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:

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#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");

}

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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?"

It prints that because the code is broken.

Code:

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else if (a = 2)

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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.

Quote:

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Originally Posted by bithub

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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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.

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.

Quote:

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Originally Posted by Subsonics

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.

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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.

Quote:

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Originally Posted by MK27

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.

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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).

Quote:

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Originally Posted by BLG

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:

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Originally Posted by Subsonics

after that, you test if it is 2, which it is, so "two" is printed.

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Not quite. The test is if it is non-zero, not two.

Yes, I realized this and changed my post. lol

Quote:

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Originally Posted by bithub

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.

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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.

Quote:

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Originally Posted by Subsonics

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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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.

Quote:

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Originally Posted by MK27

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.

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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.

Quote:

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Originally Posted by Subsonics

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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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.

Quote:

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Originally Posted by bithub

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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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.

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

Thank You For Your Help Guys.