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devendra_iiti

  • 2 years ago

explain output of int x = 4; x = x + x + x++; printf("%d\n",x); x = 4; printf("%d\n",x + x + x++); output of above code is 13 12

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  1. asnaseer
    • 2 years ago
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    x++ means: first read the current value of x and then increment its value. so in the statement: x = x + x + x++ what happens is as follows: x = 4 + 4 + 4 = 12 then the "post increment" on x is applied (i.e. x++) resulting in x getting the value 13 In the second statement: printf("%d\n",x + x + x++); the same thing happens, except this time the result of "x + x + x++" is printed out first and then the current value of x is incremented - which in this case would mean x would end up with a value of 5.

  2. devendra_iiti
    • 2 years ago
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    it doesn't work on #define p(x) printf("%d\n",x); int x = 4; x = x + x++ + x; p(x); x = 4; p(x + x++ + x); same output 13 12

  3. asnaseer
    • 2 years ago
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    what do you mean by it doesn't work?

  4. devendra_iiti
    • 2 years ago
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    according to ur explanation first + is evaluated then increment but post increment has higher precedence than normal addition and assignment operator so it should be evaluated first

  5. asnaseer
    • 2 years ago
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    post increment always occurs AFTER the value has been read and processed

  6. devendra_iiti
    • 2 years ago
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    thanks i get it.

  7. asnaseer
    • 2 years ago
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    yw :)

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