl_report(void) {
  printf("malloc_total\t%lld\n", malloc_count_total());
  printf("malloc_peak\t%lld\n", malloc_count_peak());
  printf("malloc_current\t%lld\n", malloc_count_current());
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<malloc_pasl_report>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <malloc_count_total@plt>
  lea    D9(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <malloc_count_peak@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <malloc_count_current@plt>
  lea    D1(%rip),%rdi
  add    $0x8,%rsp
  mov    %eax,%esi
  jmp    10a0 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
type_infered: string
name: retrieve_data
arguments: label: D7
type_infered: string
name: retrieve_data
arguments: label: D1
type_infered: string

tool: D9: malloc_total	%lld


tool: D7: malloc_peak	%lld


tool: D1: malloc_current	%lld


assistant: ```c
extern void malloc_pasl_report(void) {
  printf("malloc_total\t%lld\n", malloc_count_total());
  printf("malloc_peak\t%lld\n", malloc_count_peak());
  printf("malloc_current\t%lld\n", malloc_count_current());
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<malloc_pasl_report>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <malloc_count_total@plt>
  lea    D14(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <malloc_count_peak@plt>
  lea    D15(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  call   <malloc_count_current@plt>
  lea    D2(%rip),%rdi
  add    $0x8,%rsp
  mov    %eax,%esi
  jmp    10a0 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D14
type_infered: string
name: read_data
arguments: label: D15
type_infered: string
name: read_data
arguments: label: D2
type_infered: string

tool: D14: malloc_total	%lld


tool: D15: malloc_peak	%lld


tool: D2: malloc_current	%lld


assistant: ```c
extern void malloc_pasl_report(void) {
  printf("malloc_total\t%lld\n", malloc_count_total());
  printf("malloc_peak\t%lld\n", malloc_count_peak());
  printf("malloc_current\t%lld\n", malloc_count_current());
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gao>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xc,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
type_guessed: string

tool: D8: %d


assistant: ```c
int gao() {
  int i = 12;
  char buffer[4] = {0};
  gets(buffer);
  printf("%d\n", i);
  printf("%d\n", (int)strlen(buffer));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<gao>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    $0xc,%esi
  lea    D2(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D2
datatype: string

tool: D2: %d


assistant: ```c
int gao() {
  int i = 12;
  char buffer[4] = {0};
  gets(buffer);
  printf("%d\n", i);
  printf("%d\n", (int)strlen(buffer));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gao>:
  endbr64
  push   %r12
  push   %rbp
  lea    D6(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  movl   $0x0,0x4(%rsp)
  mov    %r12,%rdi
  call   <gets@plt>
  mov    $0xc,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
type: string

tool: D6: %d


assistant: ```c
int gao() {
  int i = 12;
  char buffer[4] = {0};
  gets(buffer);
  printf("%d\n", i);
  printf("%d\n", (int)strlen(buffer));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gao>:
  endbr64
  push   %r12
  push   %rbp
  lea    D7(%rip),%rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  movl   $0x0,0x4(%rsp)
  mov    %r12,%rdi
  call   <gets@plt>
  mov    $0xc,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
data_type: string

tool: D7: %d


assistant: ```c
int gao() {
  int i = 12;
  char buffer[4] = {0};
  gets(buffer);
  printf("%d\n", i);
  printf("%d\n", (int)strlen(buffer));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    -0x10(%rbp),%rbx
  jmp    L16
L15:
  add    $0x1,%rbx
L16:
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L15
  mov    %rbx,%rax
  sub    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  register const char *s;
  for (s = str; *s; ++s)
    ;
  return (s - str);
}
```
user: What is the c source code of the assembly code below?

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L14
  mov    %rdi,%rax
L3:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L3
L4:
  sub    %rdi,%rax
  ret
L14:
  mov    %rdi,%rax
  jmp    L4
```

assistant: ```c
size_t strlen(const char *str) {
  register const char *s;
  for (s = str; *s; ++s)
    ;
  return (s - str);
}
```