ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mysnprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xd8(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L36
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L36:
  movl   $0x0,-0xb8(%rbp)
  cmpq   $0x0,-0xc8(%rbp)
  je     L1
  cmpq   $0x2,-0xd0(%rbp)
  jbe    L1
  mov    -0xd8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%ecx
  mov    -0xd8(%rbp),%rdx
  mov    -0xd0(%rbp),%rsi
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  mov    %eax,-0xb8(%rbp)
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb8(%rbp),%eax
  cltq
  cmp    %rax,-0xd0(%rbp)
  ja     L1
  mov    -0xd0(%rbp),%rax
  sub    $0x1,%eax
  mov    %eax,-0xb8(%rbp)
L1:
  mov    -0xb8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int mysnprintf(char *buf, size_t len, const char *fmt, ...) {
  va_list ap;
  int n = 0;
  if (buf && (len > 2)) {
    va_start(ap, fmt);
    n = vsnprintf(buf, len, fmt, ap);
    va_end(ap);
    if (n >= len) {
      n = len - 1;
    }
  }
  return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mysnprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L28
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r12
  cmp    $0x2,%rsi
  ja     L1
L28:
  mov    $0x0,%ebp
L4:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  mov    %rdx,%rsi
  mov    $0x0,%r13d
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <vsnprintf@plt>
  mov    %eax,%ebp
  mov    %r13d,%edi
  call   <va_end@plt>
  movslq %ebp,%rdx
  lea    -0x1(%rbx),%eax
  cmp    %rbx,%rdx
  cmovae %eax,%ebp
  jmp    L4
```

assistant: ```c
int mysnprintf(char *buf, size_t len, const char *fmt, ...) {
  va_list ap;
  int n = 0;
  if (buf && (len > 2)) {
    va_start(ap, fmt);
    n = vsnprintf(buf, len, fmt, ap);
    va_end(ap);
    if (n >= len) {
      n = len - 1;
    }
  }
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mysnprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L14
  mov    %rsi,%rbx
  cmp    $0x2,%rsi
  ja     L16
L14:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L16:
  xor    %r13d,%r13d
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %rdx,%rsi
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r12,%rdx
  mov    %r13d,%ecx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <vsnprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  movslq %r12d,%rdx
  lea    -0x1(%rbx),%eax
  cmp    %rbx,%rdx
  cmovae %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int mysnprintf(char *buf, size_t len, const char *fmt, ...) {
  va_list ap;
  int n = 0;
  if (buf && (len > 2)) {
    va_start(ap, fmt);
    n = vsnprintf(buf, len, fmt, ap);
    va_end(ap);
    if (n >= len) {
      n = len - 1;
    }
  }
  return n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mysnprintf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L9
  mov    %rsi,%rbx
  cmp    $0x2,%rsi
  ja     L15
L9:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L15:
  xor    %r13d,%r13d
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %rdx,%rsi
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r12,%rdx
  mov    %r13d,%ecx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <vsnprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  movslq %r12d,%rdx
  lea    -0x1(%rbx),%eax
  cmp    %rbx,%rdx
  cmovae %eax,%r12d
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int mysnprintf(char *buf, size_t len, const char *fmt, ...) {
  va_list ap;
  int n = 0;
  if (buf && (len > 2)) {
    va_start(ap, fmt);
    n = vsnprintf(buf, len, fmt, ap);
    va_end(ap);
    if (n >= len) {
      n = len - 1;
    }
  }
  return n;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<concat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L13
L7:
  mov    -0x10(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L13:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_guessed: string

tool: D2: out of memory.


assistant: ```c
char *concat(const char *a, const char *b) {
  int la = strlen(a);
  int lb = strlen(b);
  char *tmp = malloc(la + lb + 1);
  if (!tmp) {
    printf("out of memory.\n");
    return NULL;
  }
  memcpy(tmp, a, la);
  memcpy(&tmp[la], b, lb);
  tmp[la + lb] = '\0';
  return tmp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<concat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  lea    (%rbx,%rax,1),%r14d
  lea    0x1(%r14),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L13
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rdi
  add    %rbp,%rdi
  mov    %r13d,%edx
  mov    %r12,%rsi
  call   <memcpy@plt>
  movslq %r14d,%r14
  movb   $0x0,0x0(%rbp,%r14,1)
L19:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  lea    D0(%rip),%rdi
  call   <printf@plt>
  jmp    L19
```

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

tool: D0: out of memory.


assistant: ```c
char *concat(const char *a, const char *b) {
  int la = strlen(a);
  int lb = strlen(b);
  char *tmp = malloc(la + lb + 1);
  if (!tmp) {
    printf("out of memory.\n");
    return NULL;
  }
  memcpy(tmp, a, la);
  memcpy(&tmp[la], b, lb);
  tmp[la + lb] = '\0';
  return tmp;
}
```
user: What is the c source code of the assembly code below:

```asm
<concat>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  lea    (%rbx,%rax,1),%r14d
  mov    %eax,%r13d
  lea    0x1(%r14),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  movslq %r14d,%r14
  call   <memcpy@plt>
  movslq %ebx,%rdi
  mov    %r13d,%edx
  mov    %rbp,%rsi
  add    %r12,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r14,1)
L18:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L18
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
datatype: string

tool: D7: out of memory.


assistant: ```c
char *concat(const char *a, const char *b) {
  int la = strlen(a);
  int lb = strlen(b);
  char *tmp = malloc(la + lb + 1);
  if (!tmp) {
    printf("out of memory.\n");
    return NULL;
  }
  memcpy(tmp, a, la);
  memcpy(&tmp[la], b, lb);
  tmp[la + lb] = '\0';
  return tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<concat>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  lea    (%rbx,%rax,1),%r14d
  mov    %eax,%r13d
  lea    0x1(%r14),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L15
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %rax,%rdi
  movslq %r14d,%r14
  call   <memcpy@plt>
  movslq %ebx,%rdi
  mov    %r13d,%edx
  mov    %rbp,%rsi
  add    %r12,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r14,1)
L17:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L17
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
datatype: string

tool: D4: out of memory.


assistant: ```c
char *concat(const char *a, const char *b) {
  int la = strlen(a);
  int lb = strlen(b);
  char *tmp = malloc(la + lb + 1);
  if (!tmp) {
    printf("out of memory.\n");
    return NULL;
  }
  memcpy(tmp, a, la);
  memcpy(&tmp[la], b, lb);
  tmp[la + lb] = '\0';
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L13
L50:
  addl   $0x1,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L13:
  cmpl   $0x3ff,-0x10(%rbp)
  jle    L50
  mov    -0x14(%rbp),%eax
  mov    $0x400,%esi
  mov    %eax,%edi
  call   <expecti@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L99
L60:
  subl   $0x1,-0x14(%rbp)
  subl   $0x1,-0xc(%rbp)
L99:
  cmpl   $0x0,-0xc(%rbp)
  jg     L60
  mov    -0x14(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <expecti@plt>
  movl   $0x400,-0x8(%rbp)
  jmp    L30
L41:
  subl   $0x1,-0x14(%rbp)
  subl   $0x1,-0x8(%rbp)
L30:
  cmpl   $0x0,-0x8(%rbp)
  jg     L41
  mov    -0x14(%rbp),%eax
  mov    $0xfffffc00,%esi
  mov    %eax,%edi
  call   <expecti@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L52
L87:
  cmpl   $0x63,-0x4(%rbp)
  je     L80
  addl   $0x1,-0x4(%rbp)
L52:
  cmpl   $0x63,-0x4(%rbp)
  jle    L87
  jmp    L14
L80:
  nop
L14:
  mov    -0x4(%rbp),%eax
  mov    $0x63,%esi
  mov    %eax,%edi
  call   <expecti@plt>
  nop
  leave
  ret
```

assistant: ```c
void test() {
  int accumulate = 0;
  for (int i = 0; i < 1024; i++)
    accumulate++;
  expecti(accumulate, 1024);
  for (int i = accumulate; i > 0; i--)
    accumulate--;
  expecti(accumulate, 0);
  for (int i = 1024; i > 0; i--)
    accumulate--;
  expecti(accumulate, -1024);
  int i;
  for (i = 0; i < 100; i++) {
    if (i == 99)
      break;
  }
  expecti(i, 99);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x400,%eax
L0:
  sub    $0x1,%eax
  jne    L0
  mov    $0x400,%esi
  mov    $0x400,%edi
  call   <expecti@plt>
  mov    $0x400,%eax
L6:
  sub    $0x1,%eax
  jne    L6
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <expecti@plt>
  mov    $0x400,%eax
L14:
  sub    $0x1,%eax
  jne    L14
  mov    $0xfffffc00,%esi
  mov    $0xfffffc00,%edi
  call   <expecti@plt>
  mov    $0x63,%eax
L24:
  sub    $0x1,%eax
  jne    L24
  mov    $0x63,%esi
  mov    $0x63,%edi
  call   <expecti@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test() {
  int accumulate = 0;
  for (int i = 0; i < 1024; i++)
    accumulate++;
  expecti(accumulate, 1024);
  for (int i = accumulate; i > 0; i--)
    accumulate--;
  expecti(accumulate, 0);
  for (int i = 1024; i > 0; i--)
    accumulate--;
  expecti(accumulate, -1024);
  int i;
  for (i = 0; i < 100; i++) {
    if (i == 99)
      break;
  }
  expecti(i, 99);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x400,%esi
  mov    $0x400,%edi
  call   <expecti@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <expecti@plt>
  mov    $0xfffffc00,%esi
  mov    $0xfffffc00,%edi
  call   <expecti@plt>
  mov    $0x63,%esi
  mov    $0x63,%edi
  add    $0x8,%rsp
  jmp    1050 <expecti@plt>
```

assistant: ```c
void test() {
  int accumulate = 0;
  for (int i = 0; i < 1024; i++)
    accumulate++;
  expecti(accumulate, 1024);
  for (int i = accumulate; i > 0; i--)
    accumulate--;
  expecti(accumulate, 0);
  for (int i = 1024; i > 0; i--)
    accumulate--;
  expecti(accumulate, -1024);
  int i;
  for (i = 0; i < 100; i++) {
    if (i == 99)
      break;
  }
  expecti(i, 99);
}
```
user: What is the c source code of the assembly code below?

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x400,%esi
  mov    $0x400,%edi
  call   <expecti@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <expecti@plt>
  mov    $0xfffffc00,%esi
  mov    $0xfffffc00,%edi
  call   <expecti@plt>
  mov    $0x63,%esi
  mov    $0x63,%edi
  add    $0x8,%rsp
  jmp    1050 <expecti@plt>
```

assistant: ```c
void test() {
  int accumulate = 0;
  for (int i = 0; i < 1024; i++)
    accumulate++;
  expecti(accumulate, 1024);
  for (int i = accumulate; i > 0; i--)
    accumulate--;
  expecti(accumulate, 0);
  for (int i = 1024; i > 0; i--)
    accumulate--;
  expecti(accumulate, -1024);
  int i;
  for (i = 0; i < 100; i++) {
    if (i == 99)
      break;
  }
  expecti(i, 99);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<selection_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L23
L17:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L48
L22:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  comisd %xmm1,%xmm0
  jbe    L21
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L21:
  addl   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L22
  mov    -0x4(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0xc(%rbp)
L23:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void selection_sort(double array[], int n) {
  int i, j;
  for (i = 0; i < n - 1; i++) {
    int min = i;
    for (j = i + 1; j < n; j++) {
      if (array[j] < array[min])
        min = j;
    }
    swap(array, i, min);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<selection_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L9
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  lea    -0x2(%rsi),%r13d
  add    $0x2,%r13
  mov    $0x1,%r12d
L43:
  lea    -0x1(%r12),%esi
  cmp    %r12d,%ebp
  jle    L7
  mov    %r12,%rax
  mov    %esi,%edx
L25:
  movslq %edx,%rcx
  movsd  (%rbx,%rcx,8),%xmm0
  ucomisd (%rbx,%rax,8),%xmm0
  cmova  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L25
L12:
  mov    %rbx,%rdi
  call   <swap@plt>
  add    $0x1,%r12
  cmp    %r13,%r12
  jne    L43
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  mov    %esi,%edx
  jmp    L12
L9:
  ret
```

assistant: ```c
void selection_sort(double array[], int n) {
  int i, j;
  for (i = 0; i < n - 1; i++) {
    int min = i;
    for (j = i + 1; j < n; j++) {
      if (array[j] < array[min])
        min = j;
    }
    swap(array, i, min);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<selection_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L2
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  lea    -0x2(%rsi),%ebp
  push   %rbx
  add    $0x2,%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L24:
  lea    -0x1(%r13),%esi
  cmp    %r13d,%r12d
  jle    L23
  mov    %r13,%rax
  mov    %esi,%edx
  xchg   %ax,%ax
L31:
  movslq %edx,%rcx
  movsd  (%rbx,%rcx,8),%xmm0
  ucomisd (%rbx,%rax,8),%xmm0
  cmova  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%r12d
  jg     L31
L3:
  mov    %rbx,%rdi
  add    $0x1,%r13
  call   <swap@plt>
  cmp    %rbp,%r13
  jne    L24
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L23:
  mov    %esi,%edx
  jmp    L3
L2:
  ret
```

assistant: ```c
void selection_sort(double array[], int n) {
  int i, j;
  for (i = 0; i < n - 1; i++) {
    int min = i;
    for (j = i + 1; j < n; j++) {
      if (array[j] < array[min])
        min = j;
    }
    swap(array, i, min);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<selection_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L24
  push   %r13
  lea    -0x2(%rsi),%r13d
  push   %r12
  add    $0x2,%r13
  mov    $0x1,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopw   0x0(%rax,%rax,1)
L12:
  lea    -0x1(%r12),%esi
  cmp    %r12d,%ebp
  jle    L7
  movsd  -0x8(%rbx,%r12,8),%xmm0
  mov    %r12,%rax
  mov    %esi,%edx
  cs nopw 0x0(%rax,%rax,1)
L36:
  movsd  (%rbx,%rax,8),%xmm1
  comisd %xmm1,%xmm0
  minsd  %xmm0,%xmm1
  cmova  %eax,%edx
  add    $0x1,%rax
  movapd %xmm1,%xmm0
  cmp    %eax,%ebp
  jg     L36
L41:
  mov    %rbx,%rdi
  add    $0x1,%r12
  call   <swap@plt>
  cmp    %r13,%r12
  jne    L12
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %esi,%edx
  jmp    L41
L24:
  ret
```

assistant: ```c
void selection_sort(double array[], int n) {
  int i, j;
  for (i = 0; i < n - 1; i++) {
    int min = i;
    for (j = i + 1; j < n; j++) {
      if (array[j] < array[min])
        min = j;
    }
    swap(array, i, min);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L8:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string
name: read_data
arguments: tag: D11
data_type: string

tool: D1: Enter the number and see the reverse: 

tool: D11: %d

assistant: ```c
int main(void) {
  int value, r_digit;
  printf("Enter the number and see the reverse: ");
  scanf("%d", &value);
  do {
    r_digit = value % 10;
    printf("%d", r_digit);
    value = value / 10;
  } while (value != 0);
  printf("\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  lea    D22(%rip),%rbx
L7:
  mov    0x4(%rsp),%esi
  movslq %esi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,0x4(%rsp)
  jne    L7
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D26
type_infered: string
name: unpack_data
arguments: label: D22
type_infered: string

tool: D26: Enter the number and see the reverse: 

tool: D22: %d

assistant: ```c
int main(void) {
  int value, r_digit;
  printf("Enter the number and see the reverse: ");
  scanf("%d", &value);
  do {
    r_digit = value % 10;
    printf("%d", r_digit);
    value = value / 10;
  } while (value != 0);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  lea    D15(%rip),%rdi
  lea    D7(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  nopl   (%rax)
L19:
  movslq %esi,%rax
  mov    %esi,%edx
  mov    %rbx,%rdi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  movslq 0x4(%rsp),%rsi
  mov    %rsi,%rax
  imul   $0x66666667,%rsi,%rsi
  sar    $0x1f,%eax
  sar    $0x22,%rsi
  sub    %eax,%esi
  mov    %esi,0x4(%rsp)
  jne    L19
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D15
data_type: string
name: retrieve_data
arguments: data_label: D7
data_type: string

tool: D15: Enter the number and see the reverse: 

tool: D7: %d

assistant: ```c
int main(void) {
  int value, r_digit;
  printf("Enter the number and see the reverse: ");
  scanf("%d", &value);
  do {
    r_digit = value % 10;
    printf("%d", r_digit);
    value = value / 10;
  } while (value != 0);
  printf("\n");
  return 0;
}
```