C programming

Question

int main() {
		int a, b;
    /* Some code which initializes "a" here ... */
    if (a < 0) {
        a = -a;
    }
    b = sqrt (a);
}

• Above code will always work for calculating the square root of any valid integer value on  a system . [T/F]

Comments

b I'll be integer. That's obvious. You can choose appreciate type for b, and make sqrt result as precise as possible.

Other than this problem ?

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no. No problem for any valid integer on system.

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@Debashish_deka but  the validity of this statement will depend on compiler   right ? Will we able to get a general answer for this ? I really doubt this get executed correctly always. Moreover for large -ve numbers , will it work ?

Answer 1

    #include <stdio.h>
     
    int main() {
    		int a=-4, b;
        /* Some code which initializes "a" here ... */
        if (a < 0) {
            a = -a;
        }
        b = sqrt (a);
        printf("%d",b);
     
    	return 0;
    }

yes code will always work. But will not give correct output everytime. Like for - 4 it is giving square root as 2. But it is not a correct result

Comments

I guess the question assumes root of absolute value of a. Try with the largest negative integer value for a.

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Yes Sir. fails when a = INT_MIN

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means a= -32768

how it fails?

I am getting answer  as 181.http://ideone.com/tB6uLW

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You are using online compiler :O

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what is problem with that?

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You will bet better idea on your own computer's compiler.

64 bit or 32 bit have INT_MIN .

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in devcpp

running, but a warning

"9 C:\Dev-Cpp\sqrt.cpp [Warning] converting to `int' from `double' "

So, no error at all

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that compiler taking int as 4 byte. I have checked by forking your code + same version C

Now,

INT_MIN if we negate it, in 2's complement system it will result in the same number. As in the case of 4 bit, 2's complement of $-8$ is $-8$. That negative number will not work in sqrt()

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That I told previously , that -ve number not work. but then u disagree. rt?

what is the case of INT_MIN  here? I cannot understand.

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The intention of this code is to take square root the absolute value.

So, $-4$ should print $2$. and it will print $2$.

INT_MIN for $4$ bit integer number will be $-8$  (minimum representable in 2's complement). Similarly for 32 bit integer.

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But in code I am not getting where u considers 2's complement?

a= -a;

this line not anywhere means to say we have to consider 2's complement.

$\sqrt{-8}=2\sqrt{2}i$ (means when -ve stays -ve then code will not run rt?)

But in code I am not getting this

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yes, that would be the problem with negative numbers passed to sqrt().

Ans to answer your QS on 2's complement. You can verify this

#include <stdio.h>
#include <math.h>
int main() {
	int a = 90,i,b;
	b = -a; //negation
	// this loop will print the bit contents of b = -a;
	// that is 2's complement representation of a
	for(i = (sizeof(int)<<3) -1;i>=0;i--)
		(b&(1<<i)) ? printf("1"):printf("0");
	printf("\n");
	return 0;
}

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Ok,

it is not in range as http://ideone.com/YEaAih

i.e.32768

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^What has 32768 to do with int?

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Sir, I mean int range in between -32768 to 32767

But but output here showing 32768.

that is why it fails in INT_MIN

rt?

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int range in C is never that - that was in stone age. C standard does not fix the size of int, but most compilers take it as 4 bytes. So, range is $-2^{31} - (2^{31} - 1)$ if system uses 2's complement representation (this is also not mandated by C) and since there is one extra number possible for negatives,  it fails for INT_MIN.