ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<usage>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D3
type: string
name: read_data
arguments: label: D26
type: string
name: read_data
arguments: label: D24
type: string

tool: D3: Usage %s [num_threads] [total_iterations]


tool: D26: %s will call sys_getuid() total_iterations [default 2,000,000]


tool: D24: times divided across num_threads [default 1].


assistant: ```c
void usage(char *name) {
  printf("Usage %s [num_threads] [total_iterations]\n", name);
  printf("%s will call sys_getuid() total_iterations [default 2,000,000]\n", name);
  printf("times divided across num_threads [default 1].\n\n");
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  mov    %rdi,%rbp
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  pop    %rbp
  jmp    1090 <exit@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type_infered: string
name: retrieve_data
arguments: data_tag: D28
type_infered: string
name: retrieve_data
arguments: data_tag: D6
type_infered: string

tool: D18: Usage %s [num_threads] [total_iterations]


tool: D28: %s will call sys_getuid() total_iterations [default 2,000,000]


tool: D6: times divided across num_threads [default 1].


assistant: ```c
void usage(char *name) {
  printf("Usage %s [num_threads] [total_iterations]\n", name);
  printf("%s will call sys_getuid() total_iterations [default 2,000,000]\n", name);
  printf("times divided across num_threads [default 1].\n\n");
  exit(1);
}
```
user: What is the c source code of the assembly code below?

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  mov    %rdi,%rbp
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  pop    %rbp
  jmp    1090 <exit@plt>
```

tool_calls: name: unpack_data
arguments: tag: D26
datatype: string
name: unpack_data
arguments: tag: D5
datatype: string
name: unpack_data
arguments: tag: D8
datatype: string

tool: D26: Usage %s [num_threads] [total_iterations]


tool: D5: %s will call sys_getuid() total_iterations [default 2,000,000]


tool: D8: times divided across num_threads [default 1].


assistant: ```c
void usage(char *name) {
  printf("Usage %s [num_threads] [total_iterations]\n", name);
  printf("%s will call sys_getuid() total_iterations [default 2,000,000]\n", name);
  printf("times divided across num_threads [default 1].\n\n");
  exit(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x4d0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x4d0(%rbp)
  jmp    L93
L101:
  mov    -0x4d0(%rbp),%eax
  cltq
  movl   $0x1,-0x4c0(%rbp,%rax,4)
  addl   $0x1,-0x4d0(%rbp)
L93:
  cmpl   $0x12c,-0x4d0(%rbp)
  jle    L101
  movl   $0x2,-0x4cc(%rbp)
  jmp    L8
L85:
  mov    -0x4cc(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x4c8(%rbp)
  jmp    L30
L54:
  mov    -0x4c8(%rbp),%eax
  cltd
  idivl  -0x4cc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L86
  mov    -0x4c8(%rbp),%eax
  cltq
  movl   $0x0,-0x4c0(%rbp,%rax,4)
L86:
  addl   $0x1,-0x4c8(%rbp)
L30:
  cmpl   $0x12c,-0x4c8(%rbp)
  jle    L54
  addl   $0x1,-0x4cc(%rbp)
L8:
  cmpl   $0x12c,-0x4cc(%rbp)
  jle    L85
  movl   $0x1,-0x4c4(%rbp)
  jmp    L106
L47:
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  test   %eax,%eax
  setle  %al
  movzbl %al,%eax
  test   %eax,%eax
  jne    L94
  mov    -0x4c4(%rbp),%eax
  cltq
  mov    -0x4c0(%rbp,%rax,4),%eax
  imul   -0x4c4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L94:
  addl   $0x1,-0x4c4(%rbp)
L106:
  cmpl   $0x12c,-0x4c4(%rbp)
  jle    L47
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_guessed: string

tool: D5: %d


assistant: ```c
int main() {
  int primes[301];
  for (int i = 1; i <= 300; i++)
    primes[i] = 1;
  for (int i = 2; i <= 300; i++) {
    for (int j = i * 2; j <= 300; j++) {
      if (j % i == 0)
        primes[j] = 0;
    }
  }
  for (int i = 1; i <= 300; i++)
    primes[i] <= 0 ?: printf("%d\n", primes[i] * i);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x4c8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4b8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rax
  lea    0x4b4(%rsp),%rdx
L85:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L85
  mov    $0x4,%edi
  mov    $0x2,%esi
  jmp    L80
L99:
  add    $0x1,%rcx
  cmp    $0x12c,%ecx
  jg     L39
L112:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L99
  movl   $0x0,(%rsp,%rcx,4)
  jmp    L99
L39:
  add    $0x1,%esi
  add    $0x2,%rdi
  cmp    $0x12d,%esi
  je     L1
L80:
  cmp    $0x12c,%edi
  jg     L39
  mov    %rdi,%rcx
  jmp    L112
L1:
  mov    $0x1,%ebx
  lea    D1(%rip),%rbp
  jmp    L55
L56:
  add    $0x1,%rbx
  cmp    $0x12d,%rbx
  je     L7
L55:
  mov    (%rsp,%rbx,4),%esi
  test   %esi,%esi
  jle    L56
  imul   %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L56
L7:
  mov    0x4b8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  mov    $0x0,%eax
  add    $0x4c8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D1
type: string

tool: D1: %d


assistant: ```c
int main() {
  int primes[301];
  for (int i = 1; i <= 300; i++)
    primes[i] = 1;
  for (int i = 2; i <= 300; i++) {
    for (int j = i * 2; j <= 300; j++) {
      if (j % i == 0)
        primes[j] = 0;
    }
  }
  for (int i = 1; i <= 300; i++)
    primes[i] <= 0 ?: printf("%d\n", primes[i] * i);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x4c0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4b8(%rsp)
  xor    %eax,%eax
  lea    0x4b4(%rsp),%rdx
  lea    0x4(%rsp),%rax
  nop
L23:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L23
  mov    $0x4,%edi
  mov    $0x2,%esi
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L38:
  cmp    $0x12c,%rdi
  ja     L71
  mov    %rdi,%rcx
  nopl   0x0(%rax)
L44:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L0
  movl   $0x0,(%rbx,%rcx,4)
L0:
  add    $0x1,%rcx
  cmp    $0x12c,%ecx
  jle    L44
L71:
  add    $0x1,%esi
  add    $0x2,%rdi
  cmp    $0x12d,%esi
  jne    L38
  mov    $0x1,%ebp
  lea    D2(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L60:
  mov    (%rbx,%rbp,4),%esi
  test   %esi,%esi
  jle    L42
  imul   %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
L42:
  add    $0x1,%rbp
  cmp    $0x12d,%rbp
  jne    L60
  mov    0x4b8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x4c0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
type: string

tool: D2: %d


assistant: ```c
int main() {
  int primes[301];
  for (int i = 1; i <= 300; i++)
    primes[i] = 1;
  for (int i = 2; i <= 300; i++) {
    for (int j = i * 2; j <= 300; j++) {
      if (j % i == 0)
        primes[j] = 0;
    }
  }
  for (int i = 1; i <= 300; i++)
    primes[i] <= 0 ?: printf("%d\n", primes[i] * i);
  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
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x4c0,%rsp
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x4b8(%rsp)
  xor    %eax,%eax
  lea    0x4b4(%rsp),%rdx
  lea    0x4(%rsp),%rax
  nopw   0x0(%rax,%rax,1)
L21:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L21
  mov    $0x4,%edi
  mov    $0x2,%esi
  mov    %rsp,%rbx
  nopl   0x0(%rax)
L5:
  cmp    $0x12c,%rdi
  ja     L47
  mov    %rdi,%rcx
  nopl   0x0(%rax)
L54:
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L10
  movl   $0x0,(%rbx,%rcx,4)
L10:
  add    $0x1,%rcx
  cmp    $0x12c,%ecx
  jle    L54
L47:
  add    $0x1,%esi
  add    $0x2,%rdi
  cmp    $0x12d,%esi
  jne    L5
  mov    $0x1,%ebp
  lea    D12(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L19:
  mov    (%rbx,%rbp,4),%esi
  test   %esi,%esi
  jle    L18
  imul   %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
L18:
  add    $0x1,%rbp
  cmp    $0x12d,%rbp
  jne    L19
  mov    0x4b8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x4c0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D19
datatype: byte[16]
name: retrieve_data
arguments: tag: D12
datatype: string

tool: {"D19": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D12: %d


assistant: ```c
int main() {
  int primes[301];
  for (int i = 1; i <= 300; i++)
    primes[i] = 1;
  for (int i = 2; i <= 300; i++) {
    for (int j = i * 2; j <= 300; j++) {
      if (j % i == 0)
        primes[j] = 0;
    }
  }
  for (int i = 1; i <= 300; i++)
    primes[i] <= 0 ?: printf("%d\n", primes[i] * i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  shl    $0x14,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L33
  mov    $0x1,%edi
  call   <exit@plt>
L33:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L6:
  movl   $0x0,-0x14(%rbp)
  jmp    L27
L8:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  cltq
  mov    -0x10(%rbp),%rdx
  shr    $0x2,%rdx
  cmp    %rdx,%rax
  jb     L8
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
datatype: string
name: extract_symbolic_data
arguments: label: D6
datatype: string

tool: D10: Allocated %zu bytes at %p


tool: D6: Looped

assistant: ```c
int main(int argc, char *argv[]) {
  size_t nbytes = atoi(argv[1]) * 1024l * 1024;
  int *a = malloc(nbytes);
  int i;
  if (a == NULL)
    exit(1);
  printf("Allocated %zu bytes at %p\n", nbytes, a);
  for (;;) {
    for (i = 0; i < nbytes / sizeof(int); i++)
      a[i] = i;
    printf("Looped\n");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  shl    $0x14,%rax
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L25
L19:
  mov    %rbx,%rdx
  mov    %r12,%rsi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rbp
  shr    $0x2,%rbp
  lea    D12(%rip),%r13
  jmp    L51
L25:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L19
L47:
  mov    %r13,%rdi
  call   <puts@plt>
L51:
  cmp    $0x3,%r12
  jbe    L47
  mov    $0x0,%eax
L42:
  mov    %eax,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    %rbp,%rax
  jb     L42
  jmp    L47
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type_guessed: string
name: retrieve_data
arguments: data_tag: D12
type_guessed: string

tool: D18: Allocated %zu bytes at %p


tool: D12: Looped

assistant: ```c
int main(int argc, char *argv[]) {
  size_t nbytes = atoi(argv[1]) * 1024l * 1024;
  int *a = malloc(nbytes);
  int i;
  if (a == NULL)
    exit(1);
  printf("Allocated %zu bytes at %p\n", nbytes, a);
  for (;;) {
    for (i = 0; i < nbytes / sizeof(int); i++)
      a[i] = i;
    printf("Looped\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  shl    $0x14,%rax
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L18
L45:
  mov    %rbx,%rsi
  mov    %rbp,%rdx
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  shr    $0x2,%rbx
  lea    D1(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L24:
  test   %rbx,%rbx
  je     L42
  xor    %eax,%eax
L35:
  mov    %eax,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %rbx,%rax
  jne    L35
L42:
  mov    %r12,%rdi
  call   <puts@plt>
  jmp    L24
L18:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L45
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D18
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D1
type_guessed: string

tool: D18: Allocated %zu bytes at %p


tool: D1: Looped

assistant: ```c
int main(int argc, char *argv[]) {
  size_t nbytes = atoi(argv[1]) * 1024l * 1024;
  int *a = malloc(nbytes);
  int i;
  if (a == NULL)
    exit(1);
  printf("Allocated %zu bytes at %p\n", nbytes, a);
  for (;;) {
    for (i = 0; i < nbytes / sizeof(int); i++)
      a[i] = i;
    printf("Looped\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  shl    $0x14,%rax
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L14
L3:
  mov    %rbx,%rsi
  mov    %rbx,%r12
  mov    %r13,%rdx
  xor    %eax,%eax
  lea    D33(%rip),%rdi
  shr    $0x2,%r12
  add    %r13,%rbx
  call   <printf@plt>
  lea    D6(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L49:
  test   %r12,%r12
  je     L41
  movdqa D19(%rip),%xmm0
  mov    %r13,%rax
L17:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddd  D23(%rip),%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rbx,%rax
  jne    L17
L41:
  mov    %rbp,%rdi
  call   <puts@plt>
  jmp    L49
L14:
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L3
```

tool_calls: name: read_data
arguments: data_label: D33
datatype: string
name: read_data
arguments: data_label: D6
datatype: string
name: read_data
arguments: data_label: D19
datatype: byte[16]
name: read_data
arguments: data_label: D23
datatype: byte[16]

tool: D33: Allocated %zu bytes at %p


tool: D6: Looped

tool: {"D19": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D23": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char *argv[]) {
  size_t nbytes = atoi(argv[1]) * 1024l * 1024;
  int *a = malloc(nbytes);
  int i;
  if (a == NULL)
    exit(1);
  printf("Allocated %zu bytes at %p\n", nbytes, a);
  for (;;) {
    for (i = 0; i < nbytes / sizeof(int); i++)
      a[i] = i;
    printf("Looped\n");
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<generate_macro_add_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D14(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D14
datatype: string
name: parse_data
arguments: data_tag: D19
datatype: string

tool: D14: void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { 
 	int i, j; 
 	if(strcmp("int", param_type) == 0) { 
 		int** mat; 
 		int** mat1 = (int**) mat_1; 
 		int** mat2 = (int**) mat_2; 
  		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) {  
 			mat[i] = calloc(columns, sizeof(int)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	else { 
 		double** mat; 
 		double** mat1 = (double**)mat_1; 
 		double** mat2 = (double**)mat_2; 
 		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) { 
 			mat[i] = calloc(columns, sizeof(double)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	void** mat = 0; 
 	return mat; 
 }


tool: D19: %s

assistant: ```c
void generate_macro_add_matrix() {
  char *macro = "void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { \n 	int i, j; \n 	if(strcmp(\"int\", param_type) == 0) { \n 		int** mat; \n 		int** mat1 = (int**) mat_1; \n 		int** mat2 = (int**) mat_2; \n  		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) {  \n 			mat[i] = calloc(columns, sizeof(int)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	else { \n 		double** mat; \n 		double** mat1 = (double**)mat_1; \n 		double** mat2 = (double**)mat_2; \n 		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) { \n 			mat[i] = calloc(columns, "
                "sizeof(double)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	void** mat = 0; \n 	return mat; \n }\n";
  printf("%s", macro);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<generate_macro_add_matrix>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D10
datatype: string
name: unpack_data
arguments: label: D19
datatype: string

tool: D10: void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { 
 	int i, j; 
 	if(strcmp("int", param_type) == 0) { 
 		int** mat; 
 		int** mat1 = (int**) mat_1; 
 		int** mat2 = (int**) mat_2; 
  		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) {  
 			mat[i] = calloc(columns, sizeof(int)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	else { 
 		double** mat; 
 		double** mat1 = (double**)mat_1; 
 		double** mat2 = (double**)mat_2; 
 		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) { 
 			mat[i] = calloc(columns, sizeof(double)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	void** mat = 0; 
 	return mat; 
 }


tool: D19: %s

assistant: ```c
void generate_macro_add_matrix() {
  char *macro = "void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { \n 	int i, j; \n 	if(strcmp(\"int\", param_type) == 0) { \n 		int** mat; \n 		int** mat1 = (int**) mat_1; \n 		int** mat2 = (int**) mat_2; \n  		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) {  \n 			mat[i] = calloc(columns, sizeof(int)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	else { \n 		double** mat; \n 		double** mat1 = (double**)mat_1; \n 		double** mat2 = (double**)mat_2; \n 		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) { \n 			mat[i] = calloc(columns, "
                "sizeof(double)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	void** mat = 0; \n 	return mat; \n }\n";
  printf("%s", macro);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<generate_macro_add_matrix>:
  endbr64
  lea    D9(%rip),%rsi
  lea    D4(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_label: D9
type_guessed: string
name: read_data
arguments: data_label: D4
type_guessed: string

tool: D9: void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { 
 	int i, j; 
 	if(strcmp("int", param_type) == 0) { 
 		int** mat; 
 		int** mat1 = (int**) mat_1; 
 		int** mat2 = (int**) mat_2; 
  		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) {  
 			mat[i] = calloc(columns, sizeof(int)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	else { 
 		double** mat; 
 		double** mat1 = (double**)mat_1; 
 		double** mat2 = (double**)mat_2; 
 		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) { 
 			mat[i] = calloc(columns, sizeof(double)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	void** mat = 0; 
 	return mat; 
 }


tool: D4: %s

assistant: ```c
void generate_macro_add_matrix() {
  char *macro = "void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { \n 	int i, j; \n 	if(strcmp(\"int\", param_type) == 0) { \n 		int** mat; \n 		int** mat1 = (int**) mat_1; \n 		int** mat2 = (int**) mat_2; \n  		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) {  \n 			mat[i] = calloc(columns, sizeof(int)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	else { \n 		double** mat; \n 		double** mat1 = (double**)mat_1; \n 		double** mat2 = (double**)mat_2; \n 		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) { \n 			mat[i] = calloc(columns, "
                "sizeof(double)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	void** mat = 0; \n 	return mat; \n }\n";
  printf("%s", macro);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<generate_macro_add_matrix>:
  endbr64
  lea    D12(%rip),%rsi
  lea    D10(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type: string
name: retrieve_data
arguments: data_label: D10
type: string

tool: D12: void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { 
 	int i, j; 
 	if(strcmp("int", param_type) == 0) { 
 		int** mat; 
 		int** mat1 = (int**) mat_1; 
 		int** mat2 = (int**) mat_2; 
  		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) {  
 			mat[i] = calloc(columns, sizeof(int)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	else { 
 		double** mat; 
 		double** mat1 = (double**)mat_1; 
 		double** mat2 = (double**)mat_2; 
 		mat = calloc(rows, sizeof(void*)); 
 		for(i=0; i<rows; i++) { 
 			mat[i] = calloc(columns, sizeof(double)); 
 			for(j=0; j<columns; j++) { 
 				mat[i][j] = mat1[i][j] + mat2[i][j]; 
 			} 
 		} 
 		return mat; 
 	} 
 	void** mat = 0; 
 	return mat; 
 }


tool: D10: %s

assistant: ```c
void generate_macro_add_matrix() {
  char *macro = "void** add_matrix(char const* param_type, void** mat_1, void**  mat_2, int rows, int columns) { \n 	int i, j; \n 	if(strcmp(\"int\", param_type) == 0) { \n 		int** mat; \n 		int** mat1 = (int**) mat_1; \n 		int** mat2 = (int**) mat_2; \n  		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) {  \n 			mat[i] = calloc(columns, sizeof(int)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	else { \n 		double** mat; \n 		double** mat1 = (double**)mat_1; \n 		double** mat2 = (double**)mat_2; \n 		mat = calloc(rows, sizeof(void*)); \n 		for(i=0; i<rows; i++) { \n 			mat[i] = calloc(columns, "
                "sizeof(double)); \n 			for(j=0; j<columns; j++) { \n 				mat[i][j] = mat1[i][j] + mat2[i][j]; \n 			} \n 		} \n 		return mat; \n 	} \n 	void** mat = 0; \n 	return mat; \n }\n";
  printf("%s", macro);
}
```