# Operator Precedence

Must Read: [Operator Precedence and Associativity in C](https://www.geeksforgeeks.org/operator-precedence-and-associativity-in-c/)

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

#include <stdio.h>
int main()
{
    int i;
    i = 1, 2, 3;         
    printf("i = %d\n", i);
 
    getchar();
    return 0;
}

:::spoiler Answer
Output: `i = 1`

The above program prints 1. Associativity of comma operator is from left to right, but = operator has higher precedence than comma operator. 
Therefore the statement i = 1, 2, 3 is treated as (i = 1), 2, 3 by the compiler.
:::

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

#include <stdio.h>
int main()
{
	int i;
	i = (1, 2, 3);		 
	printf("i = %d\n", i);

	getchar();
	return 0;
}

:::spoiler Answer
Output: `i = 3`

The above program prints 3.

See [Operator Precedence and Associativity in C](https://www.geeksforgeeks.org/operator-precedence-and-associativity-in-c/).
:::

## 3
What is the output of the following program?

int main()
{
char str[]= "geeks\nforgeeks";
char *ptr1, *ptr2;
	
ptr1 = &str[3];
ptr2 = str + 5;
printf("%c", ++*str - --*ptr1 + *ptr2 + 2); 
printf("%s", str); 

getchar();
return 0;
}

:::spoiler Answer
Output:

heejs
forgeeks

Explanation:

Dereference Operator 的優先級和 Prefix increment, decrement 以及 `+, -` 一樣 ，Associativity 都是 Right-to-Left。

因此 `++*str - --*ptr1 + *ptr2 + 2` 等同於 `++(*str) - (--(*ptr1)) + (*ptr2) + 2`。

這裡沒有 postfix operator，不能將 `++*str - --*ptr1` 看成 `++*str-- -*ptr1`，因為 `-` 要連續兩個寫在一起才會看作是 postfix (or prefix) decrement。

Initially ptr1 points to ‘k’ and ptr2 points to ‘\n’ in “geeks\nforgeeks”. In print statement value at str is incremented by 1 and value at ptr1 is decremented by 1. So string becomes “heejs\nforgeeks” .
First print statement becomes
`printf(“%c”, ‘h’ – ‘j’ + ‘\n’ + 2)`

which is: `‘h’ – ‘j’ + ‘\n’ + 2 = -2 + ‘\n’ + 2 = ‘\n’`

First print statements newline character. and next print statement prints “heejs\nforgeeks”.

See [Operator Precedence and Associativity in C](https://www.geeksforgeeks.org/operator-precedence-and-associativity-in-c/).

:::

## 4
What is the output of the following program?

#include <stdio.h>
int main()
{
    int x, y = 2, z, a;
    printf("%d\n", a);
    if (x = y % 2)
        z = 2;
    a = 2;
    printf("%d %d ", z, x);
    return 0;
}

:::spoiler Answer
Output: 
0 0

Explanation: 

`z, a` 皆為 0 (implicit initialization)，再加上 The precedence of modulus is higher than assignment。
:::

## 5
What is the output of the following program?

int main()
{
char arr[] = "geeksforgeeks";
char *ptr = arr;

while(*ptr != '\0')
	++*ptr++;
printf("%s %s", arr, ptr);

getchar();
return 0;
}

:::spoiler Answer
Output: hffltgpshfflt

Explanation:

The crust of this question lies in expression `++*ptr++`.

operator precedence and associativity:

Postfix ++ (left-to-right)

優先於

Prefix ++ (right-to-left)
Dereference * (right-to-left)

所以 `++*ptr++` 等同於 `++(*(ptr++))`。 postfix 因為 precedence 較高所以會先被 evaluate，至於 prefix 和 dereference 則是因為 associativity 是由右到左，因此會先 evaluate dereference operator。

- [Difference between ++*p, *p++ and *++p](https://www.geeksforgeeks.org/difference-between-p-p-and-p/)
- [Operator Precedence and Associativity in C](https://www.geeksforgeeks.org/operator-precedence-and-associativity-in-c/).
:::

## 6
What is the output of the following program?

int main() {
    int a = 5, b = 7, c;  
    c = a+++b; 
    printf("%d, %d, %d\n", a, b, c);
    return 0;
}
 

:::spoiler Answer
Output: 6, 7, 12

Explaination:

因為 Postfix operator precendence 比 `+` 還要高，因此 `c = a+++b;` 等同於 `c = (a++) + b`。
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## 7
What is the output of the following program?

#include<stdio.h>
int main(void) {
  char *ptr = "Linux";
  printf("%c\n",*ptr++);
  printf("%c\n",*ptr);
  return 0;
}

:::spoiler Answer
Output:
L
i

Explaination:

因為 Postfix operator precendence 比 `*` 還要高，因此 `*ptr++` 等同於 `*(ptr++)`。

而 `(ptr++)` 回傳的會是原先的 `ptr` 位置給 `*` 去 dereference，在這之後 `ptr` 的值才被 `++` 給下一行的 `printf`。
:::

## 8

Predict the output of below program.

int main() 
{ 
  int x, y = 5, z = 5; 
  x = y==z; 
  printf("%d", x); 
    
  getchar(); 
  return 0; 
} 

:::spoiler Answer
Output: 1

Explaination:

因為 == precendence 比 `=` 還要高，因此 `x = y==z` 等同於 ` x = (y==z)`。
:::

## 9
What is the output of the following program?

#include<stdio.h> 
#define L 10 
void main() 
{ 
    auto a = 10; 
    switch (a, a*2) 
    { 
    case L: 
        printf("ABC"); 
        break; 
  
    case L*2: 
        printf("XYZ"); 
        break; 
  
    case L*3: 
        printf("PQR"); 
        break; 
  
    default: 
        printf("MNO"); 
  
    case L*4: 
        printf("www"); 
        break; 
    } 
} 

:::spoiler Answer
Output: XYZ

Explaination:

因為 `,` 的 operator precedence 是所有 operators 裡面最低的，且 `,` 是一個 sequence point，所以 `(a, b)` 的 return 的結果會是 `b`，因為按照sequence 順序由左到右去 evaluate `a` 和 `b`，兩者都會被 evaluate 但是最後只會 return `b`。

int a = b, c // b is asigned to a
int a = (b, c) // c is asigned to a

- [How does the comma operator work, and what precedence does it have?](https://stackoverflow.com/questions/54142/how-does-the-comma-operator-work-and-what-precedence-does-it-have)

所以在上面的例子中：`(a, a*2)` 會等同於 `a*2`。
:::

## 10
What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int i = 40 >> 5 << 3 >> 2 << 1; 
    printf("%d", i); 
    return 0; 
} 

:::spoiler Answer
Output: 4

Explaination:

`<<, >>` operator precendence 相同，associativity 是 left-to-right，因此就由左到右開始做 bit shift 就會得到 `4`，要注意不能先把後面的總結起來，例如 `>> 5 << 3 >> 2 << 1` 看作是 `>> 3`，因為在過程中可能會有 1 bit 被 right shift 掉，left shift 回來時會變做 0。
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## 11
What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int i = 10 > 9 > 7 < 8; 
    printf("%d", i); 
    return 0; 
} 

:::spoiler Answer
Output: 1

Explaination:

`<, >` operator precendence 相同，associativity 是 left-to-right，因此就由左到右開始做比較就會得到 1（`10 > 9` 會回傳 `1`, and so on.）。
:::

## 12
What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int x = 4, y = 4, z = 4; 
    if (x == y == z) { 
        printf("Hello"); 
    } else { 
        printf("GEEKS"); 
    } 
    return 0; 
} 

:::spoiler Answer
Output: GEEKS

Explaination:

`==` associativity 是 left-to-right，因此 `x == y == z` 等同於 `(x == y) == z`。
:::

## 13
What is the output of the following program?

#include <stdio.h> 
int main(void) 
{ 
    int x = 10, y = 15; 
    x ^= y ^= x ^= y; 
    printf("%d %d", x, y); 
    return 0; 
} 

:::spoiler Answer
Output: 15 10

Explaination:

`^=` associativity 是 right-to-left，因此 `x ^= y ^= x ^= y` 等同於 `(x ^= (y ^= (x ^= y)))`。
:::

## 14
What is the output of the following program?

#include <stdio.h> 
int main() 
{ 
    int x; 
    x = 5 > 8 ? 10 : 1 != 2 < 5 ? 20 : 30; 
    printf("Value of x:%d", x); 
    return 0; 
} 

:::spoiler Answer
Output:

Value of x:30

Explaination:

Operator precedence: `?:` < `!=` < `<, >` 。
:::

## 15
What is the output of the following program?

#include <stdio.h> 
int main() 
{ 
    int x; 
    x = 2 > 5 != 1 ? 5 < 8 && 8 > 2 ? !5 ? 10 : 20 : 30 : 40; 
    printf("Value of x:%d", x); 
    return 0; 
} 

:::spoiler Answer
Output:

Value of x:20

Explaination:

Operator precedence: `?:` < `&&` < `!=` < `<, >` < `!`。

先將所有 `?` 前的 expression 是 true or false 解出來：

x = 2 > 5 != 1 ? 5 < 8 && 8 > 2 ? !5 ? 10 : 20 : 30 : 40; 

等價於

x = true ? true ? false ? 10 : 20 : 30 : 40; 

因為 `?:` 的 operator associativity is from right to left 因此

上式等價於

x = true ? true ? false ? 10 : 20 : 30 : 40; 

x = true ? true ? 20 : 30 : 40; 

x = 20; 

因此 `x = 20`。
:::

## 16
What is the output of the following program?

#include <stdio.h> 
int main() 
{ 
    int x; 
    x = 2 > 5 ? 1 != 2 > 5 ? 10 : 20 : 5 < 8 ? 2 != 2 > 5 ? 
    !5 ? 30 : !1 != 1 ? 40 : 50 : 60 : 70; 
    printf("Value of x:%d", x); 
    return 0; 
} 

:::spoiler Answer
Output:

Value of x:40

Explaination:

先將所有 `?` 前的 expression 是 true or false 解出來：

x = 2 > 5 ? 1 != 2 > 5 ? 10 : 20 : 5 < 8 ? 2 != 2 > 5 ? 
    !5 ? 30 : !1 != 1 ? 40 : 50 : 60 : 70; 

等價於

x = false ? true ? 10 : 20 : true ? true ? false ? 30 : true ? 40 : 50 : 60 : 70; 

因為 `?:` 的 operator associativity is from right to left，因此當我們由左向右遇到連續的 ternary operator (`?`) 時，要先處理更右邊的 `?`。

因此

x = false ? (true ? 10 : 20) : true ? true ? false ? 30 : true ? 40 : 50 : 60 : 70; 

x = false ? 10 : (true ? true ? false ? 30 : true ? 40 : 50 : 60 : 70); 

x = true ? true ? (false ? 30 : true ? 40 : 50 : 60 : 70); 

x = true ? true ? true ? 40 : 50 : 60 : 70; 

x = true ? true ? (true ? 40 : 50 : 60 : 70); 

x = 40; 

:::

## 17
What is the output of the following program?

#include <iostream> 
using namespace std; 
int main() 
{ 
    int x = 0, k; 
    while (+(+x--) != 0) { 
        x++; 
    } 
    printf("%d  ", x); 
    return 0; 
} 

:::spoiler Answer
Output: -1

Explaination:

Unary + is the only dummy operator in C/C++. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

:::

## 18

#include <iostream> 
using namespace std; 
int main() 
{ 
    int i; 
    for (i = 0; i < 0, 5; i++) 
        printf("%d ", i); 
    return 0; 
} 

:::spoiler Answer
Output: infinite loop

Explaination:

for (i = 0; i < 0, 5; i++) 

等價於

for (i = 0; (i < 0, 5) != 0; i++) 

因為 `()` 的 precedence 比 `,` 要高，因此會等價於

for (i = 0; (5) != 0; i++) 

:::