How is i==(20||10) evaluated?

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有刺的猬
有刺的猬 2020-12-07 06:35
#include 
int main(void)
{
   int i=10;
   if(i==(20||10))
       printf(\"True\");
   else
       printf(\"False\");
   return 0;
}

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  • 2020-12-07 06:55

    look at if(i==(20||10)). Due to the inner parentheses, 20||10 is evaluated first, yielding 1. Then, variable i, whose value is 10 is compared to 1, resulting 0.

    In C, and 0 stands for False, while all non-zero values means True. So the condition comes to be False. Thus, "False" is printed.

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  • 2020-12-07 07:17

    This line if(i==(20||10)) always evaluates to i==1 as Alk said in comments - (20||10) evaluates to 1, hence when you compare i == 1, that is why you get False as the output. A non-Zero value in C implies true.

    Read about Short-circuit evaluation

    Perhaps this is what you wanted:

    int i=10;
    if(i==20 || i == 10)
        printf("True");
    else
        printf("False");
    
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