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

```asm
<KMP_test>:
  endbr64
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movabs $0x4143424141434241,%rax
  mov    $0x0,%edx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  movabs $0x4241414141,%rax
  mov    $0x0,%edx
  mov    %rax,0x64(%rsp)
  mov    %rdx,0x6c(%rsp)
  movq   $0x0,0x74(%rsp)
  movq   $0x0,0x7c(%rsp)
  movq   $0x0,0x84(%rsp)
  movq   $0x0,0x8c(%rsp)
  movq   $0x0,0x94(%rsp)
  movq   $0x0,0x9c(%rsp)
  movq   $0x0,0xa4(%rsp)
  movq   $0x0,0xac(%rsp)
  movq   $0x0,0xb4(%rsp)
  movq   $0x0,0xbc(%rsp)
  movl   $0x0,0xc4(%rsp)
  mov    %rsp,%rdi
  mov    $0x0,%eax
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L34
  lea    D13(%rip),%rdi
  call   <puts@plt>
L6:
  lea    0x64(%rsp),%rdi
  mov    $0x0,%eax
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L39
  lea    D13(%rip),%rdi
  call   <puts@plt>
L20:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  mov    $0x0,%eax
  add    $0xd8,%rsp
  ret
L34:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L6
L39:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L20
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
type: string
name: parse_data
arguments: data_label: D4
type: string

tool: D13: FALSE

tool: D4: TRUE

assistant: ```c
int KMP_test(void) {
  char txt[][100] = {"ABCAABCA", "AAAAB"};
  int n = sizeof(txt) / sizeof(txt[0]);
  for (int i = 0; i < n; i++) {
    if (isRepeat(txt[i]) == 1) {
      printf("TRUE\n");
    } else {
      printf("FALSE\n");
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<KMP_test>:
  endbr64
  sub    $0xd8,%rsp
  xor    %edx,%edx
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  movabs $0x4143424141434241,%rax
  mov    %rsp,%rdi
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  movl   $0x0,0x60(%rsp)
  mov    %rdx,0x6c(%rsp)
  movl   $0x0,0xc4(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  mov    %rax,(%rsp)
  movabs $0x4241414141,%rax
  mov    %rax,0x64(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movups %xmm0,0x74(%rsp)
  movups %xmm0,0x84(%rsp)
  movups %xmm0,0x94(%rsp)
  movups %xmm0,0xa4(%rsp)
  movups %xmm0,0xb4(%rsp)
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L25
  lea    D2(%rip),%rdi
  call   <puts@plt>
L3:
  xor    %eax,%eax
  lea    0x64(%rsp),%rdi
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L32
  lea    D2(%rip),%rdi
  call   <puts@plt>
L48:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  xor    %eax,%eax
  add    $0xd8,%rsp
  ret
  nopl   0x0(%rax)
L25:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  jmp    L3
  cs nopw 0x0(%rax,%rax,1)
L32:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  jmp    L48
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D22
type_guessed: string

tool: D2: FALSE

tool: D22: TRUE

assistant: ```c
int KMP_test(void) {
  char txt[][100] = {"ABCAABCA", "AAAAB"};
  int n = sizeof(txt) / sizeof(txt[0]);
  for (int i = 0; i < n; i++) {
    if (isRepeat(txt[i]) == 1) {
      printf("TRUE\n");
    } else {
      printf("FALSE\n");
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<KMP_test>:
  endbr64
  sub    $0xd8,%rsp
  xor    %edx,%edx
  pxor   %xmm0,%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  movabs $0x4143424141434241,%rax
  mov    %rsp,%rdi
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  movl   $0x0,0x60(%rsp)
  mov    %rdx,0x6c(%rsp)
  movl   $0x0,0xc4(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  mov    %rax,(%rsp)
  movabs $0x4241414141,%rax
  mov    %rax,0x64(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movups %xmm0,0x74(%rsp)
  movups %xmm0,0x84(%rsp)
  movups %xmm0,0x94(%rsp)
  movups %xmm0,0xa4(%rsp)
  movups %xmm0,0xb4(%rsp)
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L22
  lea    D36(%rip),%rdi
  call   <puts@plt>
L39:
  xor    %eax,%eax
  lea    0x64(%rsp),%rdi
  call   <isRepeat@plt>
  cmp    $0x1,%eax
  je     L31
  lea    D36(%rip),%rdi
  call   <puts@plt>
L21:
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0xd8,%rsp
  ret
  nopl   0x0(%rax)
L22:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  jmp    L39
  cs nopw 0x0(%rax,%rax,1)
L31:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  jmp    L21
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D36
type: string
name: disassemble_data
arguments: tag: D37
type: string

tool: D36: FALSE

tool: D37: TRUE

assistant: ```c
int KMP_test(void) {
  char txt[][100] = {"ABCAABCA", "AAAAB"};
  int n = sizeof(txt) / sizeof(txt[0]);
  for (int i = 0; i < n; i++) {
    if (isRepeat(txt[i]) == 1) {
      printf("TRUE\n");
    } else {
      printf("FALSE\n");
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<inssort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L3
L38:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L58
L18:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x20(%rbp),%eax
  cltq
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <my_strcmp@plt>
  test   %rax,%rax
  jle    L8
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  subl   $0x1,-0x4(%rbp)
L58:
  cmpl   $0x0,-0x4(%rbp)
  jg     L18
  jmp    L47
L8:
  nop
L47:
  addl   $0x1,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L38
  nop
  nop
  leave
  ret
```

assistant: ```c
void inssort(char *a[], int n, int depth) {
  int i, j;
  for (i = 1; i < n; i++)
    for (j = i; j > 0; j--) {
      if (my_strcmp(a[j - 1] + depth, a[j] + depth) <= 0)
        break;
      swap(a, j, j - 1);
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inssort>:
  endbr64
  cmp    $0x1,%esi
  jle    L55
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r15d
  mov    $0x1,%r14d
  movslq %edx,%r12
  jmp    L59
L1:
  add    $0x1,%r14
  cmp    %r15,%r14
  je     L63
L59:
  mov    %r14,%rbx
  test   %r14d,%r14d
  jle    L1
L15:
  mov    %r12,%rsi
  add    0x0(%rbp,%rbx,8),%rsi
  mov    %r12,%rdi
  add    -0x8(%rbp,%rbx,8),%rdi
  call   <my_strcmp@plt>
  test   %rax,%rax
  jle    L1
  lea    -0x1(%rbx),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <swap@plt>
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jg     L15
  jmp    L1
L63:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  ret
```

assistant: ```c
void inssort(char *a[], int n, int depth) {
  int i, j;
  for (i = 1; i < n; i++)
    for (j = i; j > 0; j--) {
      if (my_strcmp(a[j - 1] + depth, a[j] + depth) <= 0)
        break;
      swap(a, j, j - 1);
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inssort>:
  endbr64
  cmp    $0x1,%esi
  jle    L15
  push   %r15
  push   %r14
  movslq %esi,%r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    %r13,%rbx
  nopl   0x0(%rax,%rax,1)
L30:
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    -0x8(%rbp,%rbx,8),%rdi
  add    %r12,%rsi
  add    %r12,%rdi
  call   <my_strcmp@plt>
  test   %rax,%rax
  jle    L26
  lea    -0x1(%rbx),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  call   <swap@plt>
  test   %ebx,%ebx
  jne    L30
L26:
  add    $0x1,%r13
  cmp    %r13,%r14
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  ret
```

assistant: ```c
void inssort(char *a[], int n, int depth) {
  int i, j;
  for (i = 1; i < n; i++)
    for (j = i; j > 0; j--) {
      if (my_strcmp(a[j - 1] + depth, a[j] + depth) <= 0)
        break;
      swap(a, j, j - 1);
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<inssort>:
  endbr64
  cmp    $0x1,%esi
  jle    L31
  push   %r15
  push   %r14
  movslq %esi,%r14
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L35:
  mov    %r13,%rbx
  nopl   0x0(%rax,%rax,1)
L12:
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    -0x8(%rbp,%rbx,8),%rdi
  add    %r12,%rsi
  add    %r12,%rdi
  call   <my_strcmp@plt>
  test   %rax,%rax
  jle    L26
  lea    -0x1(%rbx),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  sub    $0x1,%rbx
  call   <swap@plt>
  test   %ebx,%ebx
  jne    L12
L26:
  add    $0x1,%r13
  cmp    %r13,%r14
  jne    L35
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L31:
  ret
```

assistant: ```c
void inssort(char *a[], int n, int depth) {
  int i, j;
  for (i = 1; i < n; i++)
    for (j = i; j > 0; j--) {
      if (my_strcmp(a[j - 1] + depth, a[j] + depth) <= 0)
        break;
      swap(a, j, j - 1);
    }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<readDevices>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <readDevs@plt>
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string

tool: D4: r

assistant: ```c
struct device **readDevices(char *fn) {
  FILE *f;
  f = fopen(fn, "r");
  return readDevs(f);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readDevices>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rdi
  call   <readDevs@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
data_type: string

tool: D2: r

assistant: ```c
struct device **readDevices(char *fn) {
  FILE *f;
  f = fopen(fn, "r");
  return readDevs(f);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readDevices>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rsi
  call   <fopen@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1070 <readDevs@plt>
```

tool_calls: name: read_data
arguments: data_tag: D3
datatype: string

tool: D3: r

assistant: ```c
struct device **readDevices(char *fn) {
  FILE *f;
  f = fopen(fn, "r");
  return readDevs(f);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readDevices>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rsi
  call   <fopen@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1070 <readDevs@plt>
```

tool_calls: name: unpack_data
arguments: label: D5
type: string

tool: D5: r

assistant: ```c
struct device **readDevices(char *fn) {
  FILE *f;
  f = fopen(fn, "r");
  return readDevs(f);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getNumBits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L5:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L10
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
  jmp    L3
L10:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
L3:
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L5
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *getNumBits(int *numInputs, int n) {
  int *numBits = (int *)malloc(n * sizeof(int));
  int i;
  for (i = 0; i < n; i++) {
    if (numInputs[i] > 0) {
      numBits[i] = (int)pow(2, (double)numInputs[i]);
    } else {
      numBits[i] = 1;
    }
  }
  return (numBits);
}
```
user: Translate the below assembly code into C programming language:

```asm
<getNumBits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %ebx,%ebx
  jle    L10
  mov    %ebx,%ebx
  lea    (%rax,%rbx,4),%r12
  mov    %rax,%rbx
  sub    %rax,%rbp
  jmp    L9
L18:
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,(%rbx)
L33:
  add    $0x4,%rbx
  cmp    %r12,%rbx
  je     L10
L9:
  mov    0x0(%rbp,%rbx,1),%eax
  test   %eax,%eax
  jg     L18
  movl   $0x1,(%rbx)
  jmp    L33
L10:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *getNumBits(int *numInputs, int n) {
  int *numBits = (int *)malloc(n * sizeof(int));
  int i;
  for (i = 0; i < n; i++) {
    if (numInputs[i] > 0) {
      numBits[i] = (int)pow(2, (double)numInputs[i]);
    } else {
      numBits[i] = 1;
    }
  }
  return (numBits);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getNumBits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  lea    0x0(,%rbx,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  jle    L16
  lea    (%rax,%rbx,4),%r13
  sub    %rax,%rbp
  mov    %rax,%rbx
L19:
  mov    0x0(%rbp,%rbx,1),%edx
  mov    $0x1,%eax
  test   %edx,%edx
  jle    L29
  pxor   %xmm0,%xmm0
  mov    $0x2,%edi
  cvtsi2sd %edx,%xmm0
  call   <pow@plt>
L29:
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L19
L16:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *getNumBits(int *numInputs, int n) {
  int *numBits = (int *)malloc(n * sizeof(int));
  int i;
  for (i = 0; i < n; i++) {
    if (numInputs[i] > 0) {
      numBits[i] = (int)pow(2, (double)numInputs[i]);
    } else {
      numBits[i] = 1;
    }
  }
  return (numBits);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getNumBits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  lea    0x0(,%rbx,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %ebx,%ebx
  jle    L33
  lea    (%rax,%rbx,4),%r12
  sub    %rax,%rbp
  mov    %rax,%rbx
L29:
  mov    0x0(%rbp,%rbx,1),%eax
  test   %eax,%eax
  jg     L23
  movl   $0x1,(%rbx)
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L29
L33:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L23:
  pxor   %xmm0,%xmm0
  mov    $0x2,%edi
  add    $0x4,%rbx
  cvtsi2sd %eax,%xmm0
  call   <pow@plt>
  mov    %eax,-0x4(%rbx)
  cmp    %r12,%rbx
  jne    L29
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *getNumBits(int *numInputs, int n) {
  int *numBits = (int *)malloc(n * sizeof(int));
  int i;
  for (i = 0; i < n; i++) {
    if (numInputs[i] > 0) {
      numBits[i] = (int)pow(2, (double)numInputs[i]);
    } else {
      numBits[i] = 1;
    }
  }
  return (numBits);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lrc_debug_sources_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L17:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void lrc_debug_sources_(int n, int8_t *source) {
  ;
  for (int i = 0; i < n; i++) {
    int8_t e = source[i];
    if (e == 0) {
      ;
    } else {
      ;
    }
  };
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lrc_debug_sources_>:
  endbr64
  test   %edi,%edi
  jle    L1
  mov    $0x0,%eax
L11:
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L11
L1:
  ret
```

assistant: ```c
void lrc_debug_sources_(int n, int8_t *source) {
  ;
  for (int i = 0; i < n; i++) {
    int8_t e = source[i];
    if (e == 0) {
      ;
    } else {
      ;
    }
  };
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lrc_debug_sources_>:
  endbr64
  ret
```

assistant: ```c
void lrc_debug_sources_(int n, int8_t *source) {
  ;
  for (int i = 0; i < n; i++) {
    int8_t e = source[i];
    if (e == 0) {
      ;
    } else {
      ;
    }
  };
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<append_msg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L18
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L18:
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  je     L15
  mov    -0xc8(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
L15:
  mov    -0xd0(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  movslq %eax,%rdx
  mov    -0xc8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xb4(%rbp),%edx
  mov    -0xd0(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <vsprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D3
type: string

tool: D3: , 

assistant: ```c
void append_msg(char *buf, char *format, ...) {
  va_list ap;
  if (strlen(buf))
    strcat(buf, ", ");
  va_start(ap, format);
  vsprintf(buf + strlen(buf), format, ap);
  va_end(ap);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<append_msg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  test   %eax,%eax
  jne    L10
L4:
  mov    %rbp,%rsi
  mov    $0x0,%r12d
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  lea    (%rbx,%rax,1),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <vsprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  jmp    L4
```

tool_calls: name: unpack_data
arguments: tag: D1
type: string

tool: D1: , 

assistant: ```c
void append_msg(char *buf, char *format, ...) {
  va_list ap;
  if (strlen(buf))
    strcat(buf, ", ");
  va_start(ap, format);
  vsprintf(buf + strlen(buf), format, ap);
  va_end(ap);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<append_msg>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jne    L7
L8:
  xor    %r12d,%r12d
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  cltq
  lea    (%rbx,%rax,1),%rdi
  call   <vsprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  jmp    L8
```

tool_calls: name: disassemble_data
arguments: label: D0
datatype: string

tool: D0: , 

assistant: ```c
void append_msg(char *buf, char *format, ...) {
  va_list ap;
  if (strlen(buf))
    strcat(buf, ", ");
  va_start(ap, format);
  vsprintf(buf + strlen(buf), format, ap);
  va_end(ap);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<append_msg>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jne    L18
L8:
  xor    %r12d,%r12d
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  cltq
  lea    (%rbx,%rax,1),%rdi
  call   <vsprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L18:
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  jmp    L8
```

tool_calls: name: read_data
arguments: data_label: D5
data_type: string

tool: D5: , 

assistant: ```c
void append_msg(char *buf, char *format, ...) {
  va_list ap;
  if (strlen(buf))
    strcat(buf, ", ");
  va_start(ap, format);
  vsprintf(buf + strlen(buf), format, ap);
  va_end(ap);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mdbm_strlcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L8
  mov    -0x28(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t mdbm_strlcpy(char *dst, const char *src, size_t dstlen) {
  size_t srclen = strlen(src);
  if (srclen >= dstlen) {
    srclen = dstlen - 1;
  }
  memcpy(dst, src, srclen);
  dst[srclen] = 0;
  return srclen;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mdbm_strlcpy>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %rsi,%rdi
  call   <strlen@plt>
  lea    -0x1(%r13),%rbx
  cmp    %r13,%rax
  cmovb  %rax,%rbx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  movb   $0x0,0x0(%rbp,%rbx,1)
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
size_t mdbm_strlcpy(char *dst, const char *src, size_t dstlen) {
  size_t srclen = strlen(src);
  if (srclen >= dstlen) {
    srclen = dstlen - 1;
  }
  memcpy(dst, src, srclen);
  dst[srclen] = 0;
  return srclen;
}
```