ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<mult>:
  endbr64
  test   %edx,%edx
  je     L28
  mov    (%rdi),%r10
  movss  D1(%rip),%xmm2
  xor    %r8d,%r8d
  movss  D10(%rip),%xmm1
  mov    (%rsi),%r9
  mov    %edx,%esi
  nopl   0x0(%rax,%rax,1)
L17:
  mov    (%r10,%r8,8),%rdi
  mov    (%r9,%r8,8),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L29:
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rdx,%rax,4),%xmm0
  divss  %xmm2,%xmm0
  mulss  (%rdi,%rax,4),%xmm0
  mulss  %xmm1,%xmm0
  cvttss2si %xmm0,%ecx
  mov    %ecx,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L29
  add    $0x1,%r8
  cmp    %r8,%rsi
  jne    L17
L28:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type: byte[4]
name: extract_symbolic_data
arguments: label: D10
type: byte[4]

tool: {"D1": [0x00,0x00,0x7f,0x43]}

tool: {"D10": [0x00,0x00,0x7a,0x43]}

assistant: ```c
extern void mult(float ***fmap, int ***imap, unsigned int MAP_SIZE) {
  int i, j;
  for (i = 0; i < MAP_SIZE; ++i) {
    for (j = 0; j < MAP_SIZE; ++j) {
      (*imap)[i][j] = (int)(((*fmap)[i][j] * ((float)(*imap)[i][j] / 255.0f)) * 250.0f);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mult>:
  endbr64
  test   %edx,%edx
  je     L17
  mov    %edx,%eax
  push   %r14
  mov    %edx,%r10d
  movaps D21(%rip),%xmm2
  shr    $0x2,%eax
  push   %r13
  movaps D25(%rip),%xmm1
  mov    %edx,%r9d
  push   %r12
  movss  D21(%rip),%xmm4
  mov    %edx,%r11d
  and    $0xfffffffc,%r10d
  push   %rbp
  movss  D25(%rip),%xmm3
  push   %rbx
  mov    (%rsi),%rbp
  lea    -0x1(%rax),%esi
  lea    -0x1(%rdx),%ebx
  add    $0x1,%rsi
  mov    (%rdi),%r12
  xor    %edi,%edi
  shl    $0x4,%rsi
  nopw   0x0(%rax,%rax,1)
L32:
  mov    (%r12,%rdi,8),%rcx
  mov    0x0(%rbp,%rdi,8),%rdx
  cmp    $0x2,%ebx
  jbe    L59
  xor    %eax,%eax
  nopl   0x0(%rax)
L31:
  movdqu (%rdx,%rax,1),%xmm5
  movups (%rcx,%rax,1),%xmm6
  cvtdq2ps %xmm5,%xmm0
  divps  %xmm2,%xmm0
  mulps  %xmm6,%xmm0
  mulps  %xmm1,%xmm0
  cvttps2dq %xmm0,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L31
  mov    %r10d,%eax
  cmp    %r10d,%r9d
  je     L45
L10:
  movslq %eax,%r14
  pxor   %xmm0,%xmm0
  lea    0x0(,%r14,4),%r8
  lea    (%rdx,%r8,1),%r13
  cvtsi2ssl 0x0(%r13),%xmm0
  divss  %xmm4,%xmm0
  mulss  (%rcx,%r14,4),%xmm0
  mulss  %xmm3,%xmm0
  cvttss2si %xmm0,%r14d
  mov    %r14d,0x0(%r13)
  lea    0x1(%rax),%r13d
  cmp    %r13d,%r9d
  jbe    L45
  lea    0x4(%rdx,%r8,1),%r13
  pxor   %xmm0,%xmm0
  add    $0x2,%eax
  cvtsi2ssl 0x0(%r13),%xmm0
  divss  %xmm4,%xmm0
  mulss  0x4(%rcx,%r8,1),%xmm0
  mulss  %xmm3,%xmm0
  cvttss2si %xmm0,%r14d
  mov    %r14d,0x0(%r13)
  cmp    %eax,%r9d
  jbe    L45
  lea    0x8(%rdx,%r8,1),%rax
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rax),%xmm0
  divss  %xmm4,%xmm0
  mulss  0x8(%rcx,%r8,1),%xmm0
  mulss  %xmm3,%xmm0
  cvttss2si %xmm0,%edx
  mov    %edx,(%rax)
L45:
  add    $0x1,%rdi
  cmp    %r11,%rdi
  jne    L32
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L59:
  xor    %eax,%eax
  jmp    L10
L17:
  ret
```

tool_calls: name: retrieve_data
arguments: label: D21
type: byte[16]
name: retrieve_data
arguments: label: D25
type: byte[16]

tool: {"D21": [0x00,0x00,0x7f,0x43,0x00,0x00,0x7f,0x43,0x00,0x00,0x7f,0x43,0x00,0x00,0x7f,0x43]}

tool: {"D25": [0x00,0x00,0x7a,0x43,0x00,0x00,0x7a,0x43,0x00,0x00,0x7a,0x43,0x00,0x00,0x7a,0x43]}

assistant: ```c
extern void mult(float ***fmap, int ***imap, unsigned int MAP_SIZE) {
  int i, j;
  for (i = 0; i < MAP_SIZE; ++i) {
    for (j = 0; j < MAP_SIZE; ++j) {
      (*imap)[i][j] = (int)(((*fmap)[i][j] * ((float)(*imap)[i][j] / 255.0f)) * 250.0f);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  movl   $0x2,-0x14(%rbp)
  mov    $0x0,%eax
  call   <create_q@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <front@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <enqueue@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <front@plt>
  mov    %rax,%rdx
  lea    -0x18(%rbp),%rax
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  lea    -0x14(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <enqueue@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <front@plt>
  mov    %rax,%rdx
  lea    -0x18(%rbp),%rax
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <dequeue@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <front@plt>
  mov    %rax,%rdx
  lea    -0x14(%rbp),%rax
  cmp    %rax,%rdx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main() {
  int k1 = 1;
  int k2 = 2;
  struct q *queue = create_q();
  assert(front(queue) == 0);
  queue = enqueue(queue, &k1);
  assert((int *)front(queue) == &k1);
  queue = enqueue(queue, &k2);
  assert((int *)front(queue) == &k1);
  queue = dequeue(queue);
  assert((int *)front(queue) == &k2);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <create_q@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <front@plt>
  test   %rax,%rax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <front@plt>
  cmp    %rax,%rbp
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  lea    0x4(%rsp),%r12
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <enqueue@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <front@plt>
  cmp    %rax,%rbp
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <dequeue@plt>
  mov    %rax,%rdi
  call   <front@plt>
  cmp    %rax,%r12
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int k1 = 1;
  int k2 = 2;
  struct q *queue = create_q();
  assert(front(queue) == 0);
  queue = enqueue(queue, &k1);
  assert((int *)front(queue) == &k1);
  queue = enqueue(queue, &k2);
  assert((int *)front(queue) == &k1);
  queue = dequeue(queue);
  assert((int *)front(queue) == &k2);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  mov    %rsp,%rbx
  lea    0x4(%rsp),%r12
  movl   $0x2,0x4(%rsp)
  call   <create_q@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <enqueue@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%rbx
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <enqueue@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%rbx
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <dequeue@plt>
  mov    %rax,%rdi
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%r12
  sete   %dil
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int k1 = 1;
  int k2 = 2;
  struct q *queue = create_q();
  assert(front(queue) == 0);
  queue = enqueue(queue, &k1);
  assert((int *)front(queue) == &k1);
  queue = enqueue(queue, &k2);
  assert((int *)front(queue) == &k1);
  queue = dequeue(queue);
  assert((int *)front(queue) == &k2);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  mov    %rsp,%rbx
  lea    0x4(%rsp),%r12
  movl   $0x2,0x4(%rsp)
  call   <create_q@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  test   %rax,%rax
  sete   %dil
  call   <assert@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <enqueue@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%rbx
  sete   %dil
  call   <assert@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <enqueue@plt>
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%rbx
  sete   %dil
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <dequeue@plt>
  mov    %rax,%rdi
  call   <front@plt>
  xor    %edi,%edi
  cmp    %rax,%r12
  sete   %dil
  call   <assert@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int k1 = 1;
  int k2 = 2;
  struct q *queue = create_q();
  assert(front(queue) == 0);
  queue = enqueue(queue, &k1);
  assert((int *)front(queue) == &k1);
  queue = enqueue(queue, &k2);
  assert((int *)front(queue) == &k1);
  queue = dequeue(queue);
  assert((int *)front(queue) == &k2);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0xa,-0x2f8(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void main() {
  long a;
  long b[100];
  long c;
  b[7] = 10;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main() {
  long a;
  long b[100];
  long c;
  b[7] = 10;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main() {
  long a;
  long b[100];
  long c;
  b[7] = 10;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main() {
  long a;
  long b[100];
  long c;
  b[7] = 10;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int *yay = NULL;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int *yay = NULL;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int *yay = NULL;
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int *yay = NULL;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    %r9d,-0x40(%rbp)
  movss  %xmm0,-0x44(%rbp)
  mov    -0x38(%rbp),%eax
  sub    -0x34(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x40(%rbp),%eax
  sub    -0x3c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,%edi
  call   <MALLOC@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <fatal@plt>
L7:
  movl   $0x0,-0x18(%rbp)
  jmp    L27
L1:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0xff,(%rax)
  addl   $0x1,-0x18(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  imul   -0x10(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jl     L1
  mov    -0x44(%rbp),%ecx
  mov    -0x10(%rbp),%r8d
  mov    -0x14(%rbp),%edi
  mov    -0x3c(%rbp),%edx
  mov    -0x34(%rbp),%eax
  sub    $0x8,%rsp
  push   -0x8(%rbp)
  mov    0x18(%rbp),%esi
  push   %rsi
  mov    0x10(%rbp),%esi
  push   %rsi
  push   -0x30(%rbp)
  push   -0x28(%rbp)
  movd   %ecx,%xmm0
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <dist1@plt>
  add    $0x30,%rsp
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <FREE@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
datatype: string

tool: D1: malloc failed

assistant: ```c
int dist(struct signature *feats1, struct signature *feats2, int start1, int end1, int start2, int end2, int N1, int N2, float thr) {
  int n1 = end1 - start1;
  int n2 = end2 - start2;
  unsigned char *distances = (unsigned char *)MALLOC(sizeof(unsigned char) * n1 * n2);
  if (!distances)
    fatal("malloc failed");
  for (int i = 0; i < n1 * n2; i++) {
    distances[i] = 255;
  }
  int result = dist1(0, 0, start1, start2, n1, n2, feats1, feats2, N1, N2, thr, distances);
  FREE(distances);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r15
  mov    %edx,0x8(%rsp)
  mov    %r8d,%r13d
  movss  %xmm0,0xc(%rsp)
  sub    %edx,%ecx
  mov    %ecx,%ebp
  sub    %r8d,%r9d
  mov    %r9d,%r12d
  mov    %ecx,%edi
  imul   %r9d,%edi
  call   <MALLOC@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L12
L20:
  mov    %ebp,%edx
  imul   %r12d,%edx
  test   %edx,%edx
  jle    L29
  mov    %rbx,%rax
  mov    %edx,%edx
  add    %rbx,%rdx
L18:
  movb   $0xff,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L18
L29:
  sub    $0x8,%rsp
  push   %rbx
  mov    0x68(%rsp),%eax
  push   %rax
  mov    0x68(%rsp),%eax
  push   %rax
  push   %r15
  push   %r14
  movss  0x3c(%rsp),%xmm0
  mov    %r12d,%r9d
  mov    %ebp,%r8d
  mov    %r13d,%ecx
  mov    0x38(%rsp),%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <dist1@plt>
  mov    %eax,%ebp
  add    $0x30,%rsp
  mov    %rbx,%rdi
  call   <FREE@plt>
  mov    %ebp,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  lea    D6(%rip),%rdi
  call   <fatal@plt>
  jmp    L20
```

tool_calls: name: disassemble_data
arguments: label: D6
type_guessed: string

tool: D6: malloc failed

assistant: ```c
int dist(struct signature *feats1, struct signature *feats2, int start1, int end1, int start2, int end2, int N1, int N2, float thr) {
  int n1 = end1 - start1;
  int n2 = end2 - start2;
  unsigned char *distances = (unsigned char *)MALLOC(sizeof(unsigned char) * n1 * n2);
  if (!distances)
    fatal("malloc failed");
  for (int i = 0; i < n1 * n2; i++) {
    distances[i] = 255;
  }
  int result = dist1(0, 0, start1, start2, n1, n2, feats1, feats2, N1, N2, thr, distances);
  FREE(distances);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dist>:
  endbr64
  push   %r15
  sub    %edx,%ecx
  push   %r14
  mov    %ecx,%r14d
  mov    %ecx,%r15d
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %r9d,%ebx
  sub    %r8d,%ebx
  imul   %ebx,%r14d
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %r14d,%edi
  mov    %rsi,0x10(%rsp)
  movss  %xmm0,0x1c(%rsp)
  call   <MALLOC@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L28
L19:
  mov    %r15d,%eax
  imul   %ebx,%eax
  test   %eax,%eax
  jle    L24
  lea    -0x1(%r14),%edx
  mov    $0xff,%esi
  mov    %rbp,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
L24:
  sub    $0x8,%rsp
  mov    %ebx,%r9d
  mov    %r15d,%r8d
  mov    %r13d,%ecx
  push   %rbp
  mov    0x78(%rsp),%eax
  mov    %r12d,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  push   %rax
  mov    0x78(%rsp),%eax
  push   %rax
  push   0x30(%rsp)
  push   0x30(%rsp)
  movss  0x4c(%rsp),%xmm0
  call   <dist1@plt>
  add    $0x30,%rsp
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <FREE@plt>
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L28:
  lea    D5(%rip),%rdi
  call   <fatal@plt>
  jmp    L19
```

tool_calls: name: unpack_data
arguments: data_tag: D5
datatype: string

tool: D5: malloc failed

assistant: ```c
int dist(struct signature *feats1, struct signature *feats2, int start1, int end1, int start2, int end2, int N1, int N2, float thr) {
  int n1 = end1 - start1;
  int n2 = end2 - start2;
  unsigned char *distances = (unsigned char *)MALLOC(sizeof(unsigned char) * n1 * n2);
  if (!distances)
    fatal("malloc failed");
  for (int i = 0; i < n1 * n2; i++) {
    distances[i] = 255;
  }
  int result = dist1(0, 0, start1, start2, n1, n2, feats1, feats2, N1, N2, thr, distances);
  FREE(distances);
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dist>:
  endbr64
  push   %r15
  sub    %edx,%ecx
  push   %r14
  mov    %ecx,%r14d
  mov    %ecx,%r15d
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %r9d,%ebx
  sub    %r8d,%ebx
  imul   %ebx,%r14d
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %r14d,%edi
  mov    %rsi,0x10(%rsp)
  movss  %xmm0,0x1c(%rsp)
  call   <MALLOC@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L13
L7:
  mov    %r15d,%eax
  imul   %ebx,%eax
  test   %eax,%eax
  jle    L28
  lea    -0x1(%r14),%edx
  mov    $0xff,%esi
  mov    %rbp,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
L28:
  sub    $0x8,%rsp
  mov    %ebx,%r9d
  mov    %r15d,%r8d
  mov    %r13d,%ecx
  push   %rbp
  mov    0x78(%rsp),%eax
  mov    %r12d,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  push   %rax
  mov    0x78(%rsp),%eax
  push   %rax
  push   0x30(%rsp)
  push   0x30(%rsp)
  movss  0x4c(%rsp),%xmm0
  call   <dist1@plt>
  add    $0x30,%rsp
  mov    %rbp,%rdi
  mov    %eax,%r12d
  call   <FREE@plt>
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L13:
  lea    D9(%rip),%rdi
  call   <fatal@plt>
  jmp    L7
```

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

tool: D9: malloc failed

assistant: ```c
int dist(struct signature *feats1, struct signature *feats2, int start1, int end1, int start2, int end2, int N1, int N2, float thr) {
  int n1 = end1 - start1;
  int n2 = end2 - start2;
  unsigned char *distances = (unsigned char *)MALLOC(sizeof(unsigned char) * n1 * n2);
  if (!distances)
    fatal("malloc failed");
  for (int i = 0; i < n1 * n2; i++) {
    distances[i] = 255;
  }
  int result = dist1(0, 0, start1, start2, n1, n2, feats1, feats2, N1, N2, thr, distances);
  FREE(distances);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<can_test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x48(%rbp)
  je     L25
  movl   $0x0,-0x38(%rbp)
  jmp    L62
L60:
  cmpl   $0x20,-0x34(%rbp)
  je     L6
  cmpl   $0x29,-0x34(%rbp)
  je     L6
  cmpl   $0x28,-0x34(%rbp)
  jne    L10
L6:
  cmpl   $0x0,-0x38(%rbp)
  je     L4
  mov    -0x38(%rbp),%eax
  cltq
  movb   $0x0,-0x30(%rbp,%rax,1)
  mov    -0x4c(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mark_cap@plt>
L4:
  movl   $0x0,-0x38(%rbp)
  jmp    L77
L10:
  addl   $0x1,-0x38(%rbp)
L77:
  addq   $0x1,-0x48(%rbp)
L62:
  mov    -0x48(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x38(%rbp),%eax
  cltq
  mov    %dl,-0x30(%rbp,%rax,1)
  mov    -0x38(%rbp),%eax
  cltq
  movzbl -0x30(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,-0x34(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jne    L60
  cmpl   $0x0,-0x38(%rbp)
  je     L25
  mov    -0x4c(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <mark_cap@plt>
L25:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

assistant: ```c
void can_test(const char *s, int flags) {
  int ch, j;
  char name[32];
  if (s) {
    for (j = 0; (ch = name[j] = *s); s++) {
      if (ch == ' ' || ch == ')' || ch == '(') {
        if (j) {
          name[j] = '\0';
          mark_cap(name, flags);
        }
        j = 0;
      } else {
        j++;
      }
    }
    if (j) {
      mark_cap(name, flags);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<can_test>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L94
  mov    %rdi,%rbx
  mov    %esi,%ebp
  movzbl (%rdi),%eax
  mov    %al,(%rsp)
  movsbl %al,%eax
  test   %eax,%eax
  je     L94
  mov    $0x0,%edx
  mov    %rsp,%r12
  jmp    L72
L51:
  test   %edx,%edx
  jne    L69
L79:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  movslq %edx,%rcx
  mov    %al,(%rsp,%rcx,1)
  movsbl %al,%eax
  test   %eax,%eax
  je     L83
L72:
  lea    -0x28(%rax),%ecx
  cmp    $0x1,%ecx
  jbe    L51
  cmp    $0x20,%eax
  je     L51
  add    $0x1,%edx
  jmp    L79
L69:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <mark_cap@plt>
  mov    $0x0,%edx
  jmp    L79
L83:
  test   %edx,%edx
  jne    L23
L94:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L108
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  mov    %rsp,%rdi
  mov    %ebp,%esi
  call   <mark_cap@plt>
  jmp    L94
L108:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void can_test(const char *s, int flags) {
  int ch, j;
  char name[32];
  if (s) {
    for (j = 0; (ch = name[j] = *s); s++) {
      if (ch == ' ' || ch == ')' || ch == '(') {
        if (j) {
          name[j] = '\0';
          mark_cap(name, flags);
        }
        j = 0;
      } else {
        j++;
      }
    }
    if (j) {
      mark_cap(name, flags);
    }
  }
}
```