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
<dump_date>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    0x14(%rax),%edi
  mov    -0x18(%rbp),%rax
  mov    0x10(%rax),%esi
  mov    -0x18(%rbp),%rax
  mov    0xc(%rax),%r9d
  mov    -0x18(%rbp),%rax
  mov    0x8(%rax),%r8d
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%ecx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  push   %rdi
  push   %rsi
  lea    D4(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x10,%rsp
  mov    -0x8(%rbp),%rax
  leave
  ret
```

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

tool: D4: { %d, %d, %d, %d, %d, %d }

assistant: ```c
char *dump_date(const struct tm *date) {
  char *dump = malloc(80 * sizeof(char));
  sprintf(dump, "{ %d, %d, %d, %d, %d, %d }", date->tm_sec, date->tm_min, date->tm_hour, date->tm_mday, date->tm_mon, date->tm_year);
  return dump;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dump_date>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    0x4(%rbx),%ecx
  mov    0x14(%rbx),%eax
  push   %rax
  mov    0x10(%rbx),%eax
  push   %rax
  mov    0xc(%rbx),%r9d
  mov    0x8(%rbx),%r8d
  mov    (%rbx),%edx
  lea    D5(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: string

tool: D5: { %d, %d, %d, %d, %d, %d }

assistant: ```c
char *dump_date(const struct tm *date) {
  char *dump = malloc(80 * sizeof(char));
  sprintf(dump, "{ %d, %d, %d, %d, %d, %d }", date->tm_sec, date->tm_min, date->tm_hour, date->tm_mday, date->tm_mon, date->tm_year);
  return dump;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dump_date>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x50,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    0x4(%rbx),%ecx
  mov    (%rbx),%edx
  lea    D2(%rip),%rsi
  mov    %rax,%r12
  mov    0x14(%rbx),%eax
  mov    0xc(%rbx),%r9d
  mov    0x8(%rbx),%r8d
  mov    %r12,%rdi
  push   %rax
  mov    0x10(%rbx),%eax
  push   %rax
  call   <sprintf@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
data_type: string

tool: D2: { %d, %d, %d, %d, %d, %d }

assistant: ```c
char *dump_date(const struct tm *date) {
  char *dump = malloc(80 * sizeof(char));
  sprintf(dump, "{ %d, %d, %d, %d, %d, %d }", date->tm_sec, date->tm_min, date->tm_hour, date->tm_mday, date->tm_mon, date->tm_year);
  return dump;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dump_date>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x50,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    0x4(%rbx),%ecx
  mov    (%rbx),%edx
  lea    D9(%rip),%rsi
  mov    %rax,%r12
  mov    0x14(%rbx),%eax
  mov    0xc(%rbx),%r9d
  mov    0x8(%rbx),%r8d
  mov    %r12,%rdi
  push   %rax
  mov    0x10(%rbx),%eax
  push   %rax
  call   <sprintf@plt>
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type: string

tool: D9: { %d, %d, %d, %d, %d, %d }

assistant: ```c
char *dump_date(const struct tm *date) {
  char *dump = malloc(80 * sizeof(char));
  sprintf(dump, "{ %d, %d, %d, %d, %d, %d }", date->tm_sec, date->tm_min, date->tm_hour, date->tm_mday, date->tm_mon, date->tm_year);
  return dump;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<tcp_sum_calc>:
  endbr64
  mov    %rdi,%r10
  mov    %rsi,%rdi
  mov    %ecx,%eax
  mov    D4(%rip),%rcx
  mov    (%rcx),%r11
  mov    $0x0,%ecx
  test   $0x1,%al
  je     L5
  movq   $0x0,(%r8,%r10,8)
  mov    $0x1,%ecx
L5:
  add    %r10,%rcx
  je     L20
  mov    %r8,%rsi
  lea    -0x8(,%rcx,8),%rax
  and    $0xfffffffffffffff0,%rax
  lea    0x10(%r8,%rax,1),%r9
  mov    $0x0,%ecx
L6:
  mov    (%rsi),%rax
  shl    $0x8,%rax
  movzwl %ax,%eax
  movzbl 0x8(%rsi),%r8d
  add    %r8,%rax
  add    %rax,%rcx
  add    $0x10,%rsi
  cmp    %r9,%rsi
  jne    L6
L20:
  movzbl 0x8(%rdi),%eax
  add    %r10,%rax
  add    %r11,%rax
  movzbl 0x18(%rdi),%esi
  add    %rsi,%rax
  mov    (%rdi),%rsi
  shl    $0x8,%rsi
  movzwl %si,%esi
  add    %rsi,%rax
  mov    0x10(%rdi),%rsi
  shl    $0x8,%rsi
  movzwl %si,%esi
  add    %rsi,%rax
  movzbl 0x8(%rdx),%esi
  add    %rsi,%rax
  movzbl 0x18(%rdx),%esi
  add    %rsi,%rax
  mov    (%rdx),%rsi
  shl    $0x8,%rsi
  movzwl %si,%esi
  add    %rsi,%rax
  mov    0x10(%rdx),%rdx
  shl    $0x8,%rdx
  movzwl %dx,%edx
  add    %rdx,%rax
  add    %rcx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  je     L12
L26:
  movzwl %ax,%eax
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  jne    L26
L12:
  not    %rax
  ret
```

assistant: ```c
uint16_t tcp_sum_calc(uint16_t len_tcp, uint16_t src_addr[], uint16_t dest_addr[], uint8_t padding, uint16_t buff[]) {
  uint32_t i;
  uint16_t padd = 0;
  uint16_t word16;
  uint32_t sum = 0;
  uint16_t prot_tcp = IPPROTO_TCP;
  if ((padding & 1) == 1) {
    padd = 1;
    buff[len_tcp] = 0;
  }
  for (i = 0; i < len_tcp + padd; i = i + 2) {
    word16 = ((buff[i] << 8) & 0xFF00) + (buff[i + 1] & 0xFF);
    sum += (unsigned long)word16;
  }
  for (i = 0; i < 4; i = i + 2) {
    word16 = ((src_addr[i] << 8) & 0xFF00) + (src_addr[i + 1] & 0xFF);
    sum += word16;
  }
  for (i = 0; i < 4; i = i + 2) {
    word16 = ((dest_addr[i] << 8) & 0xFF00) + (dest_addr[i + 1] & 0xFF);
    sum += word16;
  }
  sum += (prot_tcp + len_tcp);
  while (sum >> 16)
    sum = (sum & 0xFFFF) + (sum >> 16);
  sum = ~sum;
  return (uint16_t)sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tcp_sum_calc>:
  endbr64
  mov    %ecx,%r9d
  mov    D5(%rip),%rax
  mov    %rdi,%rcx
  and    $0x1,%r9d
  mov    (%rax),%r11
  je     L47
  movq   $0x0,(%r8,%rdi,8)
  lea    0x1(%rdi),%rcx
L47:
  test   %rcx,%rcx
  je     L36
  lea    -0x8(,%rcx,8),%rax
  xor    %ecx,%ecx
  and    $0xfffffffffffffff0,%rax
  lea    0x10(%r8,%rax,1),%r10
  xchg   %ax,%ax
L1:
  mov    (%r8),%rax
  movzbl 0x8(%r8),%r9d
  add    $0x10,%r8
  shl    $0x8,%rax
  movzwl %ax,%eax
  add    %r9,%rax
  add    %rax,%rcx
  cmp    %r8,%r10
  jne    L1
L36:
  movzbl 0x8(%rsi),%eax
  movzbl 0x18(%rsi),%r8d
  add    %r8,%rax
  add    %r11,%rax
  add    %rdi,%rax
  mov    (%rsi),%rdi
  mov    0x10(%rsi),%rsi
  shl    $0x8,%rdi
  shl    $0x8,%rsi
  movzwl %di,%edi
  movzwl %si,%esi
  add    %rdi,%rax
  add    %rsi,%rax
  movzbl 0x8(%rdx),%esi
  add    %rsi,%rax
  movzbl 0x18(%rdx),%esi
  add    %rsi,%rax
  mov    (%rdx),%rsi
  mov    0x10(%rdx),%rdx
  shl    $0x8,%rsi
  shl    $0x8,%rdx
  movzwl %si,%esi
  movzwl %dx,%edx
  add    %rsi,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  je     L15
L28:
  movzwl %ax,%eax
  add    %rdx,%rax
  mov    %rax,%rdx
  shr    $0x10,%rdx
  jne    L28
L15:
  not    %rax
  ret
```

assistant: ```c
uint16_t tcp_sum_calc(uint16_t len_tcp, uint16_t src_addr[], uint16_t dest_addr[], uint8_t padding, uint16_t buff[]) {
  uint32_t i;
  uint16_t padd = 0;
  uint16_t word16;
  uint32_t sum = 0;
  uint16_t prot_tcp = IPPROTO_TCP;
  if ((padding & 1) == 1) {
    padd = 1;
    buff[len_tcp] = 0;
  }
  for (i = 0; i < len_tcp + padd; i = i + 2) {
    word16 = ((buff[i] << 8) & 0xFF00) + (buff[i + 1] & 0xFF);
    sum += (unsigned long)word16;
  }
  for (i = 0; i < 4; i = i + 2) {
    word16 = ((src_addr[i] << 8) & 0xFF00) + (src_addr[i + 1] & 0xFF);
    sum += word16;
  }
  for (i = 0; i < 4; i = i + 2) {
    word16 = ((dest_addr[i] << 8) & 0xFF00) + (dest_addr[i + 1] & 0xFF);
    sum += word16;
  }
  sum += (prot_tcp + len_tcp);
  while (sum >> 16)
    sum = (sum & 0xFFFF) + (sum >> 16);
  sum = ~sum;
  return (uint16_t)sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f_acc_routine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void f_acc_routine(void) {
#pragma acc parallel
  ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<f_acc_routine>:
  endbr64
  ret
```

assistant: ```c
void f_acc_routine(void) {
#pragma acc parallel
  ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f_acc_routine>:
  endbr64
  ret
```

assistant: ```c
void f_acc_routine(void) {
#pragma acc parallel
  ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f_acc_routine>:
  endbr64
  ret
```

assistant: ```c
void f_acc_routine(void) {
#pragma acc parallel
  ;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fillDynamicGrid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %ax,-0x1c(%rbp)
  movw   $0x0,-0x4(%rbp)
  jmp    L3
L1:
  movw   $0x0,-0x2(%rbp)
  jmp    L5
L34:
  movswq -0x4(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  movswq -0x2(%rbp),%rax
  add    %rax,%rax
  add    %rax,%rdx
  movzwl -0x1c(%rbp),%eax
  mov    %ax,(%rdx)
  movzwl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x2(%rbp)
L5:
  cmpw   $0x1c,-0x2(%rbp)
  jle    L34
  movzwl -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x4(%rbp)
L3:
  cmpw   $0x4e,-0x4(%rbp)
  jle    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void fillDynamicGrid(short **grid, short fillValue) {
  short i, j;
  for (i = 0; i < (100 - 20 - 1); i++) {
    for (j = 0; j < ((31 + 3) - 3 - 2); j++) {
      grid[i][j] = fillValue;
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<fillDynamicGrid>:
  endbr64
  mov    %rdi,%rcx
  add    $0x278,%rdi
  jmp    L12
L18:
  add    $0x8,%rcx
  cmp    %rdi,%rcx
  je     L21
L12:
  mov    $0x0,%eax
L17:
  mov    (%rcx),%rdx
  mov    %si,(%rdx,%rax,1)
  add    $0x2,%rax
  cmp    $0x3a,%rax
  jne    L17
  jmp    L18
L21:
  ret
```

assistant: ```c
void fillDynamicGrid(short **grid, short fillValue) {
  short i, j;
  for (i = 0; i < (100 - 20 - 1); i++) {
    for (j = 0; j < ((31 + 3) - 3 - 2); j++) {
      grid[i][j] = fillValue;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fillDynamicGrid>:
  endbr64
  mov    %esi,%ecx
  mov    %rdi,%rsi
  lea    0x278(%rdi),%rdi
L7:
  mov    (%rsi),%rax
  lea    0x3a(%rax),%rdx
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %cx,(%rax)
  add    $0x2,%rax
  cmp    %rdx,%rax
  jne    L13
  add    $0x8,%rsi
  cmp    %rdi,%rsi
  jne    L7
  ret
```

assistant: ```c
void fillDynamicGrid(short **grid, short fillValue) {
  short i, j;
  for (i = 0; i < (100 - 20 - 1); i++) {
    for (j = 0; j < ((31 + 3) - 3 - 2); j++) {
      grid[i][j] = fillValue;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fillDynamicGrid>:
  endbr64
  mov    %esi,%ecx
  mov    %rdi,%rdx
  lea    0x278(%rdi),%rsi
  movd   %ecx,%xmm0
  movd   %ecx,%xmm2
  punpcklwd %xmm0,%xmm0
  pshuflw $0x0,%xmm2,%xmm1
  pshufd $0x0,%xmm0,%xmm0
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    (%rdx),%rax
  add    $0x8,%rdx
  mov    %cx,0x38(%rax)
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  movq   %xmm1,0x30(%rax)
  cmp    %rsi,%rdx
  jne    L4
  ret
```

assistant: ```c
void fillDynamicGrid(short **grid, short fillValue) {
  short i, j;
  for (i = 0; i < (100 - 20 - 1); i++) {
    for (j = 0; j < ((31 + 3) - 3 - 2); j++) {
      grid[i][j] = fillValue;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x1,-0xc(%rbp)
  jmp    L8
L77:
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jge    L72
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L62
L72:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jle    L78
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L62
L78:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  cmp    %eax,%edx
  jne    L62
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L62:
  addl   $0x1,-0xc(%rbp)
L8:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L77
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
type_guessed: string
name: read_data
arguments: data_tag: D47
type_guessed: string
name: read_data
arguments: data_tag: D41
type_guessed: string
name: read_data
arguments: data_tag: D33
type_guessed: string
name: read_data
arguments: data_tag: D0
type_guessed: string

tool: D1: %d

tool: D47: %d %d

tool: D41: <


tool: D33: >


tool: D0: =


assistant: ```c
int main() {
  int i, n, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    if (a < b) {
      printf("<\n");
    } else if (a > b) {
      printf(">\n");
    } else if (a == b) {
      printf("=\n");
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L89
  mov    $0x1,%ebx
  lea    D0(%rip),%rbp
  lea    D12(%rip),%r14
  lea    D14(%rip),%r13
  lea    D20(%rip),%r12
  jmp    L28
L69:
  mov    %r12,%rdi
  call   <printf@plt>
L14:
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L89
L28:
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%eax
  cmp    %eax,%edx
  jl     L69
  jg     L73
  jne    L14
  mov    %r14,%rdi
  call   <printf@plt>
  jmp    L14
L73:
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L14
L89:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D35
type: string
name: read_data
arguments: data_tag: D0
type: string
name: read_data
arguments: data_tag: D12
type: string
name: read_data
arguments: data_tag: D14
type: string
name: read_data
arguments: data_tag: D20
type: string

tool: D35: %d

tool: D0: %d %d

tool: D12: =


tool: D14: >


tool: D20: <


assistant: ```c
int main() {
  int i, n, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    if (a < b) {
      printf("<\n");
    } else if (a > b) {
      printf(">\n");
    } else if (a == b) {
      printf("=\n");
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D34(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L17
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D30(%rip),%rbp
  lea    D14(%rip),%r15
  lea    D4(%rip),%r14
  jmp    L89
  nopl   0x0(%rax)
L1:
  jg     L93
  je     L12
L88:
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L17
L89:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    %eax,0x10(%rsp)
  jge    L1
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jge    L89
L17:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L93:
  mov    %r15,%rdi
  call   <printf@plt>
  jmp    L88
  nopw   0x0(%rax,%rax,1)
L12:
  lea    D41(%rip),%rdi
  call   <printf@plt>
  jmp    L88
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D34
type_infered: string
name: extract_symbolic_data
arguments: data_label: D30
type_infered: string
name: extract_symbolic_data
arguments: data_label: D14
type_infered: string
name: extract_symbolic_data
arguments: data_label: D4
type_infered: string
name: extract_symbolic_data
arguments: data_label: D41
type_infered: string

tool: D34: %d

tool: D30: %d %d

tool: D14: >


tool: D4: <


tool: D41: =


assistant: ```c
int main() {
  int i, n, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    if (a < b) {
      printf("<\n");
    } else if (a > b) {
      printf(">\n");
    } else if (a == b) {
      printf("=\n");
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D41(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L41
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D18(%rip),%rbp
  lea    D42(%rip),%r15
  lea    D22(%rip),%r14
  jmp    L48
  nopl   0x0(%rax)
L26:
  jg     L6
  je     L84
L59:
  add    $0x1,%ebx
  cmp    %ebx,0xc(%rsp)
  jl     L41
L48:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    %eax,0x10(%rsp)
  jge    L26
  mov    %r14,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jge    L48
L41:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L6:
  mov    %r15,%rdi
  call   <printf@plt>
  jmp    L59
  nopw   0x0(%rax,%rax,1)
L84:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D41
type: string
name: unpack_data
arguments: tag: D18
type: string
name: unpack_data
arguments: tag: D42
type: string
name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D5
type: string

tool: D41: %d

tool: D18: %d %d

tool: D42: >


tool: D22: <


tool: D5: =


assistant: ```c
int main() {
  int i, n, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &a, &b);
    if (a < b) {
      printf("<\n");
    } else if (a > b) {
      printf(">\n");
    } else if (a == b) {
      printf("=\n");
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<saferead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L14:
  mov    -0x18(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L86
  mov    D8(%rip),%rax
  mov    (%rax),%rdx
  mov    D2(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L86
  jmp    L29
L86:
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L51
  mov    $0xffffffff,%eax
  jmp    L57
L51:
  cmpl   $0x0,-0x4(%rbp)
  jne    L36
  mov    -0x8(%rbp),%eax
  jmp    L57
L36:
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
L29:
  cmpl   $0x0,-0x18(%rbp)
  jg     L14
  mov    -0x8(%rbp),%eax
L57:
  leave
  ret
```

assistant: ```c
int saferead(int fd, void *buffer, int count) {
  int actual = 0;
  while (count > 0) {
    int n = read(fd, buffer, count);
    if (n == -1 && errno == EINTR)
      continue;
    if (n == -1)
      return -1;
    if (n == 0)
      return actual;
    count -= n;
    actual += n;
    buffer = (char *)buffer + n;
  }
  return actual;
}
```
user: What is the c source code of the assembly code below:

```asm
<saferead>:
  endbr64
  test   %edx,%edx
  jle    L0
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%rbp
  mov    %edx,%ebx
  mov    $0x0,%r12d
  jmp    L30
L69:
  test   %eax,%eax
  je     L47
  sub    %eax,%ebx
  add    %eax,%r12d
  cltq
  add    %rax,%rbp
L21:
  test   %ebx,%ebx
  jle    L9
L30:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  jne    L69
  mov    D15(%rip),%rcx
  mov    D17(%rip),%rdx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L21
L35:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    %r12d,%eax
  jmp    L35
L0:
  mov    $0x0,%eax
  ret
L47:
  mov    %r12d,%eax
  jmp    L35
```

assistant: ```c
int saferead(int fd, void *buffer, int count) {
  int actual = 0;
  while (count > 0) {
    int n = read(fd, buffer, count);
    if (n == -1 && errno == EINTR)
      continue;
    if (n == -1)
      return -1;
    if (n == 0)
      return actual;
    count -= n;
    actual += n;
    buffer = (char *)buffer + n;
  }
  return actual;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<saferead>:
  endbr64
  test   %edx,%edx
  jle    L58
  push   %r13
  mov    %rsi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L34
  xchg   %ax,%ax
L57:
  test   %eax,%eax
  je     L47
  sub    %eax,%ebx
  add    %eax,%r12d
  cltq
  add    %rax,%r13
  test   %ebx,%ebx
  jle    L47
L34:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  jne    L57
  mov    D18(%rip),%rdx
  mov    D19(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L34
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L47:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L58:
  xor    %eax,%eax
  ret
```

assistant: ```c
int saferead(int fd, void *buffer, int count) {
  int actual = 0;
  while (count > 0) {
    int n = read(fd, buffer, count);
    if (n == -1 && errno == EINTR)
      continue;
    if (n == -1)
      return -1;
    if (n == 0)
      return actual;
    count -= n;
    actual += n;
    buffer = (char *)buffer + n;
  }
  return actual;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<saferead>:
  endbr64
  test   %edx,%edx
  jle    L56
  push   %r13
  mov    %rsi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  jmp    L15
  xchg   %ax,%ax
L47:
  test   %eax,%eax
  je     L37
  sub    %eax,%ebx
  add    %eax,%r12d
  cltq
  add    %rax,%r13
  test   %ebx,%ebx
  jle    L37
L15:
  mov    %ebx,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  cmp    $0xffffffff,%eax
  jne    L47
  mov    D18(%rip),%rdx
  mov    D4(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L15
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L37:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L56:
  xor    %eax,%eax
  ret
```

assistant: ```c
int saferead(int fd, void *buffer, int count) {
  int actual = 0;
  while (count > 0) {
    int n = read(fd, buffer, count);
    if (n == -1 && errno == EINTR)
      continue;
    if (n == -1)
      return -1;
    if (n == 0)
      return actual;
    count -= n;
    actual += n;
    buffer = (char *)buffer + n;
  }
  return actual;
}
```