ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<read_into_matrix>:
  endbr64
  test   %esi,%esi
  jle    L57
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%r14d
  lea    -0x1(%rdx),%r13d
  lea    0x4(%rdi,%r13,4),%rbp
  lea    -0x1(%rsi),%eax
  lea    (%rax,%rax,4),%rax
  lea    0x0(%r13,%rax,2),%rax
  lea    0x2c(%rdi,%rax,4),%r15
  not    %r13
  shl    $0x2,%r13
  lea    D6(%rip),%r12
  jmp    L44
L1:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L1
L38:
  add    $0x28,%rbp
  cmp    %r15,%rbp
  je     L41
L44:
  lea    0x0(%r13,%rbp,1),%rbx
  test   %r14d,%r14d
  jg     L1
  jmp    L38
L41:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type: string

tool: D6: %d

assistant: ```c
void read_into_matrix(int a[][10], int m, int n) {
  for (int i = 0; i <= m - 1; i++) {
    for (int j = 0; j <= n - 1; j++) {
      scanf("%d", &a[i][j]);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_into_matrix>:
  endbr64
  test   %esi,%esi
  jle    L13
  push   %r15
  lea    -0x1(%rsi),%ecx
  push   %r14
  lea    (%rcx,%rcx,4),%rcx
  mov    $0xfffffffffffffffc,%r14
  push   %r13
  mov    %edx,%r13d
  lea    -0x1(%rdx),%edx
  push   %r12
  lea    0x0(,%rdx,4),%rax
  lea    (%rdx,%rcx,2),%rdx
  push   %rbp
  lea    0x2c(%rdi,%rdx,4),%r15
  lea    0x4(%rdi,%rax,1),%rbp
  sub    %rax,%r14
  push   %rbx
  lea    D9(%rip),%r12
  sub    $0x8,%rsp
  xchg   %ax,%ax
L19:
  lea    (%r14,%rbp,1),%rbx
  test   %r13d,%r13d
  jle    L18
  nopl   0x0(%rax)
L20:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L20
L18:
  add    $0x28,%rbp
  cmp    %r15,%rbp
  jne    L19
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_guessed: string

tool: D9: %d

assistant: ```c
void read_into_matrix(int a[][10], int m, int n) {
  for (int i = 0; i <= m - 1; i++) {
    for (int j = 0; j <= n - 1; j++) {
      scanf("%d", &a[i][j]);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_into_matrix>:
  endbr64
  test   %esi,%esi
  jle    L36
  test   %edx,%edx
  jle    L36
  lea    -0x1(%rsi),%ecx
  push   %r14
  sub    $0x1,%edx
  push   %r13
  lea    (%rcx,%rcx,4),%rcx
  mov    $0xfffffffffffffffc,%r13
  lea    0x0(,%rdx,4),%rax
  push   %r12
  lea    (%rdx,%rcx,2),%rdx
  lea    D8(%rip),%r12
  sub    %rax,%r13
  push   %rbp
  lea    0x2c(%rdi,%rdx,4),%r14
  lea    0x4(%rdi,%rax,1),%rbp
  push   %rbx
  nopl   (%rax)
L24:
  lea    0x0(%r13,%rbp,1),%rbx
  nopl   (%rax)
L34:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    %rbp,%rbx
  jne    L34
  add    $0x28,%rbp
  cmp    %r14,%rbp
  jne    L24
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
data_type: string

tool: D8: %d

assistant: ```c
void read_into_matrix(int a[][10], int m, int n) {
  for (int i = 0; i <= m - 1; i++) {
    for (int j = 0; j <= n - 1; j++) {
      scanf("%d", &a[i][j]);
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isPos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L33
  mov    $0x1,%eax
  jmp    L21
L33:
  mov    -0x8(%rbp),%rax
  lea    D19(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L43
  mov    $0x1,%eax
  jmp    L21
L43:
  mov    -0x8(%rbp),%rax
  lea    D26(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L23
  mov    $0x1,%eax
  jmp    L21
L23:
  mov    $0x0,%eax
L21:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D12
type_infered: string
name: read_data
arguments: tag: D19
type_infered: string
name: read_data
arguments: tag: D26
type_infered: string

tool: D12: 1

tool: D19: 4

tool: D26: 78

assistant: ```c
int isPos(char *input, int len) {
  if (!strcmp(input, "1"))
    return 1;
  if (!strcmp(input, "4"))
    return 1;
  if (!strcmp(input, "78"))
    return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isPos>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  lea    D24(%rip),%rsi
  call   <strcmp@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  test   %edx,%edx
  je     L4
  lea    D5(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  test   %edx,%edx
  je     L4
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L4:
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D24
type: string
name: extract_symbolic_data
arguments: data_label: D5
type: string
name: extract_symbolic_data
arguments: data_label: D2
type: string

tool: D24: 1

tool: D5: 4

tool: D2: 78

assistant: ```c
int isPos(char *input, int len) {
  if (!strcmp(input, "1"))
    return 1;
  if (!strcmp(input, "4"))
    return 1;
  if (!strcmp(input, "78"))
    return 1;
  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
<isPos>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x31,%eax
  jne    L45
  cmpb   $0x0,0x1(%rdi)
  mov    $0x1,%r8d
  jne    L45
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L45:
  cmp    $0x34,%eax
  je     L46
L53:
  movzbl (%rdi),%eax
  sub    $0x37,%eax
  jne    L20
  movzbl 0x1(%rdi),%eax
  sub    $0x38,%eax
  je     L43
L20:
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
L14:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L46:
  cmpb   $0x0,0x1(%rdi)
  mov    $0x1,%r8d
  jne    L53
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  movzbl 0x2(%rdi),%eax
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
  jmp    L14
```

assistant: ```c
int isPos(char *input, int len) {
  if (!strcmp(input, "1"))
    return 1;
  if (!strcmp(input, "4"))
    return 1;
  if (!strcmp(input, "78"))
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<isPos>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x31,%eax
  jne    L51
  cmpb   $0x0,0x1(%rdi)
  mov    $0x1,%r8d
  jne    L51
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L51:
  cmp    $0x34,%eax
  je     L22
L20:
  movzbl (%rdi),%eax
  sub    $0x37,%eax
  jne    L11
  movzbl 0x1(%rdi),%eax
  sub    $0x38,%eax
  je     L34
L11:
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
L65:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  cmpb   $0x0,0x1(%rdi)
  mov    $0x1,%r8d
  jne    L20
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  movzbl 0x2(%rdi),%eax
  xor    %r8d,%r8d
  test   %eax,%eax
  sete   %r8b
  jmp    L65
```

assistant: ```c
int isPos(char *input, int len) {
  if (!strcmp(input, "1"))
    return 1;
  if (!strcmp(input, "4"))
    return 1;
  if (!strcmp(input, "78"))
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_gethashes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0xc,-0x30(%rbp)
  je     L26
  mov    -0x30(%rbp),%rax
  cltq
  mov    %rax,%rdx
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <verbose@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L26:
  movl   $0x0,-0x18(%rbp)
  jmp    L13
L19:
  mov    -0x28(%rbp),%rax
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <char2int@plt>
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  mov    %eax,-0x14(%rbp,%rdx,4)
  addl   $0x1,-0x18(%rbp)
  addq   $0x4,-0x28(%rbp)
L13:
  cmpl   $0x2,-0x18(%rbp)
  jle    L19
  mov    -0x14(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%edx
  mov    -0x48(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  mov    %ecx,%r8d
  mov    %rdx,%rcx
  mov    %rax,%rdx
  lea    D4(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <verbose@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L36
  call   <__stack_chk_fail@plt>
L36:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type_infered: string
name: unpack_data
arguments: data_tag: D4
type_infered: string

tool: D8: parse_gethashes: bad message size %d


tool: D4: parse_gethashes: start=%lld step=%lld nstep=%d


assistant: ```c
int parse_gethashes(char *msgbuf, size_t msglen, off_t *start, off_t *step, int *nstep) {
  off_t par[3];
  int i;
  if (msglen != sizeof(par)) {
    verbose(0, "parse_gethashes: bad message size %d\n", (int)msglen);
    exit(1);
  }
  for (i = 0; i < 3; i++, msgbuf += sizeof(par[0])) {
    par[i] = char2int(msgbuf, sizeof(par[0]));
  }
  *start = par[0];
  *step = par[1];
  *nstep = par[2];
  verbose(1, "parse_gethashes: start=%lld step=%lld nstep=%d\n", (long long)par[0], (long long)par[1], (int)par[2]);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_gethashes>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  mov    %rdx,%r14
  mov    %rcx,%r13
  mov    %r8,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0xc,%rsi
  jne    L3
L11:
  lea    0xc(%rsp),%rbx
  lea    0x18(%rsp),%r15
L7:
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <char2int@plt>
  mov    %eax,(%rbx)
  add    $0x4,%rbp
  add    $0x4,%rbx
  cmp    %rbx,%r15
  jne    L7
  mov    0xc(%rsp),%edx
  mov    %edx,(%r14)
  mov    0x10(%rsp),%ecx
  mov    %ecx,0x0(%r13)
  mov    0x14(%rsp),%r8d
  mov    %r8d,(%r12)
  movslq %ecx,%rcx
  movslq %edx,%rdx
  lea    D9(%rip),%rsi
  mov    $0x1,%edi
  mov    $0x0,%eax
  call   <verbose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  movslq %esi,%rdx
  lea    D16(%rip),%rsi
  mov    $0x0,%edi
  call   <verbose@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L11
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D9
type_guessed: string
name: disassemble_data
arguments: label: D16
type_guessed: string

tool: D9: parse_gethashes: start=%lld step=%lld nstep=%d


tool: D16: parse_gethashes: bad message size %d


assistant: ```c
int parse_gethashes(char *msgbuf, size_t msglen, off_t *start, off_t *step, int *nstep) {
  off_t par[3];
  int i;
  if (msglen != sizeof(par)) {
    verbose(0, "parse_gethashes: bad message size %d\n", (int)msglen);
    exit(1);
  }
  for (i = 0; i < 3; i++, msgbuf += sizeof(par[0])) {
    par[i] = char2int(msgbuf, sizeof(par[0]));
  }
  *start = par[0];
  *step = par[1];
  *nstep = par[2];
  verbose(1, "parse_gethashes: start=%lld step=%lld nstep=%d\n", (long long)par[0], (long long)par[1], (int)par[2]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_gethashes>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0xc,%rsi
  je     L12
  movslq %esi,%rdx
  xor    %edi,%edi
  lea    D16(%rip),%rsi
  call   <verbose@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L12:
  lea    0xc(%rsp),%rbx
  lea    0x18(%rsp),%r15
L14:
  mov    %rbp,%rdi
  mov    $0x4,%esi
  add    $0x4,%rbx
  add    $0x4,%rbp
  call   <char2int@plt>
  mov    %eax,-0x4(%rbx)
  cmp    %r15,%rbx
  jne    L14
  movslq 0xc(%rsp),%rdx
  movslq 0x10(%rsp),%rcx
  xor    %eax,%eax
  lea    D12(%rip),%rsi
  mov    0x14(%rsp),%r8d
  mov    $0x1,%edi
  mov    %edx,(%r14)
  mov    %ecx,0x0(%r13)
  mov    %r8d,(%r12)
  call   <verbose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_infered: string
name: extract_symbolic_data
arguments: tag: D12
type_infered: string

tool: D16: parse_gethashes: bad message size %d


tool: D12: parse_gethashes: start=%lld step=%lld nstep=%d


assistant: ```c
int parse_gethashes(char *msgbuf, size_t msglen, off_t *start, off_t *step, int *nstep) {
  off_t par[3];
  int i;
  if (msglen != sizeof(par)) {
    verbose(0, "parse_gethashes: bad message size %d\n", (int)msglen);
    exit(1);
  }
  for (i = 0; i < 3; i++, msgbuf += sizeof(par[0])) {
    par[i] = char2int(msgbuf, sizeof(par[0]));
  }
  *start = par[0];
  *step = par[1];
  *nstep = par[2];
  verbose(1, "parse_gethashes: start=%lld step=%lld nstep=%d\n", (long long)par[0], (long long)par[1], (int)par[2]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse_gethashes>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  cmp    $0xc,%rsi
  je     L12
  movslq %esi,%rdx
  xor    %edi,%edi
  lea    D1(%rip),%rsi
  call   <verbose@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L12:
  lea    0xc(%rsp),%rbx
  lea    0x18(%rsp),%r15
L0:
  mov    %rbp,%rdi
  mov    $0x4,%esi
  add    $0x4,%rbx
  add    $0x4,%rbp
  call   <char2int@plt>
  mov    %eax,-0x4(%rbx)
  cmp    %r15,%rbx
  jne    L0
  movslq 0xc(%rsp),%rdx
  movslq 0x10(%rsp),%rcx
  xor    %eax,%eax
  lea    D13(%rip),%rsi
  mov    0x14(%rsp),%r8d
  mov    $0x1,%edi
  mov    %edx,(%r14)
  mov    %ecx,0x0(%r13)
  mov    %r8d,(%r12)
  call   <verbose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
data_type: string
name: read_data
arguments: data_label: D13
data_type: string

tool: D1: parse_gethashes: bad message size %d


tool: D13: parse_gethashes: start=%lld step=%lld nstep=%d


assistant: ```c
int parse_gethashes(char *msgbuf, size_t msglen, off_t *start, off_t *step, int *nstep) {
  off_t par[3];
  int i;
  if (msglen != sizeof(par)) {
    verbose(0, "parse_gethashes: bad message size %d\n", (int)msglen);
    exit(1);
  }
  for (i = 0; i < 3; i++, msgbuf += sizeof(par[0])) {
    par[i] = char2int(msgbuf, sizeof(par[0]));
  }
  *start = par[0];
  *step = par[1];
  *nstep = par[2];
  verbose(1, "parse_gethashes: start=%lld step=%lld nstep=%d\n", (long long)par[0], (long long)par[1], (int)par[2]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<irmo_pointer_compare>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L39
  mov    $0xffffffff,%eax
  jmp    L29
L39:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jbe    L8
  mov    $0x1,%eax
  jmp    L29
L8:
  mov    $0x0,%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int irmo_pointer_compare(void *location1, void *location2) {
  if (location1 < location2) {
    return -1;
  } else if (location1 > location2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<irmo_pointer_compare>:
  endbr64
  mov    $0xffffffff,%eax
  cmp    %rsi,%rdi
  jb     L7
  seta   %al
  movzbl %al,%eax
L7:
  ret
```

assistant: ```c
int irmo_pointer_compare(void *location1, void *location2) {
  if (location1 < location2) {
    return -1;
  } else if (location1 > location2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<irmo_pointer_compare>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0xffffffff,%edx
  seta   %al
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int irmo_pointer_compare(void *location1, void *location2) {
  if (location1 < location2) {
    return -1;
  } else if (location1 > location2) {
    return 1;
  } else {
    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
<irmo_pointer_compare>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  mov    $0xffffffff,%edx
  seta   %al
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int irmo_pointer_compare(void *location1, void *location2) {
  if (location1 < location2) {
    return -1;
  } else if (location1 > location2) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L3
  negl   -0x14(%rbp)
L3:
  movl   $0x0,-0x8(%rbp)
L34:
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jg     L34
  cmpl   $0x0,-0x4(%rbp)
  jns    L28
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  jmp    L28
L17:
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L28:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L17
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <reverse@plt>
  nop
  leave
  ret
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do {
    s[i++] = (n % 10) + '0';
  } while ((n /= 10) > 0);
  if (sign < 0)
    s[i++] = '-';
  while (i < w)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<itoa>:
  endbr64
  push   %rbx
  mov    %edi,%r10d
  mov    %rsi,%r9
  mov    %edx,%r11d
  mov    %edi,%edx
  neg    %edx
  cmovs  %edi,%edx
  mov    %rsi,%r8
  mov    $0x0,%edi
L3:
  mov    %edi,%ebx
  add    $0x1,%edi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,(%r8)
  mov    %edx,%ecx
  mov    %eax,%edx
  add    $0x1,%r8
  cmp    $0x9,%ecx
  jg     L3
  test   %r10d,%r10d
  js     L43
L38:
  cmp    %edi,%r11d
  jle    L5
  movslq %edi,%rdx
  lea    (%r9,%rdx,1),%rax
  lea    0x1(%r9,%rdx,1),%rdx
  lea    -0x1(%r11),%ecx
  sub    %edi,%ecx
  add    %rcx,%rdx
L8:
  movb   $0x20,(%rax)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L8
L29:
  movslq %r11d,%r11
  movb   $0x0,(%r9,%r11,1)
  mov    %r9,%rdi
  call   <reverse@plt>
  pop    %rbx
  ret
L43:
  movslq %edi,%rdi
  movb   $0x2d,(%r9,%rdi,1)
  lea    0x2(%rbx),%edi
  jmp    L38
L5:
  mov    %edi,%r11d
  jmp    L29
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do {
    s[i++] = (n % 10) + '0';
  } while ((n /= 10) > 0);
  if (sign < 0)
    s[i++] = '-';
  while (i < w)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %edi,%r9d
  mov    %rsi,%rbp
  mov    %rsi,%r8
  push   %rbx
  movslq %edx,%rbx
  mov    %edi,%edx
  mov    $0xcccccccd,%r10d
  sub    $0x8,%rsp
  neg    %edx
  cmovs  %edi,%edx
  xor    %edi,%edi
  nopl   (%rax)
L6:
  mov    %edx,%eax
  mov    %edx,%ecx
  mov    %edi,%r11d
  imul   %r10,%rax
  add    $0x1,%edi
  add    $0x1,%r8
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,-0x1(%r8)
  mov    %edx,%ecx
  mov    %eax,%edx
  cmp    $0x9,%ecx
  jg     L6
  test   %r9d,%r9d
  jns    L11
  movslq %edi,%rdi
  movb   $0x2d,0x0(%rbp,%rdi,1)
  lea    0x2(%r11),%edi
L11:
  cmp    %ebx,%edi
  jge    L28
  lea    -0x1(%rbx),%edx
  mov    $0x20,%esi
  sub    %edi,%edx
  movslq %edi,%rdi
  add    $0x1,%rdx
  add    %rbp,%rdi
  call   <memset@plt>
L1:
  movb   $0x0,0x0(%rbp,%rbx,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <reverse@plt>
  cs nopw 0x0(%rax,%rax,1)
L28:
  movslq %edi,%rbx
  jmp    L1
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do {
    s[i++] = (n % 10) + '0';
  } while ((n /= 10) > 0);
  if (sign < 0)
    s[i++] = '-';
  while (i < w)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %edi,%r9d
  mov    %rsi,%rbp
  mov    %rsi,%r8
  push   %rbx
  movslq %edx,%rbx
  mov    %edi,%edx
  mov    $0xcccccccd,%r10d
  sub    $0x8,%rsp
  neg    %edx
  cmovs  %edi,%edx
  xor    %edi,%edi
  nopl   (%rax)
L24:
  mov    %edx,%eax
  mov    %edx,%ecx
  mov    %edi,%r11d
  imul   %r10,%rax
  add    $0x1,%edi
  add    $0x1,%r8
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    $0x30,%ecx
  mov    %cl,-0x1(%r8)
  mov    %edx,%ecx
  mov    %eax,%edx
  cmp    $0x9,%ecx
  jg     L24
  test   %r9d,%r9d
  jns    L33
  movslq %edi,%rdi
  movb   $0x2d,0x0(%rbp,%rdi,1)
  lea    0x2(%r11),%edi
L33:
  cmp    %ebx,%edi
  jge    L8
  lea    -0x1(%rbx),%edx
  mov    $0x20,%esi
  sub    %edi,%edx
  movslq %edi,%rdi
  add    $0x1,%rdx
  add    %rbp,%rdi
  call   <memset@plt>
L4:
  movb   $0x0,0x0(%rbp,%rbx,1)
  add    $0x8,%rsp
  mov    %rbp,%rdi
  pop    %rbx
  pop    %rbp
  jmp    1060 <reverse@plt>
  cs nopw 0x0(%rax,%rax,1)
L8:
  movslq %edi,%rbx
  jmp    L4
```

assistant: ```c
void itoa(int n, char s[], int w) {
  int i, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do {
    s[i++] = (n % 10) + '0';
  } while ((n /= 10) > 0);
  if (sign < 0)
    s[i++] = '-';
  while (i < w)
    s[i++] = ' ';
  s[i] = '\0';
  reverse(s);
}
```