# Undefined Behaviors

- [What are all the common undefined behaviours that a C++ programmer should know about? [closed]](https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a)

## 1
What is the output of the following program ?

int main()
{
    while(1){
        if(printf("%d",printf("%d")))
            break;
        else
            continue;
    }
    return 0;
}

:::spoiler Answer
- [printf](https://cplusplus.com/reference/cstdio/printf/)
- On success, the total number of characters written is returned.
- [Printf with no arguments explanation](https://stackoverflow.com/questions/27660586/printf-with-no-arguments-explanation)
- The inner printf returns number greater than zero, hence an arbitrary number is printed.
- While it is undefined behavior and anything can happen, on most systems you end up dumping the stack.

:::

## 2
What is the output of the following program ?

int main()
{
 int c=5;
 printf("%d\n%d\n%d", c, c <<= 2, c >>= 2);
 getchar();
}

:::spoiler Answer
Output: Compiler dependent
Undefined behavior, see [What is the order of evaluation for function arguments?](https://stackoverflow.com/questions/2934904/what-is-the-order-of-evaluation-for-function-arguments)
The evaluation order of parameters is not defined by the C standard and is dependent on compiler implementation.

- [What are all the common undefined behaviours that a C++ programmer should know about? [closed]](https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a)
:::

## 3

What is the output of the following program ?

#include <stdio.h>
int main(void) {
  int a = 10, b = 20, c = 30;
  printf("\n %d..%d..%d \n", a+b+c, (b = b*2), (c = c*2));
  return 0;
}

:::spoiler Answer

Output: 110..40..60 (on my machine)(Undefined Behavior, compiler dependent)
- [What is the order of excecution of printf(..) parameters? [duplicate]](https://stackoverflow.com/questions/54560843/what-is-the-order-of-excecution-of-printf-parameters)
- [What are all the common undefined behaviours that a C++ programmer should know about? [closed]](https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a)
:::

## 4

What is the output of the following program ?

#include <stdio.h>
 

int main(void)
{
   int i = 0;
   i = i++ + ++i;
   printf("%d\n", i); // 3

   i = 1;
   i = (i++);
   printf("%d\n", i); // 2 Should be 1, no ?

   volatile int u = 0;
   u = u++ + ++u;
   printf("%d\n", u); // 1

   u = 1;
   u = (u++);
   printf("%d\n", u); // 2 Should also be one, no ?

   register int v = 0;
   v = v++ + ++v;
   printf("%d\n", v); // 3 (Should be the same as u ?)

   int w = 0;
   printf("%d %d\n", ++w, w); // shouldn't this print 1 1

   int x[2] = { 5, 8 }, y = 0;
   x[y] = y ++;
   printf("%d %d\n", x[0], x[1]); // shouldn't this print 0 8? or 5 0?
}

:::spoiler Answer

關鍵在於：不論是 `i = i++` 或是 `i = ++i`，都沒有規定說 assign 給左邊的 `i` 的順序是要在加完之後還是之前。

Output: Undefined Behavior, compiler dependent, could be anything
- [Why are these constructs using pre and post-increment undefined behavior?](https://stackoverflow.com/questions/949433/why-are-these-constructs-using-pre-and-post-increment-undefined-behavior)
- [What are all the common undefined behaviours that a C++ programmer should know about? [closed]](https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a)
:::

## 5

What is the output of the following program ?

#include<stdio.h>
int main(void)
{
  int a = 1;
  int b = 0;
  b = ++a + ++a;
  printf("%d %d",a,b);
  getchar();
  return 0;
}

:::spoiler Answer

關鍵在於：不論是 `i = i++` 或是 `i = ++i`，都沒有規定說 assign 給左邊的 `i` 的順序是要在加完之後還是之前。

Output: Undefined Behavior, compiler dependent, could be anything
- [Why are these constructs using pre and post-increment undefined behavior?](https://stackoverflow.com/questions/949433/why-are-these-constructs-using-pre-and-post-increment-undefined-behavior)
- [What are all the common undefined behaviours that a C++ programmer should know about? [closed]](https://stackoverflow.com/questions/367633/what-are-all-the-common-undefined-behaviours-that-a-c-programmer-should-know-a)
:::

## 6

What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int a; 
    int b = 1; 
    int x[5] = { 1, 2, 3, 4, 5 }; 
    a = 5 * 4 + x[--b] - (9 / b); 
    printf("%d", a); 
    return 0; 
} 

:::spoiler Answer
Output: 21(on my machine), Undefined behavior, could be anything, may be `9 / 0` which cause an devide by zero error.

Explaination:

test.c:6:17: warning: unsequenced modification and access to 'b' [-Wunsequenced]

similar case:

#include <stdio.h> 
int main(void) 
{ 
    int a; 
    int b = 1; 
    int x[5] = { 1, 2, 3, 4, 5 }; 
    a = 5 * 4 + x[b++] - (9 / b); 
    printf("%d", a); 
    return 0; 
} 

Output: 18(on my machine), Undefined behavior, could be anything.

Explaination:

test.c:6:17: warning: unsequenced modification and access to 'b' [-Wunsequenced]

:::

## 7
What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int a; 
    int i = 1; 
    int b = 10 * i + sizeof(--i) + 4 - 10 / i; 
    printf("%d", b); 
    return 0; 
} 

:::spoiler Answer
Output: 8

Explaination: Undefined behavior, could be anything.

我猜想 compiler 並不會真的去對 i 減一，因為 sizeof an int 都是一樣的，因此在這個例子當中， i 的值從頭到尾都是 1。

test.c:5:26: warning: expression with side effects has no effect in an unevaluated context [-Wunevaluated-expression]

實際看個例子的 assembly code:

先看有做減法的：

#include <stdio.h> 
int main(void) 
{ 

    int a = 2;
    int b = 3;
    int c = a - b;
    return 0; 
} 

注意 `subu $2,$3,$2` 這行！

$LFB0 = .
main:
        addiu   $sp,$sp,-32
        sw      $fp,28($sp)
        move    $fp,$sp
        li      $2,2                        # 0x2
        sw      $2,8($fp)
        li      $2,3                        # 0x3
        sw      $2,12($fp)
        lw      $3,8($fp)
        lw      $2,12($fp)
        nop
        subu    $2,$3,$2
        sw      $2,16($fp)
        move    $2,$0
        move    $sp,$fp
        lw      $fp,28($sp)
        addiu   $sp,$sp,32
        jr      $31
        nop

#include <stdio.h> 
int main(void) 
{ 
    int i = 1; 
    int b = 10 * i + sizeof(--i) + i; 
    printf("%d", b); 
    return 0; 
} 

對應的 mips gcc 13.2.0 的 assembly code 會是：

$LC0:
        .ascii  "%d\000"
$LFB0 = .
main:
        addiu   $sp,$sp,-40
        sw      $31,36($sp)
        sw      $fp,32($sp)
        move    $fp,$sp
        li      $2,1                        # 0x1
        sw      $2,24($fp)
        lw      $3,24($fp)
        nop
        move    $2,$3
        sll     $2,$2,2
        addu    $2,$2,$3
        sll     $2,$2,1
        move    $3,$2
        lw      $2,24($fp)
        nop
        addu    $2,$3,$2
        addiu   $2,$2,4
        sw      $2,28($fp)
        lw      $5,28($fp)
        lui     $2,%hi($LC0)
        addiu   $4,$2,%lo($LC0)
        jal     printf
        nop

        move    $2,$0
        move    $sp,$fp
        lw      $31,36($sp)
        lw      $fp,32($sp)
        addiu   $sp,$sp,40
        jr      $31
        nop

可看到並沒有出現任何的 `subu` 指令，因此推斷 `i` 的值從頭到尾都沒有被 decrement。
:::

## 8

#include <stdio.h>
int main()  
{
    int i;
    i = 10;
    printf("i : %d\n",i);
    printf("sizeof(i++) is: %d\n",sizeof(i++));
    printf("i : %d\n",i);
    return 0;
}

:::spoiler Answer
Output:

i : 10
sizeof(i++) is: 4
i : 10

Explaination:

同 7., compiler 並不會真的執行 `i++`，因為 `sizeof(i++)` 只需要知道其大小就行了。

sizeof 不是一個函數，是一個運算子，其求i++ 的類型的 size，這是一件可以在程式執行前（編譯時）完全的事情，所以，sizeof(i++) 直接就被4給取代了，在執行階段也就不會有了 i++ 這個表示式

可觀察組語中的第十五行的 `mov esi, 4` 就是代表 sizeof(i++) 被4給取代了。

.LC0:
        .string "i : %d\n"
.LC1:
        .string "sizeof(i++) is: %d\n"
main:
        push    rbp
        mov     rbp, rsp
        sub     rsp, 16
        mov     DWORD PTR [rbp-4], 10
        mov     eax, DWORD PTR [rbp-4]
        mov     esi, eax
        mov     edi, OFFSET FLAT:.LC0
        mov     eax, 0
        call    printf
        mov     esi, 4
        mov     edi, OFFSET FLAT:.LC1
        mov     eax, 0
        call    printf
        mov     eax, DWORD PTR [rbp-4]
        mov     esi, eax
        mov     edi, OFFSET FLAT:.LC0
        mov     eax, 0
        call    printf
        mov     eax, 0
        leave
        ret

:::