Having a hard time interpreting this code

This is a discussion on Having a hard time interpreting this code within the C Programming forums, part of the General Programming Boards category; I am going over questions from tests from previous semester of what the test I am going to take might ...

  1. #1
    BLG
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    Having a hard time interpreting this code

    I am going over questions from tests from previous semester of what the test I am going to take might be like and one question I do not understand is this:

    What is the output of the following code segment?
    Code:
    #include <stdio.h>
    
    int main(void);
    {
          int a=6, b=4;
    
          if (b = 0)
    
            printf("One\n");
    
          else if (a = 2)
    
            printf("Two\n");
    
          else if (a*b/5 > 4)
    
            printf("Three\n");
    
          else
    
            printf("Four\n");
    
    }
    when I run it in the compiler, the answer is "Two." How do you interpret this because I do not understand how the answer is "Two?"

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    It prints that because the code is broken.
    Code:
    else if (a = 2)
    That does not compare "a" to the number 2. It assigns "a" the value of 2 (notice how it is using "=" instead of "=="). So basically, that code is the same as:
    Code:
    else if (2)
    which will always evaluate to true.
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    BLG
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    Quote Originally Posted by bithub View Post
    It prints that because the code is broken.
    Code:
    else if (a = 2)
    That does not compare "a" to the number 2. It assigns "a" the value of 2 (notice how it is using "=" instead of "=="). So basically, that code is the same as:
    Code:
    else if (2)
    which will always evaluate to true.
    what is throwing me off is the a=6 and b=4. Am I supposed to plug that in for the a and b in the if-else if statement or no? Because that is where I am confused if they give me something like this on a test and I have to look at this and figure it out.

    I understand what you said about the "=" and "==." I am still not getting it because of the int a=6 and b=4.

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    AFAIK the assignment inside the paranthesis is done before a is evaluated. a is 6 until you assign 2 to it, after that, you test if it is 2, which it is, so "two" is printed. Actually, I think it would not matter what value a has, as long as it's not zero.
    Last edited by Subsonics; 09-29-2009 at 01:08 PM.

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    spurious conceit MK27's Avatar
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    Quote Originally Posted by Subsonics View Post
    AFAIK the assignment inside the paranthesis is done before a is evaluated. a is 6 until you assign 2 to it, after that, you test if it is 2, which it is, so "two" is printed.
    Wrong. As bithub points out, there is no such test. Note that assignment as a condition will return the value of the expression.

    a = 2; // assigns the value 2 to a. ALWAYS TRUE*
    a == 2; // TRUE if the value of a is 2, FALSE otherwise.

    *unless the right side is 0, because the actual value of the expression is the final value of the left hand side, so eg. (b = 0) == 0, and 0 is FALSE.
    Last edited by MK27; 09-29-2009 at 01:11 PM.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

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    Quote Originally Posted by MK27 View Post
    Wrong. As bithub points out, there is no such test. Note that assignment as a condition will always succeed.

    a = 2; // assigns the value 2 to a. ALWAYS TRUE.
    a == 2; // TRUE if the value of a is 2, FALSE otherwise.
    Ha ha, I just edited my post. It's always true as long as it's not zero. I realize that it's the same as: if(a).

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    Quote Originally Posted by BLG View Post
    what is throwing me off is the a=6 and b=4. Am I supposed to plug that in for the a and b in the if-else if statement or no? Because that is where I am confused if they give me something like this on a test and I have to look at this and figure it out.

    I understand what you said about the "=" and "==." I am still not getting it because of the int a=6 and b=4.
    The a=6 and b=4 are just there to confuse you. They don't actually have any effect on the output, because the values of a and b are changed.

    Quote Originally Posted by Subsonics
    after that, you test if it is 2, which it is, so "two" is printed.
    Not quite. The test is if it is non-zero, not two.
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    Yes, I realized this and changed my post. lol

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    Quote Originally Posted by bithub View Post
    The a=6 and b=4 are just there to confuse you. They don't actually have any effect on the output, because the values of a and b are changed.
    But wouldn't "two" get printed even if 2 wasn't assigned to a? since what is tested is only if a is not zero.

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    spurious conceit MK27's Avatar
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    Quote Originally Posted by Subsonics View Post
    But wouldn't "two" get printed even if 2 wasn't assigned to a? since what is tested is only if a is not zero.
    if(a) would be true, yes, cause it is non-zero. Unless a is 0.

    I have an old keyboard with a giant

    ==

    at the top in permanent marker because of this.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

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    Quote Originally Posted by MK27 View Post
    if(a) would be true, yes, cause it is non-zero. Unless a is 0.

    I have an old keyboard with a giant

    ==

    at the top in permanent marker because of this.
    Yes, but my point was that the reason that the initial value of a doesn't have an effect on the output is not that it is changed to 2.

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    Quote Originally Posted by Subsonics View Post
    Yes, but my point was that the reason that the initial value of a doesn't have an effect on the output is not that it is changed to 2.
    Your point is wrong. The reason the initial value of a doesn't have any effect is precisely due to the fact that it was changed to 2. If the conditional was just:
    Code:
    if (a)
    instead of
    Code:
    if (a=2)
    then the initial value of a is what would determine the output of the conditional.
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    Quote Originally Posted by bithub View Post
    Your point is wrong. The reason the initial value of a doesn't have any effect is precisely due to the fact that it was changed to 2. If the conditional was just:
    Code:
    if (a)
    instead of
    Code:
    if (a=2)
    then the initial value of a is what would determine the output of the conditional.
    I did not say "doesn't have any effect" I said "doesn't have any effect on the output", if a is 2 or 6 wouldn't matter the printf statement would still get executed.

  14. #14
    BLG
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    It is a little confusing but I think I am starting to get it.

    Thank You For Your Help Guys.

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