ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x9,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  movl   $0x3,0x10(%rsp)
  movl   $0x7,0x14(%rsp)
  movl   $0x4,0x18(%rsp)
  movl   $0x6,0x1c(%rsp)
  movl   $0x5,0x20(%rsp)
  movl   $0x0,0x24(%rsp)
  mov    %rsp,%rbx
  mov    $0x9,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <quick_sort@plt>
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[] = {1, 9, 2, 8, 3, 7, 4, 6, 5, 0};
  int n = sizeof(a) / sizeof(int);
  quick_sort(a, 0, 9);
  print(a, n);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    $0x9,%edx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x900000001,%rax
  mov    %rsp,%rbp
  movq   $0x5,0x20(%rsp)
  mov    %rbp,%rdi
  mov    %rax,(%rsp)
  movabs $0x800000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x700000003,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x600000004,%rax
  mov    %rax,0x18(%rsp)
  call   <quick_sort@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int a[] = {1, 9, 2, 8, 3, 7, 4, 6, 5, 0};
  int n = sizeof(a) / sizeof(int);
  quick_sort(a, 0, 9);
  print(a, n);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  mov    $0x9,%edx
  sub    $0x30,%rsp
  movdqa D3(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    D16(%rip),%rax
  movaps %xmm0,(%rsp)
  movdqa D11(%rip),%xmm0
  mov    %rbp,%rdi
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <quick_sort@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
type_infered: byte[16]
name: unpack_data
arguments: tag: D16
type_infered: byte[8]
name: unpack_data
arguments: tag: D11
type_infered: byte[16]

tool: {"D3": [0x01,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D16": [0x05,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D11": [0x03,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int a[] = {1, 9, 2, 8, 3, 7, 4, 6, 5, 0};
  int n = sizeof(a) / sizeof(int);
  quick_sort(a, 0, 9);
  print(a, n);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kosraju_sharir_scc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    %r8,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x34(%rbp),%eax
  lea    0x2(%rax),%edx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  mov    %edx,%edi
  call   <u_alloc@plt>
  mov    %rax,-0x20(%rbp)
  lea    -0x28(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <u_alloc@plt>
  mov    %rax,-0x18(%rbp)
  lea    -0x28(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <u_alloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdi
  mov    -0x20(%rbp),%rcx
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rsi
  mov    -0x34(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <graph_reverse@plt>
  mov    -0x58(%rbp),%rdi
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x34(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <topological_order@plt>
  mov    -0x28(%rbp),%r8
  mov    -0x50(%rbp),%rdi
  mov    -0x10(%rbp),%rcx
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rsi
  mov    -0x34(%rbp),%eax
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <graph_connected_components_perm@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int kosraju_sharir_scc(int n, int *xadj, int *adj, int *part, int *aux) {
  int nnz = xadj[n];
  int *auxptr = aux;
  int *rxadj = u_alloc(n + 2, &auxptr);
  int *radj = u_alloc(nnz, &auxptr);
  int *perm = u_alloc(n, &auxptr);
  graph_reverse(n, xadj, adj, rxadj, radj);
  topological_order(n, rxadj, radj, perm, aux);
  return graph_connected_components_perm(n, xadj, adj, perm, part, auxptr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kosraju_sharir_scc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %rdx,(%rsp)
  mov    %rcx,0x8(%rsp)
  mov    %r8,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    (%rsi,%rax,4),%r13d
  mov    %r8,0x10(%rsp)
  lea    0x10(%rsp),%r15
  lea    0x2(%rdi),%edi
  mov    %r15,%rsi
  call   <u_alloc@plt>
  mov    %rax,%r12
  mov    %r15,%rsi
  mov    %r13d,%edi
  call   <u_alloc@plt>
  mov    %rax,%r13
  mov    %r15,%rsi
  mov    %ebx,%edi
  call   <u_alloc@plt>
  mov    %rax,%r15
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    (%rsp),%rdx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <graph_reverse@plt>
  mov    %r14,%r8
  mov    %r15,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <topological_order@plt>
  mov    0x10(%rsp),%r9
  mov    0x8(%rsp),%r8
  mov    %r15,%rcx
  mov    (%rsp),%rdx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <graph_connected_components_perm@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int kosraju_sharir_scc(int n, int *xadj, int *adj, int *part, int *aux) {
  int nnz = xadj[n];
  int *auxptr = aux;
  int *rxadj = u_alloc(n + 2, &auxptr);
  int *radj = u_alloc(nnz, &auxptr);
  int *perm = u_alloc(n, &auxptr);
  graph_reverse(n, xadj, adj, rxadj, radj);
  topological_order(n, rxadj, radj, perm, aux);
  return graph_connected_components_perm(n, xadj, adj, perm, part, auxptr);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<kosraju_sharir_scc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rcx,0x8(%rsp)
  lea    0x10(%rsp),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  add    $0x2,%edi
  mov    %r8,0x10(%rsp)
  mov    (%rsi,%rax,4),%ebx
  mov    %r14,%rsi
  mov    %r8,(%rsp)
  call   <u_alloc@plt>
  mov    %r14,%rsi
  mov    %ebx,%edi
  mov    %rax,%r15
  call   <u_alloc@plt>
  mov    %r14,%rsi
  mov    %ebp,%edi
  mov    %rax,%rbx
  call   <u_alloc@plt>
  mov    %rbx,%r8
  mov    %r15,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %rax,%r14
  call   <graph_reverse@plt>
  mov    (%rsp),%r8
  mov    %r14,%rcx
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <topological_order@plt>
  mov    0x10(%rsp),%r9
  mov    %r13,%rdx
  mov    %r14,%rcx
  mov    0x8(%rsp),%r8
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <graph_connected_components_perm@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int kosraju_sharir_scc(int n, int *xadj, int *adj, int *part, int *aux) {
  int nnz = xadj[n];
  int *auxptr = aux;
  int *rxadj = u_alloc(n + 2, &auxptr);
  int *radj = u_alloc(nnz, &auxptr);
  int *perm = u_alloc(n, &auxptr);
  graph_reverse(n, xadj, adj, rxadj, radj);
  topological_order(n, rxadj, radj, perm, aux);
  return graph_connected_components_perm(n, xadj, adj, perm, part, auxptr);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<kosraju_sharir_scc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rcx,0x8(%rsp)
  lea    0x10(%rsp),%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  add    $0x2,%edi
  mov    %r8,0x10(%rsp)
  mov    (%rsi,%rax,4),%ebx
  mov    %r14,%rsi
  mov    %r8,(%rsp)
  call   <u_alloc@plt>
  mov    %r14,%rsi
  mov    %ebx,%edi
  mov    %rax,%r15
  call   <u_alloc@plt>
  mov    %r14,%rsi
  mov    %ebp,%edi
  mov    %rax,%rbx
  call   <u_alloc@plt>
  mov    %rbx,%r8
  mov    %r15,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %rax,%r14
  call   <graph_reverse@plt>
  mov    (%rsp),%r8
  mov    %r14,%rcx
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %ebp,%edi
  call   <topological_order@plt>
  mov    0x10(%rsp),%r9
  mov    %r13,%rdx
  mov    %r14,%rcx
  mov    0x8(%rsp),%r8
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <graph_connected_components_perm@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int kosraju_sharir_scc(int n, int *xadj, int *adj, int *part, int *aux) {
  int nnz = xadj[n];
  int *auxptr = aux;
  int *rxadj = u_alloc(n + 2, &auxptr);
  int *radj = u_alloc(nnz, &auxptr);
  int *perm = u_alloc(n, &auxptr);
  graph_reverse(n, xadj, adj, rxadj, radj);
  topological_order(n, rxadj, radj, perm, aux);
  return graph_connected_components_perm(n, xadj, adj, perm, part, auxptr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  fldt   0x10(%rbp)
  fstpl  -0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  push   0x18(%rbp)
  push   0x10(%rbp)
  movapd %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <math@plt>
  add    $0x10,%rsp
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  push   0x18(%rbp)
  push   0x10(%rbp)
  movapd %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <math@plt>
  add    $0x10,%rsp
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  fldl   -0x8(%rbp)
  leave
  ret
```

assistant: ```c
long double foo(float f, double d, long double ld) {
  double r = ld;
  r += math(f, d, ld);
#pragma omp target map(r)
  { r += math(f, d, ld); }
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<foo>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  movq   %xmm1,%rbx
  fldt   0x30(%rsp)
  fstl   0x10(%rsp)
  fstpt  (%rsp)
  movss  %xmm0,0x1c(%rsp)
  call   <math@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movapd %xmm0,%xmm2
  addsd  0x10(%rsp),%xmm2
  movsd  %xmm2,0x10(%rsp)
  push   0x38(%rsp)
  push   0x38(%rsp)
  movq   %rbx,%xmm1
  movss  0x2c(%rsp),%xmm0
  call   <math@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  0x20(%rsp),%xmm0
  movsd  %xmm0,0x20(%rsp)
  fldl   0x20(%rsp)
  add    $0x30,%rsp
  pop    %rbx
  ret
```

assistant: ```c
long double foo(float f, double d, long double ld) {
  double r = ld;
  r += math(f, d, ld);
#pragma omp target map(r)
  { r += math(f, d, ld); }
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  sub    $0x38,%rsp
  fldt   0x40(%rsp)
  movsd  %xmm1,0x28(%rsp)
  movss  %xmm0,0x24(%rsp)
  fstl   0x18(%rsp)
  fstpt  (%rsp)
  call   <math@plt>
  pxor   %xmm2,%xmm2
  sub    $0x10,%rsp
  cvtsi2sd %rax,%xmm2
  addsd  0x28(%rsp),%xmm2
  fldt   0x50(%rsp)
  movsd  %xmm2,0x28(%rsp)
  movss  0x34(%rsp),%xmm0
  movsd  0x38(%rsp),%xmm1
  fstpt  (%rsp)
  call   <math@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  0x28(%rsp),%xmm0
  movsd  %xmm0,0x28(%rsp)
  fldl   0x28(%rsp)
  add    $0x48,%rsp
  ret
```

assistant: ```c
long double foo(float f, double d, long double ld) {
  double r = ld;
  r += math(f, d, ld);
#pragma omp target map(r)
  { r += math(f, d, ld); }
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  sub    $0x38,%rsp
  fldt   0x40(%rsp)
  movsd  %xmm1,0x28(%rsp)
  movss  %xmm0,0x24(%rsp)
  fstl   0x18(%rsp)
  fstpt  (%rsp)
  call   <math@plt>
  pxor   %xmm2,%xmm2
  sub    $0x10,%rsp
  cvtsi2sd %rax,%xmm2
  addsd  0x28(%rsp),%xmm2
  fldt   0x50(%rsp)
  movsd  %xmm2,0x28(%rsp)
  movss  0x34(%rsp),%xmm0
  movsd  0x38(%rsp),%xmm1
  fstpt  (%rsp)
  call   <math@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  0x28(%rsp),%xmm0
  movsd  %xmm0,0x28(%rsp)
  fldl   0x28(%rsp)
  add    $0x48,%rsp
  ret
```

assistant: ```c
long double foo(float f, double d, long double ld) {
  double r = ld;
  r += math(f, d, ld);
#pragma omp target map(r)
  { r += math(f, d, ld); }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fact>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L4
  mov    $0x1,%eax
  jmp    L10
L4:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fact>
  imul   -0x4(%rbp),%eax
L10:
  leave
  ret
```

assistant: ```c
int fact(int n) {
  if (n == 1) {
    return 1;
  } else {
    return n * fact(n - 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fact>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%edi
  jne    L12
L4:
  mov    %ebx,%eax
  pop    %rbx
  ret
L12:
  lea    -0x1(%rdi),%edi
  call   <fact>
  imul   %eax,%ebx
  jmp    L4
```

assistant: ```c
int fact(int n) {
  if (n == 1) {
    return 1;
  } else {
    return n * fact(n - 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fact>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L13
  xchg   %ax,%ax
L14:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L14
L13:
  ret
```

assistant: ```c
int fact(int n) {
  if (n == 1) {
    return 1;
  } else {
    return n * fact(n - 1);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<fact>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L18
  xchg   %ax,%ax
L10:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L10
L18:
  ret
```

assistant: ```c
int fact(int n) {
  if (n == 1) {
    return 1;
  } else {
    return n * fact(n - 1);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Resize_Description>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type: string

tool: D8: The `Resize` class can be implemented by objects which can be resized in some way. Resizing to a larger size than the current may allow for some resource or other to be preallocated or reserved. For example this class is implemented by `Array` and `Table` to either remove a number of items quickly or to preallocate memory space if it is known that many items are going to be added at a later date.

assistant: ```c
const char *Resize_Description(void) {
  return "The `Resize` class can be implemented by objects which can be resized in "
         "some way. Resizing to a larger size than the current may allow for some "
         "resource or other to be preallocated or reserved. For example this class "
         "is implemented by `Array` and `Table` to either remove a number of items "
         "quickly or to preallocate memory space if it is known that many items are "
         "going to be added at a later date.";
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Resize_Description>:
  endbr64
  lea    D7(%rip),%rax
  ret
```

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

tool: D7: The `Resize` class can be implemented by objects which can be resized in some way. Resizing to a larger size than the current may allow for some resource or other to be preallocated or reserved. For example this class is implemented by `Array` and `Table` to either remove a number of items quickly or to preallocate memory space if it is known that many items are going to be added at a later date.

assistant: ```c
const char *Resize_Description(void) {
  return "The `Resize` class can be implemented by objects which can be resized in "
         "some way. Resizing to a larger size than the current may allow for some "
         "resource or other to be preallocated or reserved. For example this class "
         "is implemented by `Array` and `Table` to either remove a number of items "
         "quickly or to preallocate memory space if it is known that many items are "
         "going to be added at a later date.";
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Resize_Description>:
  endbr64
  lea    D4(%rip),%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_infered: string

tool: D4: The `Resize` class can be implemented by objects which can be resized in some way. Resizing to a larger size than the current may allow for some resource or other to be preallocated or reserved. For example this class is implemented by `Array` and `Table` to either remove a number of items quickly or to preallocate memory space if it is known that many items are going to be added at a later date.

assistant: ```c
const char *Resize_Description(void) {
  return "The `Resize` class can be implemented by objects which can be resized in "
         "some way. Resizing to a larger size than the current may allow for some "
         "resource or other to be preallocated or reserved. For example this class "
         "is implemented by `Array` and `Table` to either remove a number of items "
         "quickly or to preallocate memory space if it is known that many items are "
         "going to be added at a later date.";
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<Resize_Description>:
  endbr64
  lea    D7(%rip),%rax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_guessed: string

tool: D7: The `Resize` class can be implemented by objects which can be resized in some way. Resizing to a larger size than the current may allow for some resource or other to be preallocated or reserved. For example this class is implemented by `Array` and `Table` to either remove a number of items quickly or to preallocate memory space if it is known that many items are going to be added at a later date.

assistant: ```c
const char *Resize_Description(void) {
  return "The `Resize` class can be implemented by objects which can be resized in "
         "some way. Resizing to a larger size than the current may allow for some "
         "resource or other to be preallocated or reserved. For example this class "
         "is implemented by `Array` and `Table` to either remove a number of items "
         "quickly or to preallocate memory space if it is known that many items are "
         "going to be added at a later date.";
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f_asin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <asin@plt>
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
char f_asin(double *a, const double *b) {
  *a = asin(*b);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<f_asin>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsd  (%rsi),%xmm0
  call   <asin@plt>
  movsd  %xmm0,(%rbx)
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
char f_asin(double *a, const double *b) {
  *a = asin(*b);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<f_asin>:
  endbr64
  push   %rbx
  movsd  (%rsi),%xmm0
  mov    %rdi,%rbx
  call   <asin@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
char f_asin(double *a, const double *b) {
  *a = asin(*b);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f_asin>:
  endbr64
  push   %rbx
  movsd  (%rsi),%xmm0
  mov    %rdi,%rbx
  call   <asin@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
char f_asin(double *a, const double *b) {
  *a = asin(*b);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xrename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rename@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int xrename(const char *oldpath, const char *newpath) {
  int ret = rename(oldpath, newpath);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的?

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

assistant: ```c
int xrename(const char *oldpath, const char *newpath) {
  int ret = rename(oldpath, newpath);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<xrename>:
  endbr64
  jmp    1050 <rename@plt>
```

assistant: ```c
int xrename(const char *oldpath, const char *newpath) {
  int ret = rename(oldpath, newpath);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<xrename>:
  endbr64
  jmp    1050 <rename@plt>
```

assistant: ```c
int xrename(const char *oldpath, const char *newpath) {
  int ret = rename(oldpath, newpath);
  if (ret == -1) {
    ;
  }
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inc_t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L16
L20:
  subl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L27
  cmpl   $0x0,-0x4(%rbp)
  jne    L20
L27:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void inc_t(uint8_t *t, int tlength) {
  int i;
  i = tlength - 1;
  while (!(++t[i]) && i)
    i--;
}
```
user: What is the c source code of the assembly code below?

```asm
<inc_t>:
  endbr64
  lea    -0x1(%rsi),%edx
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%rcx
  mov    (%rcx),%eax
  add    $0x1,%eax
  mov    %eax,(%rcx)
  test   %edx,%edx
  je     L9
  test   %eax,%eax
  jne    L9
  sub    $0x2,%esi
  movslq %esi,%rax
L6:
  mov    (%rdi,%rax,4),%esi
  lea    0x1(%rsi),%edx
  mov    %edx,(%rdi,%rax,4)
  test   %eax,%eax
  setne  %cl
  test   %edx,%edx
  sete   %dl
  sub    $0x1,%rax
  test   %dl,%cl
  jne    L6
L9:
  ret
```

assistant: ```c
void inc_t(uint8_t *t, int tlength) {
  int i;
  i = tlength - 1;
  while (!(++t[i]) && i)
    i--;
}
```