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
<shuffle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L10
L17:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %rax,%rcx
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  mov    -0x30(%rbp),%rdx
  mov    %rdx,%rdi
  sub    -0x18(%rbp),%rdi
  mov    $0x0,%edx
  div    %rdi
  lea    0x1(%rax),%rsi
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  imul   -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x38(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  imul   -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x8(%rbp),%rax
  imul   -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  imul   -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x38(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  addq   $0x1,-0x18(%rbp)
L10:
  mov    -0x30(%rbp),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x18(%rbp)
  jb     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void shuffle(void *arr, size_t n, size_t size) {
  size_t i;
  void *swp = calloc(1, size);
  for (i = 0; i < n - 1; ++i) {
    size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
    memcpy(swp, arr + (j * size), size);
    memcpy(arr + (j * size), arr + (i * size), size);
    memcpy(arr + (i * size), swp, size);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<shuffle>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %rdi,(%rsp)
  mov    %rsi,%rbx
  mov    %rsi,0x8(%rsp)
  mov    %rdx,%rbp
  mov    %rdx,%rsi
  mov    $0x1,%edi
  call   <calloc@plt>
  sub    $0x1,%rbx
  je     L17
  mov    %rax,%r14
  mov    %rbx,%r15
  mov    $0x0,%r12d
L15:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %rax,%rbx
  mov    0x8(%rsp),%rcx
  sub    %r12,%rcx
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  mov    $0x0,%edx
  div    %rcx
  lea    0x1(%rax),%rcx
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rax,%rbx
  add    %r12,%rbx
  imul   %rbp,%rbx
  add    (%rsp),%rbx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  add    $0x1,%r12
  add    %rbp,%r13
  cmp    %r15,%r12
  jne    L15
L17:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void shuffle(void *arr, size_t n, size_t size) {
  size_t i;
  void *swp = calloc(1, size);
  for (i = 0; i < n - 1; ++i) {
    size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
    memcpy(swp, arr + (j * size), size);
    memcpy(arr + (j * size), arr + (i * size), size);
    memcpy(arr + (i * size), swp, size);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<shuffle>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    $0x1,%edi
  mov    %rsi,0x18(%rsp)
  mov    %rdx,%rsi
  call   <calloc@plt>
  sub    $0x1,%r15
  mov    %r15,0x8(%rsp)
  je     L15
  mov    D7(%rip),%r15
  mov    %rax,%r12
  xor    %r13d,%r13d
  xchg   %ax,%ax
L5:
  xor    %eax,%eax
  call   <rand@plt>
  mov    0x18(%rsp),%rsi
  xor    %edx,%edx
  mov    %r12,%rdi
  mov    %rax,%r8
  mov    (%r15),%rax
  sub    %r13,%rsi
  div    %rsi
  xor    %edx,%edx
  lea    0x1(%rax),%rsi
  mov    %r8,%rax
  div    %rsi
  mov    %rbx,%rdx
  mov    %rax,%r8
  mov    0x10(%rsp),%rax
  add    %r13,%r8
  add    $0x1,%r13
  imul   %rbx,%r8
  lea    (%rax,%r8,1),%r14
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdx
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rdx
  mov    %r12,%rsi
  call   <memcpy@plt>
  add    %rbx,%rbp
  cmp    0x8(%rsp),%r13
  jne    L5
L15:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void shuffle(void *arr, size_t n, size_t size) {
  size_t i;
  void *swp = calloc(1, size);
  for (i = 0; i < n - 1; ++i) {
    size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
    memcpy(swp, arr + (j * size), size);
    memcpy(arr + (j * size), arr + (i * size), size);
    memcpy(arr + (i * size), swp, size);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<shuffle>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    $0x1,%edi
  mov    %rsi,0x18(%rsp)
  mov    %rdx,%rsi
  call   <calloc@plt>
  sub    $0x1,%r15
  mov    %r15,0x8(%rsp)
  je     L15
  mov    D1(%rip),%r15
  mov    %rax,%r12
  xor    %r13d,%r13d
  xchg   %ax,%ax
L1:
  xor    %eax,%eax
  call   <rand@plt>
  mov    0x18(%rsp),%rsi
  xor    %edx,%edx
  mov    %r12,%rdi
  mov    %rax,%r8
  mov    (%r15),%rax
  sub    %r13,%rsi
  div    %rsi
  xor    %edx,%edx
  lea    0x1(%rax),%rsi
  mov    %r8,%rax
  div    %rsi
  mov    %rbx,%rdx
  mov    %rax,%r8
  mov    0x10(%rsp),%rax
  add    %r13,%r8
  add    $0x1,%r13
  imul   %rbx,%r8
  lea    (%rax,%r8,1),%r14
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdx
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbp,%rdi
  mov    %rbx,%rdx
  mov    %r12,%rsi
  call   <memcpy@plt>
  add    %rbx,%rbp
  cmp    0x8(%rsp),%r13
  jne    L1
L15:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void shuffle(void *arr, size_t n, size_t size) {
  size_t i;
  void *swp = calloc(1, size);
  for (i = 0; i < n - 1; ++i) {
    size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
    memcpy(swp, arr + (j * size), size);
    memcpy(arr + (j * size), arr + (i * size), size);
    memcpy(arr + (i * size), swp, size);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<exec3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <logInfoMessage@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <backup_folder@plt>
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <logInfoMessage@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D19
type: string
name: read_data
arguments: data_label: D6
type: string

tool: D19: backup starting

tool: D6: updating and backing completed

assistant: ```c
void exec3(char *source, char *target) {
  logInfoMessage("backup starting");
  backup_folder(source, target);
  logInfoMessage("updating and backing completed");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<exec3>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  lea    D12(%rip),%rdi
  call   <logInfoMessage@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <backup_folder@plt>
  lea    D17(%rip),%rdi
  call   <logInfoMessage@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D12
data_type: string
name: read_data
arguments: label: D17
data_type: string

tool: D12: backup starting

tool: D17: updating and backing completed

assistant: ```c
void exec3(char *source, char *target) {
  logInfoMessage("backup starting");
  backup_folder(source, target);
  logInfoMessage("updating and backing completed");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<exec3>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D7(%rip),%rdi
  sub    $0x8,%rsp
  call   <logInfoMessage@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <backup_folder@plt>
  add    $0x8,%rsp
  lea    D1(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <logInfoMessage@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D7
type_guessed: string
name: disassemble_data
arguments: tag: D1
type_guessed: string

tool: D7: backup starting

tool: D1: updating and backing completed

assistant: ```c
void exec3(char *source, char *target) {
  logInfoMessage("backup starting");
  backup_folder(source, target);
  logInfoMessage("updating and backing completed");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<exec3>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D0(%rip),%rdi
  sub    $0x8,%rsp
  call   <logInfoMessage@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <backup_folder@plt>
  add    $0x8,%rsp
  lea    D14(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <logInfoMessage@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_infered: string
name: unpack_data
arguments: data_tag: D14
type_infered: string

tool: D0: backup starting

tool: D14: updating and backing completed

assistant: ```c
void exec3(char *source, char *target) {
  logInfoMessage("backup starting");
  backup_folder(source, target);
  logInfoMessage("updating and backing completed");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cftmd0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    -0x38(%rbp),%eax
  shl    $0x2,%eax
  mov    %eax,-0x2c(%rbp)
  mov    -0x48(%rbp),%rax
  movsd  0x10(%rax),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  add    $0x8,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L35
L10:
  mov    -0x8(%rbp),%r8
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x40(%rbp),%rsi
  movsd  D14(%rip),%xmm1
  movsd  D14(%rip),%xmm0
  mov    D14(%rip),%rdi
  mov    -0x30(%rbp),%eax
  pxor   %xmm5,%xmm5
  movapd %xmm1,%xmm4
  pxor   %xmm3,%xmm3
  movapd %xmm0,%xmm2
  pxor   %xmm1,%xmm1
  movq   %rdi,%xmm0
  mov    %eax,%edi
  call   <btrfly@plt>
  addl   $0x2,-0x30(%rbp)
L35:
  mov    -0x30(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L10
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  jmp    L26
L18:
  movsd  -0x28(%rbp),%xmm0
  movq   D27(%rip),%xmm1
  movapd %xmm0,%xmm2
  xorpd  %xmm1,%xmm2
  mov    -0x8(%rbp),%r8
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x40(%rbp),%rsi
  movsd  -0x28(%rbp),%xmm3
  movsd  D14(%rip),%xmm1
  movsd  -0x28(%rbp),%xmm0
  mov    -0x28(%rbp),%rdi
  mov    -0x30(%rbp),%eax
  movapd %xmm3,%xmm5
  movapd %xmm2,%xmm4
  movapd %xmm1,%xmm3
  pxor   %xmm2,%xmm2
  movapd %xmm0,%xmm1
  movq   %rdi,%xmm0
  mov    %eax,%edi
  call   <btrfly@plt>
  addl   $0x2,-0x30(%rbp)
L26:
  mov    -0x38(%rbp),%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x30(%rbp)
  jl     L18
  nop
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D14
type: f64
name: read_data
arguments: label: D27
type: f32[2]

tool: D14: 1

tool: {"D27": [0.0,-0.0]}

assistant: ```c
void cftmd0(int n, int l, double *a, double *w) {
  int j, m = l << 2;
  double wk1r = w[2];
  double *v = w + 1;
  double *b = a + l;
  double *c = a + l + l;
  double *d = a + l + l + l;
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = 0; j < l; j += 2)
    btrfly(j, 1.0, 0.0, 1.0, 0.0, 1.0, 0.0, a, b, c, d);
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = m; j < l + m; j += 2)
    btrfly(j, wk1r, wk1r, 0.0, 1.0, -wk1r, wk1r, a, b, c, d);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cftmd0>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0x1c(%rsp)
  mov    %rdx,%r12
  lea    0x0(,%rsi,4),%ebx
  movsd  0x10(%rcx),%xmm7
  movsd  %xmm7,0x8(%rsp)
  movslq %esi,%rbp
  shl    $0x3,%rbp
  lea    (%rdx,%rbp,1),%r13
  lea    0x0(%r13,%rbp,1),%r14
  add    %r14,%rbp
  test   %esi,%esi
  jle    L8
  mov    %esi,%eax
  sub    $0x1,%eax
  and    $0xfffffffe,%eax
  add    $0x2,%eax
  mov    %eax,0x10(%rsp)
  mov    $0x0,%r15d
L37:
  mov    %rbp,%r8
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  pxor   %xmm5,%xmm5
  mov    D11(%rip),%rax
  movq   %rax,%xmm4
  movapd %xmm5,%xmm3
  movq   %rax,%xmm2
  movapd %xmm5,%xmm1
  movq   %rax,%xmm0
  mov    %r15d,%edi
  call   <btrfly@plt>
  add    $0x2,%r15d
  cmp    %r15d,0x10(%rsp)
  jne    L37
L8:
  mov    0x1c(%rsp),%r15d
  add    %ebx,%r15d
  cmp    %r15d,%ebx
  jge    L27
  movsd  0x8(%rsp),%xmm6
  xorpd  D1(%rip),%xmm6
  movsd  %xmm6,0x10(%rsp)
L5:
  mov    %rbp,%r8
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  movsd  0x8(%rsp),%xmm5
  movsd  0x10(%rsp),%xmm4
  mov    D11(%rip),%rax
  movq   %rax,%xmm3
  pxor   %xmm2,%xmm2
  movapd %xmm5,%xmm1
  movapd %xmm5,%xmm0
  mov    %ebx,%edi
  call   <btrfly@plt>
  add    $0x2,%ebx
  cmp    %r15d,%ebx
  jl     L5
L27:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: tag: D11
type_infered: f64
name: unpack_data
arguments: tag: D1
type_infered: f32[4]

tool: D11: 1

tool: {"D1": [0.0,-0.0,0.0,0.0]}

assistant: ```c
void cftmd0(int n, int l, double *a, double *w) {
  int j, m = l << 2;
  double wk1r = w[2];
  double *v = w + 1;
  double *b = a + l;
  double *c = a + l + l;
  double *d = a + l + l + l;
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = 0; j < l; j += 2)
    btrfly(j, 1.0, 0.0, 1.0, 0.0, 1.0, 0.0, a, b, c, d);
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = m; j < l + m; j += 2)
    btrfly(j, wk1r, wk1r, 0.0, 1.0, -wk1r, wk1r, a, b, c, d);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cftmd0>:
  endbr64
  push   %r15
  movslq %esi,%r8
  push   %r14
  mov    %r8,%rax
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    0x0(,%r8,4),%ebx
  sub    $0x28,%rsp
  movsd  0x10(%rcx),%xmm7
  mov    %r8d,0x1c(%rsp)
  shl    $0x3,%r8
  lea    (%rdx,%r8,1),%r12
  lea    (%r12,%r8,1),%r13
  lea    0x0(%r13,%r8,1),%r15
  test   %eax,%eax
  jle    L20
  mov    0x1c(%rsp),%eax
  movsd  D18(%rip),%xmm6
  xor    %r14d,%r14d
  pxor   %xmm1,%xmm1
  sub    $0x1,%eax
  and    $0xfffffffe,%eax
  add    $0x2,%eax
  mov    %eax,0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L12:
  movapd %xmm1,%xmm5
  movapd %xmm6,%xmm4
  movapd %xmm1,%xmm3
  mov    %r14d,%edi
  movapd %xmm6,%xmm2
  movapd %xmm6,%xmm0
  mov    %r15,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  movsd  %xmm7,0x8(%rsp)
  add    $0x2,%r14d
  movsd  %xmm6,(%rsp)
  call   <btrfly@plt>
  cmp    %r14d,0x10(%rsp)
  movsd  (%rsp),%xmm6
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm7
  jne    L12
L20:
  mov    0x1c(%rsp),%r14d
  add    %ebx,%r14d
  cmp    %r14d,%ebx
  jge    L39
  movapd %xmm7,%xmm4
  movsd  D18(%rip),%xmm6
  xorpd  D4(%rip),%xmm4
  cs nopw 0x0(%rax,%rax,1)
L30:
  movapd %xmm7,%xmm5
  movapd %xmm6,%xmm3
  movapd %xmm7,%xmm1
  mov    %ebx,%edi
  movapd %xmm7,%xmm0
  mov    %r15,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  pxor   %xmm2,%xmm2
  add    $0x2,%ebx
  movsd  %xmm4,0x10(%rsp)
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm7,(%rsp)
  call   <btrfly@plt>
  cmp    %r14d,%ebx
  movsd  (%rsp),%xmm7
  movsd  0x8(%rsp),%xmm6
  movsd  0x10(%rsp),%xmm4
  jl     L30
L39:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D18
type_infered: f64
name: disassemble_data
arguments: data_label: D4
type_infered: f32[4]

tool: D18: 1

tool: {"D4": [0.0,-0.0,0.0,0.0]}

assistant: ```c
void cftmd0(int n, int l, double *a, double *w) {
  int j, m = l << 2;
  double wk1r = w[2];
  double *v = w + 1;
  double *b = a + l;
  double *c = a + l + l;
  double *d = a + l + l + l;
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = 0; j < l; j += 2)
    btrfly(j, 1.0, 0.0, 1.0, 0.0, 1.0, 0.0, a, b, c, d);
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = m; j < l + m; j += 2)
    btrfly(j, wk1r, wk1r, 0.0, 1.0, -wk1r, wk1r, a, b, c, d);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cftmd0>:
  endbr64
  push   %r15
  movslq %esi,%r8
  push   %r14
  mov    %r8,%rax
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  lea    0x0(,%r8,4),%ebx
  sub    $0x28,%rsp
  movsd  0x10(%rcx),%xmm7
  mov    %r8d,0x1c(%rsp)
  shl    $0x3,%r8
  lea    (%rdx,%r8,1),%r12
  lea    (%r12,%r8,1),%r13
  lea    0x0(%r13,%r8,1),%r15
  test   %eax,%eax
  jle    L3
  sub    $0x1,%eax
  movsd  D17(%rip),%xmm6
  xor    %r14d,%r14d
  pxor   %xmm1,%xmm1
  and    $0xfffffffe,%eax
  add    $0x2,%eax
  mov    %eax,0x10(%rsp)
  xchg   %ax,%ax
L32:
  movapd %xmm1,%xmm5
  movapd %xmm6,%xmm4
  movapd %xmm1,%xmm3
  mov    %r14d,%edi
  movapd %xmm6,%xmm2
  movapd %xmm6,%xmm0
  mov    %r15,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  movsd  %xmm7,0x8(%rsp)
  add    $0x2,%r14d
  movsd  %xmm6,(%rsp)
  call   <btrfly@plt>
  cmp    %r14d,0x10(%rsp)
  movsd  (%rsp),%xmm6
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm7
  jne    L32
L3:
  mov    0x1c(%rsp),%r14d
  add    %ebx,%r14d
  cmp    %r14d,%ebx
  jge    L5
  movapd %xmm7,%xmm4
  movsd  D17(%rip),%xmm6
  xorpd  D3(%rip),%xmm4
  xchg   %ax,%ax
L18:
  movapd %xmm7,%xmm5
  movapd %xmm6,%xmm3
  movapd %xmm7,%xmm1
  mov    %ebx,%edi
  movapd %xmm7,%xmm0
  mov    %r15,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  pxor   %xmm2,%xmm2
  add    $0x2,%ebx
  movsd  %xmm4,0x10(%rsp)
  movsd  %xmm6,0x8(%rsp)
  movsd  %xmm7,(%rsp)
  call   <btrfly@plt>
  cmp    %r14d,%ebx
  movsd  (%rsp),%xmm7
  movsd  0x8(%rsp),%xmm6
  movsd  0x10(%rsp),%xmm4
  jl     L18
L5:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D17
type_guessed: f64
name: parse_data
arguments: tag: D3
type_guessed: f32[4]

tool: D17: 1

tool: {"D3": [0.0,-0.0,0.0,0.0]}

assistant: ```c
void cftmd0(int n, int l, double *a, double *w) {
  int j, m = l << 2;
  double wk1r = w[2];
  double *v = w + 1;
  double *b = a + l;
  double *c = a + l + l;
  double *d = a + l + l + l;
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = 0; j < l; j += 2)
    btrfly(j, 1.0, 0.0, 1.0, 0.0, 1.0, 0.0, a, b, c, d);
#pragma mta use 100 streams
#pragma mta assert no dependence
  for (j = m; j < l + m; j += 2)
    btrfly(j, wk1r, wk1r, 0.0, 1.0, -wk1r, wk1r, a, b, c, d);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ft_range>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L46
  mov    $0x0,%eax
  jmp    L16
L46:
  movl   $0x0,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%eax
  sub    -0x24(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L28
  mov    $0x0,%eax
  jmp    L16
L28:
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L47
  mov    $0x0,%eax
  jmp    L16
L8:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  lea    0x1(%rdx),%ecx
  mov    %ecx,-0x14(%rbp)
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
L47:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L8
  mov    -0x8(%rbp),%rax
L16:
  leave
  ret
```

assistant: ```c
int *ft_range(int min, int max) {
  int array_len;
  int i;
  int nb;
  int *range;
  if (min >= max)
    return (NULL);
  i = 0;
  nb = min;
  array_len = max - min;
  if (array_len < 0)
    return (NULL);
  range = malloc(array_len * sizeof(int));
  if (range == NULL)
    return (NULL);
  while (i < array_len)
    range[i++] = nb++;
  return (range);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_range>:
  endbr64
  cmp    %esi,%edi
  jge    L17
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %esi,%edi
  sub    %ebx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L12
  mov    %rax,%rdx
L10:
  mov    %ebx,%ecx
  add    $0x1,%ebx
  mov    %ecx,(%rdx)
  add    $0x4,%rdx
  cmp    %ebx,%ebp
  jne    L10
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L17:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int *ft_range(int min, int max) {
  int array_len;
  int i;
  int nb;
  int *range;
  if (min >= max)
    return (NULL);
  i = 0;
  nb = min;
  array_len = max - min;
  if (array_len < 0)
    return (NULL);
  range = malloc(array_len * sizeof(int));
  if (range == NULL)
    return (NULL);
  while (i < array_len)
    range[i++] = nb++;
  return (range);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_range>:
  endbr64
  cmp    %esi,%edi
  jge    L27
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%edi
  sub    %ebx,%edi
  shl    $0x2,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %rax,%rax
  je     L0
  mov    %rax,%rdx
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    %ebx,%ecx
  add    $0x1,%ebx
  add    $0x4,%rdx
  mov    %ecx,-0x4(%rdx)
  cmp    %ebx,%ebp
  jne    L20
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
int *ft_range(int min, int max) {
  int array_len;
  int i;
  int nb;
  int *range;
  if (min >= max)
    return (NULL);
  i = 0;
  nb = min;
  array_len = max - min;
  if (array_len < 0)
    return (NULL);
  range = malloc(array_len * sizeof(int));
  if (range == NULL)
    return (NULL);
  while (i < array_len)
    range[i++] = nb++;
  return (range);
}
```