ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L79:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L79
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%r14d
  mov    $0x0,%r13d
  lea    D15(%rip),%r12
  mov    $0x0,%r8d
  mov    $0x0,%esi
  cmpl   $0x0,0x8(%rsp)
  jg     L94
L39:
  sub    %r8d,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  mov    $0x0,%eax
  add    $0x9c60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L73:
  lea    0x10(%rsp,%r14,4),%rbp
  mov    $0x0,%ebx
L15:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L15
L54:
  add    $0x1,%r13d
  mov    0x8(%rsp),%r11d
  add    $0x64,%r14
  cmp    %r13d,%r11d
  jle    L31
L94:
  cmpl   $0x0,0xc(%rsp)
  jg     L73
  jmp    L54
L31:
  test   %r11d,%r11d
  jle    L148
  mov    0xc(%rsp),%ecx
  lea    0x10(%rsp),%r10
  mov    $0x1,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%esi
  jmp    L150
L2:
  add    $0x4,%rdx
  cmp    %ecx,%eax
  je     L81
L66:
  add    $0x1,%eax
  test   $0x1,%al
  je     L165
  add    (%rdx),%esi
L165:
  test   %edi,%edi
  jne    L2
  add    (%rdx),%r8d
  jmp    L2
L81:
  lea    0x1(%r9),%eax
  add    $0x190,%r10
  cmp    %r11d,%r9d
  je     L39
  mov    %eax,%r9d
L150:
  mov    %r10,%rdx
  mov    $0x0,%eax
  mov    %r9d,%edi
  and    $0x1,%edi
  test   %ecx,%ecx
  jg     L66
  jmp    L81
L148:
  mov    $0x0,%r8d
  mov    $0x0,%esi
  jmp    L39
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D21
type_infered: string
name: parse_data
arguments: label: D15
type_infered: string

tool: D21: %d %d

tool: D15: %d

assistant: ```c
int main() {
  int a[100][100], n, m;
  int i, j, sumCols = 0, sumRows = 0;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j)
      scanf("%d", &a[i][j]);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j) {
      if ((j + 1) % 2)
        sumCols += a[i][j];
      if (!((i + 1) % 2))
        sumRows += a[i][j];
    }
  printf("%d", sumCols - sumRows);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L37:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L37
  sub    $0xc68,%rsp
  lea    D5(%rip),%rdi
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  lea    D2(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%r15
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L5
  nopl   (%rax)
L38:
  mov    0xc(%rsp),%eax
  lea    (%r15,%r14,4),%rbp
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L54
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L15
L54:
  mov    0x8(%rsp),%r11d
  add    $0x1,%r13d
  add    $0x64,%r14
  cmp    %r13d,%r11d
  jg     L38
  test   %r11d,%r11d
  jle    L5
  mov    0xc(%rsp),%ecx
  lea    0x10(%rsp),%r10
  xor    %r8d,%r8d
  xor    %esi,%esi
  xor    %r9d,%r9d
  cs nopw 0x0(%rax,%rax,1)
L74:
  add    $0x1,%r9d
  mov    %r10,%rdx
  xor    %eax,%eax
  mov    %r9d,%edi
  and    $0x1,%edi
  test   %ecx,%ecx
  jle    L36
  nopl   0x0(%rax,%rax,1)
L33:
  add    $0x1,%eax
  test   $0x1,%al
  je     L128
  add    (%rdx),%esi
L128:
  test   %edi,%edi
  jne    L3
  add    (%rdx),%r8d
L3:
  add    $0x4,%rdx
  cmp    %ecx,%eax
  jne    L33
L36:
  add    $0x190,%r10
  cmp    %r9d,%r11d
  jne    L74
L29:
  sub    %r8d,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x9c58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L127
  add    $0x9c68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  xor    %r8d,%r8d
  xor    %esi,%esi
  jmp    L29
L127:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string

tool: D5: %d %d

tool: D2: %d

assistant: ```c
int main() {
  int a[100][100], n, m;
  int i, j, sumCols = 0, sumRows = 0;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j)
      scanf("%d", &a[i][j]);
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j) {
      if ((j + 1) % 2)
        sumCols += a[i][j];
      if (!((i + 1) % 2))
        sumRows += a[i][j];
    }
  printf("%d", sumCols - sumRows);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<read_ublox>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  jmp    L3
L32:
  movb   $0x0,-0xa(%rbp)
  lea    -0xa(%rbp),%rax
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x42,%edi
  call   <i2cslowread@plt>
  movzbl -0xa(%rbp),%eax
  cmp    $0xff,%al
  je     L5
  movzbl -0xa(%rbp),%eax
  jmp    L9
L5:
  mov    $0x64,%edi
  call   <delayUS@plt>
  addl   $0x1,-0x10(%rbp)
L3:
  cmpl   $0x9,-0x10(%rbp)
  jle    L32
  mov    $0xffffffff,%eax
L9:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
unsigned char read_ublox() {
  unsigned char idat[2];
  int ix;
  for (ix = 0; ix < 10; ix++) {
    idat[0] = 0;
    i2cslowread(0x42, (unsigned char *)idat, 1, 1);
    if (idat[0] != 0xff)
      return idat[0];
    delayUS(100);
  }
  return 0xff;
}
```
user: What is the c source code of the assembly code below:

```asm
<read_ublox>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xa,%ebx
  lea    0x6(%rsp),%r12
L14:
  movb   $0x0,0x6(%rsp)
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    $0x42,%edi
  call   <i2cslowread@plt>
  movzbl 0x6(%rsp),%ebp
  cmp    $0xff,%bpl
  jne    L19
  mov    $0x64,%edi
  call   <delayUS@plt>
  sub    $0x1,%ebx
  jne    L14
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    %ebp,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char read_ublox() {
  unsigned char idat[2];
  int ix;
  for (ix = 0; ix < 10; ix++) {
    idat[0] = 0;
    i2cslowread(0x42, (unsigned char *)idat, 1, 1);
    if (idat[0] != 0xff)
      return idat[0];
    delayUS(100);
  }
  return 0xff;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_ublox>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbp
  cs nopw 0x0(%rax,%rax,1)
L20:
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    $0x42,%edi
  movb   $0x0,0x6(%rsp)
  call   <i2cslowread@plt>
  movzbl 0x6(%rsp),%r12d
  cmp    $0xff,%r12b
  jne    L12
  mov    $0x64,%edi
  call   <delayUS@plt>
  sub    $0x1,%ebx
  jne    L20
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char read_ublox() {
  unsigned char idat[2];
  int ix;
  for (ix = 0; ix < 10; ix++) {
    idat[0] = 0;
    i2cslowread(0x42, (unsigned char *)idat, 1, 1);
    if (idat[0] != 0xff)
      return idat[0];
    delayUS(100);
  }
  return 0xff;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_ublox>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbp
  cs nopw 0x0(%rax,%rax,1)
L24:
  mov    $0x1,%ecx
  mov    $0x1,%edx
  mov    %rbp,%rsi
  mov    $0x42,%edi
  movb   $0x0,0x6(%rsp)
  call   <i2cslowread@plt>
  movzbl 0x6(%rsp),%r12d
  cmp    $0xff,%r12b
  jne    L6
  mov    $0x64,%edi
  call   <delayUS@plt>
  sub    $0x1,%ebx
  jne    L24
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned char read_ublox() {
  unsigned char idat[2];
  int ix;
  for (ix = 0; ix < 10; ix++) {
    idat[0] = 0;
    i2cslowread(0x42, (unsigned char *)idat, 1, 1);
    if (idat[0] != 0xff)
      return idat[0];
    delayUS(100);
  }
  return 0xff;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<printArr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D18
datatype: string
name: extract_symbolic_data
arguments: data_tag: D1
datatype: string

tool: D18: Vertex   Distance from Source

tool: D1: %d 		 %d


assistant: ```c
void printArr(int dist[], int n) {
  printf("Vertex   Distance from Source\n");
  for (int i = 0; i < n; ++i)
    printf("%d \t\t %d\n", i, dist[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printArr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  lea    D18(%rip),%rdi
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L4
  mov    %ebp,%ebp
  mov    $0x0,%ebx
  lea    D19(%rip),%r13
L7:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L7
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: tag: D18
type: string
name: read_data
arguments: tag: D19
type: string

tool: D18: Vertex   Distance from Source

tool: D19: %d 		 %d


assistant: ```c
void printArr(int dist[], int n) {
  printf("Vertex   Distance from Source\n");
  for (int i = 0; i < n; ++i)
    printf("%d \t\t %d\n", i, dist[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printArr>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  lea    D14(%rip),%rdi
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L15
  xor    %ebx,%ebx
  lea    D1(%rip),%r13
  nopl   (%rax)
L19:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L19
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
data_type: string
name: disassemble_data
arguments: data_tag: D1
data_type: string

tool: D14: Vertex   Distance from Source

tool: D1: %d 		 %d


assistant: ```c
void printArr(int dist[], int n) {
  printf("Vertex   Distance from Source\n");
  for (int i = 0; i < n; ++i)
    printf("%d \t\t %d\n", i, dist[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printArr>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  lea    D19(%rip),%rdi
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <puts@plt>
  test   %ebp,%ebp
  jle    L11
  xor    %ebx,%ebx
  lea    D10(%rip),%r13
  nopl   (%rax)
L10:
  mov    (%r12,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L10
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D19
type: string
name: parse_data
arguments: data_tag: D10
type: string

tool: D19: Vertex   Distance from Source

tool: D10: %d 		 %d


assistant: ```c
void printArr(int dist[], int n) {
  printf("Vertex   Distance from Source\n");
  for (int i = 0; i < n; ++i)
    printf("%d \t\t %d\n", i, dist[i]);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<DoVariable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <GetWord@plt>
  mov    $0x0,%eax
  call   <Create@plt>
  mov    $0x3,%edi
  call   <Comp@plt>
  mov    $0x0,%edi
  call   <Comp@plt>
  mov    $0x0,%eax
  call   <Reveal@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void DoVariable() {
  GetWord();
  Create();
  Comp(3);
  Comp(0);
  Reveal();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<DoVariable>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <GetWord@plt>
  mov    $0x0,%eax
  call   <Create@plt>
  mov    $0x3,%edi
  call   <Comp@plt>
  mov    $0x0,%edi
  call   <Comp@plt>
  mov    $0x0,%eax
  call   <Reveal@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void DoVariable() {
  GetWord();
  Create();
  Comp(3);
  Comp(0);
  Reveal();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<DoVariable>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <GetWord@plt>
  xor    %eax,%eax
  call   <Create@plt>
  mov    $0x3,%edi
  call   <Comp@plt>
  xor    %edi,%edi
  call   <Comp@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10a0 <Reveal@plt>
```

assistant: ```c
void DoVariable() {
  GetWord();
  Create();
  Comp(3);
  Comp(0);
  Reveal();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<DoVariable>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <GetWord@plt>
  xor    %eax,%eax
  call   <Create@plt>
  mov    $0x3,%edi
  call   <Comp@plt>
  xor    %edi,%edi
  call   <Comp@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10a0 <Reveal@plt>
```

assistant: ```c
void DoVariable() {
  GetWord();
  Create();
  Comp(3);
  Comp(0);
  Reveal();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<counting_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r15
  mov    %edx,%ebx
  mov    %ecx,%r14d
  mov    %r8d,%ebp
  lea    0x0(,%rdx,4),%eax
  mov    %eax,0xc(%rsp)
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0xc(%rsp),%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  test   %ebx,%ebx
  jle    L50
  mov    %r13,%rcx
  lea    -0x1(%rbx),%eax
  lea    0x4(%r13,%rax,4),%rsi
L18:
  mov    (%rcx),%eax
  cltd
  idiv   %ebp
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%edi
  sar    $0x1f,%edi
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  addl   $0x1,(%r12,%rax,4)
  add    $0x4,%rcx
  cmp    %rsi,%rcx
  jne    L18
  test   %r14d,%r14d
  jle    L75
L35:
  lea    0x4(%r12),%rax
  mov    $0x1,%edx
L85:
  mov    -0x4(%rax),%ecx
  add    %ecx,(%rax)
  add    $0x1,%edx
  add    $0x4,%rax
  cmp    %edx,%r14d
  jge    L85
  sub    $0x1,%ebx
  js     L82
L10:
  movslq %ebx,%rbx
L47:
  mov    0x0(%r13,%rbx,4),%ecx
  mov    %ecx,%eax
  cltd
  idiv   %ebp
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  lea    (%r12,%rax,4),%rax
  movslq (%rax),%rdx
  mov    %ecx,-0x4(%r15,%rdx,4)
  subl   $0x1,(%rax)
  sub    $0x1,%rbx
  test   %ebx,%ebx
  jns    L47
L82:
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  test   %r14d,%r14d
  jg     L35
  jmp    L82
L75:
  sub    $0x1,%ebx
  jmp    L10
```

assistant: ```c
void counting_sort(int a[], int *b, int size, int k, int unit) {
  int *c;
  c = (int *)malloc(sizeof(int) * size);
  memset(c, 0, (sizeof(int) * size));
  for (int j = 0; j < size; j++) {
    int offset = (a[j] / unit) % 10;
    c[offset] = c[offset] + 1;
  }
  for (int i = 1; i <= k; i++) {
    c[i] = c[i] + c[(i - 1)];
  }
  for (int j = (size - 1); j >= 0; j--) {
    int offset = (a[j] / unit) % 10;
    b[(c[offset] - 1)] = a[j];
    c[offset] = c[offset] - 1;
  }
  free(c);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<counting_sort>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  shl    $0x2,%edx
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %edx,%edi
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memset@plt>
  test   %r15d,%r15d
  jle    L58
  lea    -0x1(%r15),%eax
  mov    %rbx,%rsi
  mov    %rax,%rcx
  lea    0x4(%rbx,%rax,4),%rdi
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%rsi),%eax
  add    $0x4,%rsi
  cltd
  idiv   %r12d
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x22,%rdx
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  addl   $0x1,0x0(%r13,%rax,4)
  cmp    %rsi,%rdi
  jne    L14
  test   %r14d,%r14d
  jle    L34
L39:
  lea    0x4(%r13),%rax
  mov    $0x1,%edx
  nopl   0x0(%rax)
L75:
  add    $0x1,%edx
  mov    -0x4(%rax),%esi
  add    %esi,(%rax)
  add    $0x4,%rax
  cmp    %edx,%r14d
  jge    L75
  test   %ecx,%ecx
  js     L38
L34:
  movslq %ecx,%rcx
L17:
  mov    (%rbx,%rcx,4),%esi
  sub    $0x1,%rcx
  mov    %esi,%eax
  cltd
  idiv   %r12d
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x22,%rdx
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  lea    0x0(%r13,%rax,4),%rax
  movslq (%rax),%rdx
  mov    %esi,-0x4(%rbp,%rdx,4)
  subl   $0x1,(%rax)
  test   %ecx,%ecx
  jns    L17
L38:
  add    $0x18,%rsp
  mov    %r13,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <free@plt>
  nopl   0x0(%rax)
L58:
  lea    -0x1(%r15),%ecx
  test   %r14d,%r14d
  jg     L39
  jmp    L38
```

assistant: ```c
void counting_sort(int a[], int *b, int size, int k, int unit) {
  int *c;
  c = (int *)malloc(sizeof(int) * size);
  memset(c, 0, (sizeof(int) * size));
  for (int j = 0; j < size; j++) {
    int offset = (a[j] / unit) % 10;
    c[offset] = c[offset] + 1;
  }
  for (int i = 1; i <= k; i++) {
    c[i] = c[i] + c[(i - 1)];
  }
  for (int j = (size - 1); j >= 0; j--) {
    int offset = (a[j] / unit) % 10;
    b[(c[offset] - 1)] = a[j];
    c[offset] = c[offset] - 1;
  }
  free(c);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<counting_sort>:
  endbr64
  push   %r15
  mov    %edx,%r15d
  shl    $0x2,%edx
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %edx,%edi
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  call   <malloc@plt>
  mov    0xc(%rsp),%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memset@plt>
  test   %r15d,%r15d
  jle    L13
  lea    -0x1(%r15),%eax
  mov    %rbx,%rsi
  mov    %rax,%rcx
  lea    0x4(%rbx,%rax,4),%rdi
  nopl   0x0(%rax,%rax,1)
L29:
  mov    (%rsi),%eax
  add    $0x4,%rsi
  cltd
  idiv   %r12d
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x22,%rdx
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  addl   $0x1,0x0(%r13,%rax,4)
  cmp    %rsi,%rdi
  jne    L29
  test   %r14d,%r14d
  jle    L4
L79:
  mov    0x0(%r13),%esi
  lea    0x4(%r13),%rax
  mov    $0x1,%edx
  nopl   0x0(%rax,%rax,1)
L54:
  add    (%rax),%esi
  add    $0x1,%edx
  add    $0x4,%rax
  mov    %esi,-0x4(%rax)
  cmp    %edx,%r14d
  jge    L54
  test   %ecx,%ecx
  js     L41
L4:
  movslq %ecx,%rcx
  nopl   0x0(%rax,%rax,1)
L10:
  mov    (%rbx,%rcx,4),%esi
  sub    $0x1,%rcx
  mov    %esi,%eax
  cltd
  idiv   %r12d
  movslq %eax,%rdx
  mov    %eax,%edi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%edi
  sar    $0x22,%rdx
  sub    %edi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  lea    0x0(%r13,%rax,4),%rax
  movslq (%rax),%rdx
  mov    %esi,-0x4(%rbp,%rdx,4)
  subl   $0x1,(%rax)
  test   %ecx,%ecx
  jns    L10
L41:
  add    $0x18,%rsp
  mov    %r13,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <free@plt>
  nopl   0x0(%rax)
L13:
  lea    -0x1(%r15),%ecx
  test   %r14d,%r14d
  jg     L79
  jmp    L41
```

assistant: ```c
void counting_sort(int a[], int *b, int size, int k, int unit) {
  int *c;
  c = (int *)malloc(sizeof(int) * size);
  memset(c, 0, (sizeof(int) * size));
  for (int j = 0; j < size; j++) {
    int offset = (a[j] / unit) % 10;
    c[offset] = c[offset] + 1;
  }
  for (int i = 1; i <= k; i++) {
    c[i] = c[i] + c[(i - 1)];
  }
  for (int j = (size - 1); j >= 0; j--) {
    int offset = (a[j] / unit) % 10;
    b[(c[offset] - 1)] = a[j];
    c[offset] = c[offset] - 1;
  }
  free(c);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D10
data_type: string
name: retrieve_data
arguments: label: D13
data_type: string

tool: D10: hehe

tool: D13: %s


assistant: ```c
int main(int argc, char const *argv[]) {
  char *a;
  a = "hehe";
  printf("%s\n", a);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rsi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type: string
name: extract_symbolic_data
arguments: data_label: D15
type: string

tool: D4: hehe

tool: D15: %s


assistant: ```c
int main(int argc, char const *argv[]) {
  char *a;
  a = "hehe";
  printf("%s\n", a);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D15
type: string
name: parse_data
arguments: data_label: D6
type: string

tool: D15: hehe

tool: D6: %s


assistant: ```c
int main(int argc, char const *argv[]) {
  char *a;
  a = "hehe";
  printf("%s\n", a);
  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
  sub    $0x8,%rsp
  lea    D15(%rip),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D15
type_infered: string
name: retrieve_data
arguments: data_label: D1
type_infered: string

tool: D15: hehe

tool: D1: %s


assistant: ```c
int main(int argc, char const *argv[]) {
  char *a;
  a = "hehe";
  printf("%s\n", a);
  return 0;
}
```