ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<bench_tanh>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  mov    $0x8,%esi
  push   %r14
  mov    %rcx,%r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  imul   %edx,%r12d
  mov    %edi,%ebp
  push   %rbx
  mov    %r12d,%edi
  sub    $0x18,%rsp
  call   <calloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <get_time@plt>
  movsd  %xmm0,(%rsp)
  test   %ebp,%ebp
  jle    L8
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r13,%rdx
  mov    %r14,%rsi
  mov    %r12d,%edi
  add    $0x1,%ebx
  call   <morpheus_tanh@plt>
  cmp    %ebx,%ebp
  jne    L13
L8:
  xor    %eax,%eax
  call   <get_time@plt>
  mov    %r13,%rdi
  movsd  %xmm0,0x8(%rsp)
  call   <free@plt>
  pxor   %xmm1,%xmm1
  mov    %r15,%rsi
  movsd  0x8(%rsp),%xmm0
  cvtsi2sd %ebp,%xmm1
  subsd  (%rsp),%xmm0
  add    $0x18,%rsp
  pop    %rbx
  lea    D4(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  divsd  %xmm1,%xmm0
  jmp    10b0 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
data_type: string

tool: D4: Benchmark %s: %f ms


assistant: ```c
void bench_tanh(int n, const char *message, int size, double *x) {
  double *y = calloc(size * size, sizeof(double));
  double start = get_time();
  for (int i = 0; i < n; ++i) {
    morpheus_tanh(size * size, x, y);
  }
  double finish = get_time();
  free(y);
  double ms = (finish - start) / n;
  printf("Benchmark %s: %f ms\n", message, ms);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x64(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L23
L12:
  addl   $0x1,-0x64(%rbp)
L23:
  mov    -0x64(%rbp),%eax
  cltq
  movzbl -0x60(%rbp,%rax,1),%eax
  test   %al,%al
  je     L40
  cmpl   $0x27,-0x64(%rbp)
  jle    L12
L40:
  mov    -0x64(%rbp),%eax
  cltq
  movb   $0x0,-0x30(%rbp,%rax,1)
  mov    -0x64(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x68(%rbp)
  movl   $0x0,-0x6c(%rbp)
  jmp    L49
L6:
  mov    -0x6c(%rbp),%eax
  cltq
  movzbl -0x60(%rbp,%rax,1),%edx
  mov    -0x68(%rbp),%eax
  cltq
  mov    %dl,-0x30(%rbp,%rax,1)
  addl   $0x1,-0x6c(%rbp)
  subl   $0x1,-0x68(%rbp)
L49:
  mov    -0x6c(%rbp),%eax
  cmp    -0x64(%rbp),%eax
  jl     L6
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
data_type: string
name: extract_symbolic_data
arguments: tag: D10
data_type: string
name: extract_symbolic_data
arguments: tag: D0
data_type: string

tool: D18: Digite a string :

tool: D10: %s

tool: D0: Inversa : %s


assistant: ```c
int main() {
  char st[40], stInv[40];
  int i, j, tam = 0;
  printf("Digite a string :");
  scanf("%s", st);
  while (st[tam] != '\0' && tam < 40) {
    tam++;
  }
  stInv[tam] = '\0';
  j = tam - 1;
  i = 0;
  while (i < tam) {
    stInv[j] = st[i];
    i++;
    j--;
  }
  printf("Inversa : %s\n", stInv);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x0,(%rsp)
  je     L24
  lea    0x1(%rsp),%rdx
  mov    $0x0,%eax
L59:
  mov    %eax,%edi
  add    $0x1,%eax
  cmpb   $0x0,(%rdx)
  setne  %sil
  cmp    $0x27,%eax
  setle  %cl
  add    $0x1,%rdx
  test   %cl,%sil
  jne    L59
  movslq %eax,%rdx
  movb   $0x0,0x30(%rsp,%rdx,1)
  test   %eax,%eax
  jle    L58
  mov    %rsp,%rsi
  movslq %edi,%rax
  lea    0x30(%rsp,%rax,1),%rcx
  mov    %rsi,%rax
L1:
  movzbl (%rax),%edx
  mov    %dl,(%rcx)
  add    $0x1,%rax
  sub    $0x1,%rcx
  mov    %eax,%edx
  sub    %esi,%edx
  sub    $0x1,%edx
  cmp    %edx,%edi
  jg     L1
L58:
  lea    0x30(%rsp),%rsi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x68,%rsp
  ret
L24:
  movb   $0x0,0x30(%rsp)
  jmp    L58
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D24
type: string
name: retrieve_data
arguments: data_label: D1
type: string
name: retrieve_data
arguments: data_label: D26
type: string

tool: D24: Digite a string :

tool: D1: %s

tool: D26: Inversa : %s


assistant: ```c
int main() {
  char st[40], stInv[40];
  int i, j, tam = 0;
  printf("Digite a string :");
  scanf("%s", st);
  while (st[tam] != '\0' && tam < 40) {
    tam++;
  }
  stInv[tam] = '\0';
  j = tam - 1;
  i = 0;
  while (i < tam) {
    stInv[j] = st[i];
    i++;
    j--;
  }
  printf("Inversa : %s\n", stInv);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D27(%rip),%rdi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  movzbl (%rsp),%esi
  test   %sil,%sil
  je     L46
  lea    0x1(%rsp),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rdx
  movslq %eax,%rcx
  add    $0x1,%eax
  cmpb   $0x0,-0x1(%rdx)
  jne    L21
  cltq
  lea    0x30(%rsp),%r8
  movb   $0x0,0x30(%rsp,%rax,1)
  lea    (%r8,%rcx,1),%rdx
  mov    %rbx,%rax
  add    %rbx,%rcx
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L55:
  movzbl 0x1(%rax),%esi
  add    $0x1,%rax
L26:
  mov    %sil,(%rdx)
  sub    $0x1,%rdx
  cmp    %rcx,%rax
  jne    L55
L29:
  xor    %eax,%eax
  mov    %r8,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L46:
  movb   $0x0,0x30(%rsp)
  lea    0x30(%rsp),%r8
  jmp    L29
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D27
type: string
name: parse_data
arguments: data_label: D2
type: string
name: parse_data
arguments: data_label: D0
type: string

tool: D27: Digite a string :

tool: D2: %s

tool: D0: Inversa : %s


assistant: ```c
int main() {
  char st[40], stInv[40];
  int i, j, tam = 0;
  printf("Digite a string :");
  scanf("%s", st);
  while (st[tam] != '\0' && tam < 40) {
    tam++;
  }
  stInv[tam] = '\0';
  j = tam - 1;
  i = 0;
  while (i < tam) {
    stInv[j] = st[i];
    i++;
    j--;
  }
  printf("Inversa : %s\n", stInv);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D23(%rip),%rdi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x0,(%rsp)
  je     L47
  lea    0x1(%rsp),%rcx
  xor    %edx,%edx
  nopl   0x0(%rax)
L1:
  add    $0x1,%rcx
  movslq %edx,%rax
  add    $0x1,%edx
  cmpb   $0x0,-0x1(%rcx)
  jne    L1
  movslq %edx,%rdx
  lea    0x30(%rsp),%r8
  lea    0x2f(%rsp),%rsi
  movb   $0x0,0x30(%rsp,%rdx,1)
  add    %r8,%rax
  mov    %rbx,%rdx
L14:
  movzbl (%rdx),%ecx
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,0x1(%rax)
  cmp    %rsi,%rax
  jne    L14
L4:
  xor    %eax,%eax
  mov    %r8,%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L47:
  movb   $0x0,0x30(%rsp)
  lea    0x30(%rsp),%r8
  jmp    L4
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D23
type: string
name: parse_data
arguments: data_label: D6
type: string
name: parse_data
arguments: data_label: D19
type: string

tool: D23: Digite a string :

tool: D6: %s

tool: D19: Inversa : %s


assistant: ```c
int main() {
  char st[40], stInv[40];
  int i, j, tam = 0;
  printf("Digite a string :");
  scanf("%s", st);
  while (st[tam] != '\0' && tam < 40) {
    tam++;
  }
  stInv[tam] = '\0';
  j = tam - 1;
  i = 0;
  while (i < tam) {
    stInv[j] = st[i];
    i++;
    j--;
  }
  printf("Inversa : %s\n", stInv);
}
```
user: What is the c source code of the assembly code below:

```asm
<madd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L0:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L4:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void madd(float *A, float *B, float *C, int rows, int cols) {
  int N, i;
  N = rows * cols;
#pragma omp parallel for
  for (i = 0; i < N; ++i) {
    C[i] = A[i] + B[i];
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<madd>:
  endbr64
  imul   %r8d,%ecx
  test   %ecx,%ecx
  jle    L19
  mov    %ecx,%ecx
  mov    $0x0,%eax
L1:
  movss  (%rdi,%rax,4),%xmm0
  addss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L1
L19:
  ret
```

assistant: ```c
void madd(float *A, float *B, float *C, int rows, int cols) {
  int N, i;
  N = rows * cols;
#pragma omp parallel for
  for (i = 0; i < N; ++i) {
    C[i] = A[i] + B[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<madd>:
  endbr64
  imul   %r8d,%ecx
  test   %ecx,%ecx
  jle    L15
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L5:
  movss  (%rdi,%rax,4),%xmm0
  addss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L5
L15:
  ret
```

assistant: ```c
void madd(float *A, float *B, float *C, int rows, int cols) {
  int N, i;
  N = rows * cols;
#pragma omp parallel for
  for (i = 0; i < N; ++i) {
    C[i] = A[i] + B[i];
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<madd>:
  endbr64
  imul   %ecx,%r8d
  test   %r8d,%r8d
  jle    L47
  lea    0x4(%rdi),%rcx
  mov    %rdx,%rax
  lea    0x4(%rsi),%r9
  sub    %rcx,%rax
  cmp    $0x8,%rax
  mov    %rdx,%rax
  seta   %cl
  sub    %r9,%rax
  cmp    $0x8,%rax
  seta   %al
  test   %al,%cl
  je     L44
  cmp    $0x1,%r8d
  je     L44
  lea    -0x1(%r8),%eax
  mov    %r8d,%r9d
  cmp    $0x2,%eax
  jbe    L88
  mov    %r8d,%ecx
  xor    %eax,%eax
  shr    $0x2,%ecx
  shl    $0x4,%rcx
  nopl   0x0(%rax)
L51:
  movups (%rdi,%rax,1),%xmm0
  movups (%rsi,%rax,1),%xmm2
  addps  %xmm2,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L51
  mov    %r8d,%ecx
  and    $0xfffffffc,%ecx
  mov    %ecx,%eax
  cmp    %ecx,%r8d
  je     L47
  mov    %r8d,%r9d
  sub    %ecx,%r9d
  cmp    $0x1,%r9d
  je     L98
L75:
  movq   (%rdi,%rcx,4),%xmm0
  movq   (%rsi,%rcx,4),%xmm1
  addps  %xmm1,%xmm0
  movlps %xmm0,(%rdx,%rcx,4)
  mov    %r9d,%ecx
  and    $0xfffffffe,%ecx
  add    %ecx,%eax
  cmp    %ecx,%r9d
  je     L47
L98:
  cltq
  movss  (%rdi,%rax,4),%xmm0
  addss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  ret
  nopl   0x0(%rax)
L44:
  movslq %r8d,%r8
  xor    %eax,%eax
  nopl   (%rax)
L26:
  movss  (%rdi,%rax,4),%xmm0
  addss  (%rsi,%rax,4),%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L26
L47:
  ret
L88:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L75
```

assistant: ```c
void madd(float *A, float *B, float *C, int rows, int cols) {
  int N, i;
  N = rows * cols;
#pragma omp parallel for
  for (i = 0; i < N; ++i) {
    C[i] = A[i] + B[i];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<read_result>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  imul   %eax,%eax
  cltq
  shl    $0x2,%eax
  mov    $0x1000,%esi
  mov    %eax,%edi
  call   <xmalloc_aligned@plt>
  mov    -0x38(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x3c(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%ecx
  lea    D2(%rip),%rdx
  mov    $0x20,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x3c(%rbp),%edx
  lea    -0x30(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <read_matrix@plt>
  mov    %eax,%edi
  call   <assert@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D2
datatype: string

tool: D2: matrixmul_compare_%06d.dat

assistant: ```c
void read_result(int **res, int matrix_size) {
  *res = xmalloc_aligned(matrix_size * matrix_size * sizeof(int), 4096);
  char res_path[32];
  snprintf(res_path, sizeof(res_path), "matrixmul_compare_%06d.dat", matrix_size);
  assert(read_matrix(*res, res_path, matrix_size));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<read_result>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %esi,%edi
  imul   %esi,%edi
  shl    $0x2,%edi
  mov    $0x1000,%esi
  call   <xmalloc_aligned@plt>
  mov    %rax,0x0(%rbp)
  mov    %rsp,%r12
  mov    %ebx,%ecx
  lea    D1(%rip),%rdx
  mov    $0x20,%esi
  mov    %r12,%rdi
  call   <snprintf@plt>
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    0x0(%rbp),%rdi
  call   <read_matrix@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D1: matrixmul_compare_%06d.dat

assistant: ```c
void read_result(int **res, int matrix_size) {
  *res = xmalloc_aligned(matrix_size * matrix_size * sizeof(int), 4096);
  char res_path[32];
  snprintf(res_path, sizeof(res_path), "matrixmul_compare_%06d.dat", matrix_size);
  assert(read_matrix(*res, res_path, matrix_size));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<read_result>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  imul   %esi,%edi
  mov    $0x1000,%esi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  shl    $0x2,%edi
  mov    %rsp,%rbp
  call   <xmalloc_aligned@plt>
  mov    %r12d,%ecx
  mov    $0x20,%esi
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  lea    D3(%rip),%rdx
  call   <snprintf@plt>
  mov    (%rbx),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <read_matrix@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3: matrixmul_compare_%06d.dat

assistant: ```c
void read_result(int **res, int matrix_size) {
  *res = xmalloc_aligned(matrix_size * matrix_size * sizeof(int), 4096);
  char res_path[32];
  snprintf(res_path, sizeof(res_path), "matrixmul_compare_%06d.dat", matrix_size);
  assert(read_matrix(*res, res_path, matrix_size));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_result>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  imul   %esi,%edi
  mov    $0x1000,%esi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  shl    $0x2,%edi
  mov    %rsp,%rbp
  call   <xmalloc_aligned@plt>
  mov    %r12d,%ecx
  mov    $0x20,%esi
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  lea    D6(%rip),%rdx
  call   <snprintf@plt>
  mov    (%rbx),%rdi
  mov    %r12d,%edx
  mov    %rbp,%rsi
  call   <read_matrix@plt>
  mov    %eax,%edi
  call   <assert@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: matrixmul_compare_%06d.dat

assistant: ```c
void read_result(int **res, int matrix_size) {
  *res = xmalloc_aligned(matrix_size * matrix_size * sizeof(int), 4096);
  char res_path[32];
  snprintf(res_path, sizeof(res_path), "matrixmul_compare_%06d.dat", matrix_size);
  assert(read_matrix(*res, res_path, matrix_size));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<source_peeknc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%edx
  lea    -0x9(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <source_peekn@plt>
  test   %rax,%rax
  jg     L16
  mov    $0xffffffff,%eax
  jmp    L8
L16:
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
L8:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
int source_peeknc(unsigned pos) {
  char c;
  if (source_peekn(&c, pos) <= 0)
    return -1;
  return (unsigned int)(unsigned char)c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<source_peeknc>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdi
  call   <source_peekn@plt>
  test   %rax,%rax
  jle    L2
  movzbl 0x7(%rsp),%eax
L6:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L2:
  mov    $0xffffffff,%eax
  jmp    L6
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int source_peeknc(unsigned pos) {
  char c;
  if (source_peekn(&c, pos) <= 0)
    return -1;
  return (unsigned int)(unsigned char)c;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<source_peeknc>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdi
  call   <source_peekn@plt>
  test   %rax,%rax
  jle    L10
  movzbl 0x7(%rsp),%eax
L4:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x18,%rsp
  ret
L10:
  mov    $0xffffffff,%eax
  jmp    L4
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int source_peeknc(unsigned pos) {
  char c;
  if (source_peekn(&c, pos) <= 0)
    return -1;
  return (unsigned int)(unsigned char)c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<source_peeknc>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdi
  call   <source_peekn@plt>
  test   %rax,%rax
  jle    L11
  movzbl 0x7(%rsp),%eax
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  ret
L11:
  mov    $0xffffffff,%eax
  jmp    L7
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int source_peeknc(unsigned pos) {
  char c;
  if (source_peekn(&c, pos) <= 0)
    return -1;
  return (unsigned int)(unsigned char)c;
}
```
user: What is the c source code of the assembly code below?

```asm
<change>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x6,-0x14(%rbp)
  movl   $0x9,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap1@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D38
type_guessed: string
name: disassemble_data
arguments: data_tag: D1
type_guessed: string
name: disassemble_data
arguments: data_tag: D55
type_guessed: string
name: disassemble_data
arguments: data_tag: D50
type_guessed: string
name: disassemble_data
arguments: data_tag: D12
type_guessed: string
name: disassemble_data
arguments: data_tag: D20
type_guessed: string

tool: D38: nuchange x,y=%d,%d


tool: D1: swap changed num is: %d,%d


tool: D55: swap1 changed num is: %d,%d


tool: D50: SWAP changed num is: %d,%d


tool: D12: max is %d


tool: D20: min is %d


assistant: ```c
int change() {
  int x = 6;
  int y = 9;
  printf("nuchange x,y=%d,%d\n", x, y);
  swap(&x, &y);
  printf("swap changed num is: %d,%d\n", x, y);
  swap1(&x, &y);
  printf("swap1 changed num is: %d,%d\n", x, y);
  {
    int temp;
    temp = (x);
    (x) = (y);
    (y) = temp;
  };
  printf("SWAP changed num is: %d,%d\n", x, y);
  printf("max is %d\n", (((x) > (y)) ? (x) : (y)));
  printf("min is %d\n", (((x) < (y)) ? (x) : (y)));
  return 0;
}
```