ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D28(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L27
  mov    $0x0,%r12d
  lea    0x10(%rsp),%r14
  lea    D28(%rip),%rbp
  mov    $0x0,%r15d
  lea    D9(%rip),%r13
  jmp    L30
L34:
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x10(%rsp)
  test   %eax,%eax
  je     L87
L6:
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x14(%rsp),%edi
  call   <Isprime@plt>
  test   %rax,%rax
  je     L34
  add    0x14(%rsp),%ebx
  jmp    L34
L87:
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%r12d
  cmp    %r12d,0xc(%rsp)
  jle    L27
L30:
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x10(%rsp),%ebx
  lea    -0x1(%rbx),%eax
  mov    %eax,0x10(%rsp)
  test   %ebx,%ebx
  je     L87
  mov    %r15d,%ebx
  jmp    L6
L27:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L89:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D28
type_guessed: string
name: read_data
arguments: label: D9
type_guessed: string

tool: D28: %d

tool: D9: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int gp, i, count;
  scanf("%d", &gp);
  getchar();
  for (i = 0; i < gp; i++) {
    int num, total = 0;
    scanf("%d", &count);
    getchar();
    while (count--) {
      scanf("%d", &num);
      getchar();
      if (Isprime(num))
        total += num;
    }
    printf("%d\n", total);
  }
  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
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L4
  xor    %r13d,%r13d
  lea    0x10(%rsp),%r15
  lea    D4(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L35:
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x10(%rsp),%ebp
  lea    -0x1(%rbp),%eax
  mov    %eax,0x10(%rsp)
  test   %ebp,%ebp
  je     L14
  xor    %ebp,%ebp
  lea    0x14(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L24:
  mov    %rbx,%rdi
  mov    %r12,%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x14(%rsp),%edi
  call   <Isprime@plt>
  test   %rax,%rax
  je     L15
  add    0x14(%rsp),%ebp
L15:
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x10(%rsp)
  test   %eax,%eax
  jne    L24
L14:
  mov    %ebp,%esi
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <printf@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L35
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D11
data_type: string
name: disassemble_data
arguments: data_tag: D4
data_type: string

tool: D11: %d

tool: D4: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int gp, i, count;
  scanf("%d", &gp);
  getchar();
  for (i = 0; i < gp; i++) {
    int num, total = 0;
    scanf("%d", &count);
    getchar();
    while (count--) {
      scanf("%d", &num);
      getchar();
      if (Isprime(num))
        total += num;
    }
    printf("%d\n", total);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L52
  xor    %r13d,%r13d
  lea    0x10(%rsp),%r15
  lea    D19(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x10(%rsp),%ebp
  lea    -0x1(%rbp),%eax
  mov    %eax,0x10(%rsp)
  test   %ebp,%ebp
  je     L12
  xor    %ebp,%ebp
  lea    0x14(%rsp),%r12
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %rbx,%rdi
  mov    %r12,%rsi
  call   <scanf@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x14(%rsp),%edi
  call   <Isprime@plt>
  test   %rax,%rax
  je     L14
  add    0x14(%rsp),%ebp
L14:
  mov    0x10(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x10(%rsp)
  test   %eax,%eax
  jne    L4
L12:
  mov    %ebp,%esi
  mov    %r14,%rdi
  add    $0x1,%r13d
  call   <printf@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L11
L52:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L53
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D4
type_guessed: string
name: parse_data
arguments: label: D19
type_guessed: string

tool: D4: %d

tool: D19: %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int gp, i, count;
  scanf("%d", &gp);
  getchar();
  for (i = 0; i < gp; i++) {
    int num, total = 0;
    scanf("%d", &count);
    getchar();
    while (count--) {
      scanf("%d", &num);
      getchar();
      if (Isprime(num))
        total += num;
    }
    printf("%d\n", total);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<string_concat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L7
  cmpq   $0x0,-0x30(%rbp)
  jne    L46
L7:
  mov    $0x0,%eax
  jmp    L19
L46:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L8
  mov    $0x0,%eax
  jmp    L19
L8:
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x18(%rbp),%rax
L19:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *string_concat(const char *string1, const char *string2) {
  if (string1 == NULL || string2 == NULL) {
    return NULL;
  }
  char *result = malloc(strlen(string1) + strlen(string2) + 1);
  if (result == NULL) {
    return NULL;
  }
  strcpy(result, string1);
  strcat(result, string2);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_concat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L5
  mov    %rdi,%rbp
  mov    %rsi,%r12
  test   %rsi,%rsi
  je     L5
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L24
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
L24:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x0,%ebx
  jmp    L24
```

assistant: ```c
char *string_concat(const char *string1, const char *string2) {
  if (string1 == NULL || string2 == NULL) {
    return NULL;
  }
  char *result = malloc(strlen(string1) + strlen(string2) + 1);
  if (result == NULL) {
    return NULL;
  }
  strcpy(result, string1);
  strcat(result, string2);
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_concat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L2
  mov    %rsi,%r13
  test   %rsi,%rsi
  je     L2
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x1(%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L28
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
L28:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_concat(const char *string1, const char *string2) {
  if (string1 == NULL || string2 == NULL) {
    return NULL;
  }
  char *result = malloc(strlen(string1) + strlen(string2) + 1);
  if (result == NULL) {
    return NULL;
  }
  strcpy(result, string1);
  strcat(result, string2);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_concat>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  test   %rdi,%rdi
  je     L1
  mov    %rsi,%r13
  test   %rsi,%rsi
  je     L1
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    0x1(%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L22
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
L22:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_concat(const char *string1, const char *string2) {
  if (string1 == NULL || string2 == NULL) {
    return NULL;
  }
  char *result = malloc(strlen(string1) + strlen(string2) + 1);
  if (result == NULL) {
    return NULL;
  }
  strcpy(result, string1);
  strcat(result, string2);
  return result;
}
```
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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L14:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x10(%rbp),%eax
  imul   %eax,%edx
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x10(%rbp),%eax
  mov    %edx,%ecx
  mov    %esi,%edx
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L15:
  cmpl   $0x9,-0xc(%rbp)
  jle    L14
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D13
type_infered: string
name: retrieve_data
arguments: tag: D16
type_infered: string

tool: D13: %d

tool: D16: %d x %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) {
    printf("%d x %d = %d\n", n, i + 1, n * (i + 1));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    $0x0,%ebx
  lea    D17(%rip),%rbp
L14:
  add    $0x1,%ebx
  mov    0x4(%rsp),%esi
  mov    %ebx,%ecx
  imul   %esi,%ecx
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L14
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
datatype: string
name: disassemble_data
arguments: data_tag: D17
datatype: string

tool: D3: %d

tool: D17: %d x %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) {
    printf("%d x %d = %d\n", n, i + 1, n * (i + 1));
  }
  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   %rbp
  lea    D19(%rip),%rdi
  lea    D14(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  nopl   0x0(%rax)
L4:
  add    $0x1,%ebx
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  mov    %ebx,%ecx
  mov    %ebx,%edx
  imul   %esi,%ecx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L4
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D19
datatype: string
name: retrieve_data
arguments: tag: D14
datatype: string

tool: D19: %d

tool: D14: %d x %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) {
    printf("%d x %d = %d\n", n, i + 1, n * (i + 1));
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D10(%rip),%rdi
  lea    D5(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  nopl   0x0(%rax)
L18:
  add    $0x1,%ebx
  mov    0x4(%rsp),%esi
  mov    %rbp,%rdi
  mov    %ebx,%ecx
  mov    %ebx,%edx
  imul   %esi,%ecx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L18
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string

tool: D10: %d

tool: D5: %d x %d = %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 10; i++) {
    printf("%d x %d = %d\n", n, i + 1, n * (i + 1));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printData>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

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

tool: D0: 
Estimated area: %f sq. units



assistant: ```c
void printData(double area) {
  printf("\nEstimated area: %f sq. units\n\n", area);
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<printData>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
datatype: string

tool: D5: 
Estimated area: %f sq. units



assistant: ```c
void printData(double area) {
  printf("\nEstimated area: %f sq. units\n\n", area);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<printData>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string

tool: D5: 
Estimated area: %f sq. units



assistant: ```c
void printData(double area) {
  printf("\nEstimated area: %f sq. units\n\n", area);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<printData>:
  endbr64
  lea    D1(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
datatype: string

tool: D1: 
Estimated area: %f sq. units



assistant: ```c
void printData(double area) {
  printf("\nEstimated area: %f sq. units\n\n", area);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<palindrom>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L22:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%ebx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  sub    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,%bl
  je     L1
  mov    $0x0,%eax
  jmp    L29
L1:
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x14(%rbp)
  jl     L22
  mov    $0x1,%eax
L29:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int palindrom(char *s) {
  int i;
  for (i = 0; i < strlen(s) / 2; i++) {
    if (*(s + i) != *(s + (strlen(s) - 1 - i))) {
      return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<palindrom>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%ebx
L11:
  mov    %ebx,%r13d
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%eax
  sar    %eax
  cmp    %ebx,%eax
  jle    L4
  movzbl 0x0(%rbp,%rbx,1),%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    $0x1,%rbx
  sub    $0x1,%eax
  sub    %r13d,%eax
  cltq
  cmp    0x0(%rbp,%rax,1),%r12b
  je     L11
  mov    $0x0,%eax
  jmp    L26
L4:
  mov    $0x1,%eax
L26:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int palindrom(char *s) {
  int i;
  for (i = 0; i < strlen(s) / 2; i++) {
    if (*(s + i) != *(s + (strlen(s) - 1 - i))) {
      return 0;
    }
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<palindrom>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L6
  nopl   (%rax)
L7:
  mov    %rbp,%rdi
  movzbl 0x0(%rbp,%rbx,1),%r13d
  add    $0x1,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  sub    %r12d,%eax
  cltq
  cmp    0x0(%rbp,%rax,1),%r13b
  jne    L4
L6:
  mov    %rbp,%rdi
  mov    %ebx,%r12d
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %ebx,%eax
  jg     L7
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L4:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int palindrom(char *s) {
  int i;
  for (i = 0; i < strlen(s) / 2; i++) {
    if (*(s + i) != *(s + (strlen(s) - 1 - i))) {
      return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<palindrom>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L0
  nopl   (%rax)
L27:
  mov    %rbp,%rdi
  movzbl 0x0(%rbp,%rbx,1),%r13d
  add    $0x1,%rbx
  call   <strlen@plt>
  sub    $0x1,%eax
  sub    %r12d,%eax
  cltq
  cmp    0x0(%rbp,%rax,1),%r13b
  jne    L8
L0:
  mov    %rbp,%rdi
  mov    %ebx,%r12d
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %ebx,%eax
  jg     L27
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int palindrom(char *s) {
  int i;
  for (i = 0; i < strlen(s) / 2; i++) {
    if (*(s + i) != *(s + (strlen(s) - 1 - i))) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lummat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movss  D15(%rip),%xmm0
  movss  %xmm0,-0x5c(%rbp)
  movss  D26(%rip),%xmm0
  movss  %xmm0,-0x58(%rbp)
  movss  D19(%rip),%xmm0
  movss  %xmm0,-0x54(%rbp)
  movss  -0x5c(%rbp),%xmm0
  movss  %xmm0,-0x50(%rbp)
  movss  -0x5c(%rbp),%xmm0
  movss  %xmm0,-0x4c(%rbp)
  movss  -0x5c(%rbp),%xmm0
  movss  %xmm0,-0x48(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x44(%rbp)
  movss  -0x58(%rbp),%xmm0
  movss  %xmm0,-0x40(%rbp)
  movss  -0x58(%rbp),%xmm0
  movss  %xmm0,-0x3c(%rbp)
  movss  -0x58(%rbp),%xmm0
  movss  %xmm0,-0x38(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x34(%rbp)
  movss  -0x54(%rbp),%xmm0
  movss  %xmm0,-0x30(%rbp)
  movss  -0x54(%rbp),%xmm0
  movss  %xmm0,-0x2c(%rbp)
  movss  -0x54(%rbp),%xmm0
  movss  %xmm0,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x20(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  D9(%rip),%xmm0
  movss  %xmm0,-0x14(%rbp)
  mov    -0x68(%rbp),%rdx
  mov    -0x68(%rbp),%rcx
  lea    -0x50(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <matrixmult@plt>
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
data_type: f32
name: extract_symbolic_data
arguments: data_label: D26
data_type: f32
name: extract_symbolic_data
arguments: data_label: D19
data_type: f32
name: extract_symbolic_data
arguments: data_label: D9
data_type: f32

tool: D15: 0.3086

tool: D26: 0.6094

tool: D19: 0.082

tool: D9: 1

assistant: ```c
lummat(mat) float mat[4][4];
{
  float mmat[4][4];
  float rwgt, gwgt, bwgt;
  rwgt = (0.3086);
  gwgt = (0.6094);
  bwgt = (0.0820);
  mmat[0][0] = rwgt;
  mmat[0][1] = rwgt;
  mmat[0][2] = rwgt;
  mmat[0][3] = 0.0;
  mmat[1][0] = gwgt;
  mmat[1][1] = gwgt;
  mmat[1][2] = gwgt;
  mmat[1][3] = 0.0;
  mmat[2][0] = bwgt;
  mmat[2][1] = bwgt;
  mmat[2][2] = bwgt;
  mmat[2][3] = 0.0;
  mmat[3][0] = 0.0;
  mmat[3][1] = 0.0;
  mmat[3][2] = 0.0;
  mmat[3][3] = 1.0;
  matrixmult(mmat, mat, mat);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lummat>:
  endbr64
  sub    $0x58,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movss  D23(%rip),%xmm0
  movss  %xmm0,(%rsp)
  movss  %xmm0,0x4(%rsp)
  movss  %xmm0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movss  D11(%rip),%xmm0
  movss  %xmm0,0x10(%rsp)
  movss  %xmm0,0x14(%rsp)
  movss  %xmm0,0x18(%rsp)
  movl   $0x0,0x1c(%rsp)
  movss  D9(%rip),%xmm0
  movss  %xmm0,0x20(%rsp)
  movss  %xmm0,0x24(%rsp)
  movss  %xmm0,0x28(%rsp)
  movl   $0x0,0x2c(%rsp)
  movl   $0x0,0x30(%rsp)
  movl   $0x0,0x34(%rsp)
  movl   $0x0,0x38(%rsp)
  movl   $0x3f800000,0x3c(%rsp)
  mov    %rsp,%rdi
  mov    %rsi,%rdx
  call   <matrixmult@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D23
data_type: f32
name: retrieve_data
arguments: label: D11
data_type: f32
name: retrieve_data
arguments: label: D9
data_type: f32

tool: D23: 0.3086

tool: D11: 0.6094

tool: D9: 0.082

assistant: ```c
lummat(mat) float mat[4][4];
{
  float mmat[4][4];
  float rwgt, gwgt, bwgt;
  rwgt = (0.3086);
  gwgt = (0.6094);
  bwgt = (0.0820);
  mmat[0][0] = rwgt;
  mmat[0][1] = rwgt;
  mmat[0][2] = rwgt;
  mmat[0][3] = 0.0;
  mmat[1][0] = gwgt;
  mmat[1][1] = gwgt;
  mmat[1][2] = gwgt;
  mmat[1][3] = 0.0;
  mmat[2][0] = bwgt;
  mmat[2][1] = bwgt;
  mmat[2][2] = bwgt;
  mmat[2][3] = 0.0;
  mmat[3][0] = 0.0;
  mmat[3][1] = 0.0;
  mmat[3][2] = 0.0;
  mmat[3][3] = 1.0;
  matrixmult(mmat, mat, mat);
}
```