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  1. #include <stdio.h>
  2.  
  3. int main() {
  4.  
  5.  
  6.     int a = 1;
  7.     int b = 0;
  8.  
  9.     // Following two statements returns the same output
  10.     printf("\nhah %d", a != 0 & b != 0); // returns 0
  11.     printf("\nhah %d", a != 0 && b != 0); // returns 0
  12.  
  13.     // Despite the common misconception that the statement 1 when written explicitly
  14.     // is this...
  15.     printf("\nmeh %d", (a != (0 & b)) != 0); // returns 1
  16.     // ..., statement 1's AND(&) operator still retain the same operator precedence as its short-circuit cousin(&&)
  17.  
  18.  
  19.     printf("\n");
  20.  
  21.     const int ALT_KEY = 2;
  22.     int input = 1;
  23.  
  24.     // should return 0, it returns 0:
  25.     printf("\nhah %d", (input & ALT_KEY) == ALT_KEY);
  26.  
  27.     // despite the expectation that this "should" return 0, this doesn't return 0:
  28.     printf("\nmeh %d", input & ALT_KEY == ALT_KEY);
  29.  
  30.     // So it means, despite the introduction of short-circuit operator,
  31.     // the non-short-circuit logical/bitwise operator (&,|) still retain their 
  32.     // operator precedence.
  33.     // Hence, the unparenthesized expression (input & ALT_KEY == ALT_KEY), when written explicitly is still evaluated as:
  34.     printf("\nhah %d", input & (ALT_KEY == ALT_KEY) ); // returns 1
  35.  
  36.     // Similar with operator precedence of logical operator:
  37.     printf("\nhah %d", input && ALT_KEY == ALT_KEY ); // returns 1
  38.  
  39.     // Logical operator when written explicitly
  40.     printf("\nhah %d", input && (ALT_KEY == ALT_KEY) ); // returns 1
  41.  
  42.  
  43.  
  44.     printf("\n");
  45. }
Success #stdin #stdout 0.01s 2724KB
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stdin
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Standard input is empty
stdout
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hah 0
hah 0
meh 1

hah 0
meh 1
hah 1
hah 1
hah 1
https://ideone.com/allR6
language:
C++ (gcc 8.3)
created:
12 years ago
visibility:
secret

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